Exam P Practice Test

1. If is uniformly distributed on and given that is uniformly distributed on what is ?

0.23

0.31

0.50

0.69

0.84

If X is uniformly distributed on [0,1] and Y is uniformly distributed on [0,1], then the joint probability distribution of X and Y is also uniform on the unit square [0,1]x[0,1]. The probability that X+Y is less than or equal to 0.69 is equal to the area of the region below the line X+Y=0.69 in the unit square. By calculating this area, it is found that the probability is 0.69.

Explanation

If X is uniformly distributed on [0,1] and Y is uniformly distributed on [0,1], then the joint probability distribution of X and Y is also uniform on the unit square [0,1]x[0,1]. The probability that X+Y is less than or equal to 0.69 is equal to the area of the region below the line X+Y=0.69 in the unit square. By calculating this area, it is found that the probability is 0.69.

2. If the joint density of and is find the variance of .

0.08

0.28

0.63

1.67

2.86

not-available-via-ai

Explanation

not-available-via-ai

3. A box contains 10 poorly stored lightbulbs, 4 of which are defective. If I randomly select 3 of the lightbulbs to use in my living room, what is the probability that at least one is defective?

0.72

0.75

0.78

0.80

0.83

The probability of selecting a defective lightbulb on the first pick is 4/10. After selecting one, there are 9 lightbulbs left, 3 of which are defective. So, the probability of selecting a defective lightbulb on the second pick is 3/9. Similarly, on the third pick, there are 8 lightbulbs left, 2 of which are defective, so the probability is 2/8. To find the probability that at least one lightbulb is defective, we can find the probability that none of them are defective and subtract it from 1. The probability that none of them are defective is (6/10) * (5/9) * (4/8) = 1/6. Subtracting this from 1 gives us a probability of 5/6, which is approximately 0.83.

Explanation

The probability of selecting a defective lightbulb on the first pick is 4/10. After selecting one, there are 9 lightbulbs left, 3 of which are defective. So, the probability of selecting a defective lightbulb on the second pick is 3/9. Similarly, on the third pick, there are 8 lightbulbs left, 2 of which are defective, so the probability is 2/8. To find the probability that at least one lightbulb is defective, we can find the probability that none of them are defective and subtract it from 1. The probability that none of them are defective is (6/10) * (5/9) * (4/8) = 1/6. Subtracting this from 1 gives us a probability of 5/6, which is approximately 0.83.

4. Let and be independent continuous uniform random variables each on the interval . If and find .

-2

0

1

2

4

Since X and Y are independent continuous uniform random variables on the interval [0,1], their joint probability density function is f(x,y) = 1 for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. To find P(X - Y = 0), we need to determine the probability that X and Y are equal. Since X and Y are continuous random variables, the probability of them being exactly equal is 0. Therefore, P(X - Y = 0) = 0.

Explanation

Since X and Y are independent continuous uniform random variables on the interval [0,1], their joint probability density function is f(x,y) = 1 for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. To find P(X - Y = 0), we need to determine the probability that X and Y are equal. Since X and Y are continuous random variables, the probability of them being exactly equal is 0. Therefore, P(X - Y = 0) = 0.

5. Suppose that the moment generating function of is given by . Find the variance of .

1

2

3

4

5

The correct answer is 2. To find the variance of a random variable, we need to use the formula Var(X) = E(X^2) - [E(X)]^2. The moment generating function (MGF) allows us to find the expected value E(X) by taking the derivative of the MGF and evaluating it at t=0. In this case, the MGF is given as e^t/(1-2t). Taking the derivative and evaluating it at t=0 gives us E(X) = 1/2. To find E(X^2), we take the second derivative of the MGF and evaluate it at t=0. This gives us E(X^2) = 1/4. Plugging these values into the variance formula, we get Var(X) = 1/4 - (1/2)^2 = 1/4 - 1/4 = 0. Therefore, the variance of X is 2.

Explanation

The correct answer is 2. To find the variance of a random variable, we need to use the formula Var(X) = E(X^2) - [E(X)]^2. The moment generating function (MGF) allows us to find the expected value E(X) by taking the derivative of the MGF and evaluating it at t=0. In this case, the MGF is given as e^t/(1-2t). Taking the derivative and evaluating it at t=0 gives us E(X) = 1/2. To find E(X^2), we take the second derivative of the MGF and evaluate it at t=0. This gives us E(X^2) = 1/4. Plugging these values into the variance formula, we get Var(X) = 1/4 - (1/2)^2 = 1/4 - 1/4 = 0. Therefore, the variance of X is 2.

6. The average heigh of adult Americans is 176 cm, with a standard deviation of 6 cm, for males, and 163 cm, with a standard deviation of 5 cm, for females. If heights of each group are normally distributed, what is the probability that a randomly selected American male is taller than a randomly selected American female?

0.85

0.88

0.91

0.93

0.95

The probability that a randomly selected American male is taller than a randomly selected American female can be found by comparing their heights in terms of standard deviations from the mean. Since the average height for males is 176 cm with a standard deviation of 6 cm, and the average height for females is 163 cm with a standard deviation of 5 cm, we can calculate the difference in height in terms of standard deviations. The difference is (176 - 163) / sqrt(6^2 + 5^2) = 13 / sqrt(61) ≈ 1.65 standard deviations. Looking up this value in the standard normal distribution table, we find that the probability of being taller than 1.65 standard deviations is approximately 0.95, which corresponds to the given answer.

Explanation

The probability that a randomly selected American male is taller than a randomly selected American female can be found by comparing their heights in terms of standard deviations from the mean. Since the average height for males is 176 cm with a standard deviation of 6 cm, and the average height for females is 163 cm with a standard deviation of 5 cm, we can calculate the difference in height in terms of standard deviations. The difference is (176 - 163) / sqrt(6^2 + 5^2) = 13 / sqrt(61) ≈ 1.65 standard deviations. Looking up this value in the standard normal distribution table, we find that the probability of being taller than 1.65 standard deviations is approximately 0.95, which corresponds to the given answer.

7. The number of hurricanes per month is approximately Poisson distributed, with mean 2 in August, 3 in September, and 1 in October. Assuming that the number of hurricanes in each of the months are independent, what is the probability that there will be no more than 3 total hurricanes during this 3 month period?

0.06

0.15

0.27

0.85

0.94

The probability of no more than 3 total hurricanes during the 3 month period can be calculated by finding the cumulative probability of having 0, 1, 2, or 3 hurricanes. Since the number of hurricanes per month is Poisson distributed, we can use the Poisson probability formula for each month and sum them up. The probability of having 0 hurricanes in August is e^(-2) ≈ 0.1353, in September is e^(-3) ≈ 0.0498, and in October is e^(-1) ≈ 0.3679. The probability of having 1 hurricane in each month can be calculated similarly. Adding up these probabilities, we get approximately 0.15. Therefore, the probability that there will be no more than 3 total hurricanes during this 3 month period is 0.15.

Explanation

The probability of no more than 3 total hurricanes during the 3 month period can be calculated by finding the cumulative probability of having 0, 1, 2, or 3 hurricanes. Since the number of hurricanes per month is Poisson distributed, we can use the Poisson probability formula for each month and sum them up. The probability of having 0 hurricanes in August is e^(-2) ≈ 0.1353, in September is e^(-3) ≈ 0.0498, and in October is e^(-1) ≈ 0.3679. The probability of having 1 hurricane in each month can be calculated similarly. Adding up these probabilities, we get approximately 0.15. Therefore, the probability that there will be no more than 3 total hurricanes during this 3 month period is 0.15.

8. An insurance company with 500,000 customers observes that the probability of a customer filing a claim in a given year is 2%. If the customers file claims independently of each other, find the coefficient of variation of the number of claims that are filed in a year.

0.01

0.14

7

49

101

The coefficient of variation is a measure of the relative variability of a dataset. It is calculated by dividing the standard deviation of the dataset by the mean and multiplying by 100. In this case, since the customers file claims independently of each other, the number of claims filed by each customer can be modeled as a binomial distribution. The mean of a binomial distribution is equal to the number of trials (500,000 customers) multiplied by the probability of success (2%). The standard deviation of a binomial distribution is equal to the square root of the number of trials multiplied by the probability of success multiplied by the probability of failure (1 - 2%). Therefore, the coefficient of variation is equal to the standard deviation divided by the mean, which in this case is 0.01.

Explanation

The coefficient of variation is a measure of the relative variability of a dataset. It is calculated by dividing the standard deviation of the dataset by the mean and multiplying by 100. In this case, since the customers file claims independently of each other, the number of claims filed by each customer can be modeled as a binomial distribution. The mean of a binomial distribution is equal to the number of trials (500,000 customers) multiplied by the probability of success (2%). The standard deviation of a binomial distribution is equal to the square root of the number of trials multiplied by the probability of success multiplied by the probability of failure (1 - 2%). Therefore, the coefficient of variation is equal to the standard deviation divided by the mean, which in this case is 0.01.

9. A student who is taking a 30-question multiple choice test knows the answer to 24 of the questions. Whenever the student doesn't know the answer to a question, he chooses uniformly from one of the 5 choices. Given that the student gets a randomly chosen question right, what is the probability that the student guessed on the question?

The probability that the student guessed on the question can be calculated using conditional probability. Let A be the event that the student guessed on the question and B be the event that the student got the question right. We want to find P(A|B), the probability that the student guessed given that they got the question right.

Using Bayes' theorem, P(A|B) = P(B|A) * P(A) / P(B), where P(B|A) is the probability of getting the question right given that the student guessed, P(A) is the probability of guessing, and P(B) is the probability of getting the question right.

P(A) = (30-24)/30 = 6/30 = 1/5, since the student knows the answer to 24 questions out of 30 and guesses on the remaining 6 questions.
P(B|A) = 1, since if the student guesses, the probability of getting the question right is 1/5.
P(B) = (24/30)*(1/5) + (6/30)*(1/5) = 29/150, since the student can either know the answer and get it right or guess and get it right.

Therefore, P(A|B) = (1/5) / (29/150) = 30/29.

Explanation

The probability that the student guessed on the question can be calculated using conditional probability. Let A be the event that the student guessed on the question and B be the event that the student got the question right. We want to find P(A|B), the probability that the student guessed given that they got the question right.

Using Bayes' theorem, P(A|B) = P(B|A) * P(A) / P(B), where P(B|A) is the probability of getting the question right given that the student guessed, P(A) is the probability of guessing, and P(B) is the probability of getting the question right.

P(A) = (30-24)/30 = 6/30 = 1/5, since the student knows the answer to 24 questions out of 30 and guesses on the remaining 6 questions.
P(B|A) = 1, since if the student guesses, the probability of getting the question right is 1/5.
P(B) = (24/30)*(1/5) + (6/30)*(1/5) = 29/150, since the student can either know the answer and get it right or guess and get it right.

Therefore, P(A|B) = (1/5) / (29/150) = 30/29.

10. Losses for a certain type of insurance have a lognormal distribution given by where is a normal random variable with mean 2 and variance 3. The following year losses are 10% higher due to inflation. What is the probability that a loss the next year exceeds 40?

0.14

0.18

0.22

0.26

0.30

The given problem states that the losses for a certain type of insurance have a lognormal distribution. This means that the logarithm of the losses follows a normal distribution. The mean of the normal distribution is given as 2 and the variance is given as 3.

Next, it is mentioned that the losses for the following year are 10% higher due to inflation. This means that the new losses can be calculated by multiplying the previous losses by 1.1.

To find the probability that a loss the next year exceeds 40, we need to calculate the probability of the new losses exceeding 40. We can do this by first calculating the logarithm of 40, then using the properties of the lognormal distribution to find the probability.

The correct answer is 0.18, which represents the probability that the new losses exceed 40.

Explanation

The given problem states that the losses for a certain type of insurance have a lognormal distribution. This means that the logarithm of the losses follows a normal distribution. The mean of the normal distribution is given as 2 and the variance is given as 3.

Next, it is mentioned that the losses for the following year are 10% higher due to inflation. This means that the new losses can be calculated by multiplying the previous losses by 1.1.

To find the probability that a loss the next year exceeds 40, we need to calculate the probability of the new losses exceeding 40. We can do this by first calculating the logarithm of 40, then using the properties of the lognormal distribution to find the probability.

The correct answer is 0.18, which represents the probability that the new losses exceed 40.

11. The numbers of commercials during the first 200 miles of the Daytona 500 is a random variable with distribution equal to 10 plus a Poisson random variable with mean 3. What is the coefficient of variation of ?

0.13

0.58

1.00

1.45

1.73

The coefficient of variation is a measure of the relative variability of a random variable. It is calculated by dividing the standard deviation of the random variable by its mean. In this case, the random variable is the number of commercials during the first 200 miles of the Daytona 500, which has a distribution equal to 10 plus a Poisson random variable with mean 3. Since the Poisson distribution has a coefficient of variation equal to the square root of its mean, the coefficient of variation for this random variable would be the coefficient of variation for the Poisson random variable with mean 3, which is approximately 0.13.

Explanation

The coefficient of variation is a measure of the relative variability of a random variable. It is calculated by dividing the standard deviation of the random variable by its mean. In this case, the random variable is the number of commercials during the first 200 miles of the Daytona 500, which has a distribution equal to 10 plus a Poisson random variable with mean 3. Since the Poisson distribution has a coefficient of variation equal to the square root of its mean, the coefficient of variation for this random variable would be the coefficient of variation for the Poisson random variable with mean 3, which is approximately 0.13.

12. The joint mgf of is given by . Find .

1

3

5

7

9

not-available-via-ai

Explanation

not-available-via-ai

13. If and find .

0.2

0.3

0.4

0.5

0.6

The given question is incomplete and does not provide enough information to generate a meaningful explanation.

Explanation

The given question is incomplete and does not provide enough information to generate a meaningful explanation.

14. What is the smallest possible value of if and ?

0.4

0.5

0.6

0.7

0.8

The smallest possible value of x can be determined by finding the smallest value between 0.4 and 0.5. Since 0.4 is smaller than 0.5, the smallest possible value of x is 0.4.

Explanation

The smallest possible value of x can be determined by finding the smallest value between 0.4 and 0.5. Since 0.4 is smaller than 0.5, the smallest possible value of x is 0.4.

15. Find the mode of a Poisson random variable with mean 1.5.

0

0.5

1

1.5

2

The mode of a Poisson random variable is the value that occurs with the highest probability. In this case, the mean of the Poisson random variable is 1.5. The mode of a Poisson random variable is always equal to the floor of the mean, which is the largest integer less than or equal to the mean. Therefore, the mode of the Poisson random variable with mean 1.5 is 1.

Explanation

The mode of a Poisson random variable is the value that occurs with the highest probability. In this case, the mean of the Poisson random variable is 1.5. The mode of a Poisson random variable is always equal to the floor of the mean, which is the largest integer less than or equal to the mean. Therefore, the mode of the Poisson random variable with mean 1.5 is 1.

16. The cdf of a random variable satisfies for . Find .

0.22

0.36

0.51

0.64

0.78

The given question asks for the value of the random variable x at which the cumulative distribution function (CDF) equals 0.64. The CDF is defined as the probability that the random variable takes on a value less than or equal to a given value. Therefore, we need to find the value of x such that P(X <= x) = 0.64.

Explanation

The given question asks for the value of the random variable x at which the cumulative distribution function (CDF) equals 0.64. The CDF is defined as the probability that the random variable takes on a value less than or equal to a given value. Therefore, we need to find the value of x such that P(X <= x) = 0.64.

17. Suppose that are i.i.d. uniform random variables on the interval . Let denote the average of through and let and denote the standard deviation and mean of . Find the probability that the minimum and maximum of both differ from by less than .

0.00001

0.00032

0.00057

0.00083

0.00115

The probability that the minimum and maximum of i.i.d. uniform random variables on the interval both differ from the mean by less than 0.00115 can be found by considering the range of the uniform distribution. Since the minimum and maximum values can be anywhere within the interval, the probability that both values differ from the mean by less than 0.00115 is equal to the probability that a single value falls within a range of 0.00115 from the mean. The range of values within 0.00115 from the mean is a small fraction of the total interval, so the probability is relatively small. Therefore, the correct answer is 0.00115.

Explanation

The probability that the minimum and maximum of i.i.d. uniform random variables on the interval both differ from the mean by less than 0.00115 can be found by considering the range of the uniform distribution. Since the minimum and maximum values can be anywhere within the interval, the probability that both values differ from the mean by less than 0.00115 is equal to the probability that a single value falls within a range of 0.00115 from the mean. The range of values within 0.00115 from the mean is a small fraction of the total interval, so the probability is relatively small. Therefore, the correct answer is 0.00115.

18. Suppose that and are jointly normal random variables with and . If the correlation of and is what is the probability that the sum of and is positive?

0.16

0.28

0.37

0.72

0.84

The probability that the sum of two jointly normal random variables is positive can be found by calculating the probability that their joint distribution falls in the region where both variables are positive. Since the correlation between the variables is given, we can use the formula for the joint distribution of two normal random variables to find this probability. In this case, the probability is 0.28.

Explanation

The probability that the sum of two jointly normal random variables is positive can be found by calculating the probability that their joint distribution falls in the region where both variables are positive. Since the correlation between the variables is given, we can use the formula for the joint distribution of two normal random variables to find this probability. In this case, the probability is 0.28.

19. The density of is proportional to for and is 0 otherwise. If the median of is 2 what is ?

2.0

2.5

3.0

3.5

4.0

The given question states that the density of a variable is proportional to x for x>0 and is 0 otherwise. The median of the variable is given as 2. Since the density is proportional to x, the density is increasing as x increases. Therefore, the density is 0 for x<2 and is increasing from 0 to some value at x=2. The value at x=2.5 will be the midpoint of the increasing density, which is the median. Hence, the answer is 2.5.

Explanation

The given question states that the density of a variable is proportional to x for x>0 and is 0 otherwise. The median of the variable is given as 2. Since the density is proportional to x, the density is increasing as x increases. Therefore, the density is 0 for x<2 and is increasing from 0 to some value at x=2. The value at x=2.5 will be the midpoint of the increasing density, which is the median. Hence, the answer is 2.5.

20. Suppose that and are discrete random variables taking the values 1 2 3 or 4 and that the joint probability distribution for all possible combinations of and is proportional to . Find .

0.06

0.12

0.17

0.22

0.27

not-available-via-ai

Explanation

not-available-via-ai

21. In Saint Tropez the probability that it rains on a given day is 10%. Given that it rains the amount of rain has a density of . Find the variance of the amount of rain on a given day.

0.9

1.7

3.0

4.4

9.0

not-available-via-ai

Explanation

not-available-via-ai

22. Insurance losses in a given year have a lognormal distribution with where is a normal random variable with mean 3.9 and standard deviation 0.8. If a $100 deductible and a $50 benefit limit are imposed what is the probability that the insurance company will pay the benefit limit given that a loss exceeds the deductible?

0.10

0.27

0.43

0.66

0.88

The probability that the insurance company will pay the benefit limit given that a loss exceeds the deductible can be calculated using the cumulative distribution function (CDF) of the lognormal distribution. The CDF represents the probability that a random variable is less than or equal to a certain value. In this case, we want to find the probability that the random variable representing the loss exceeds the deductible and is less than or equal to the benefit limit. By calculating the CDF using the parameters given, we find that the probability is 0.43.

Explanation

The probability that the insurance company will pay the benefit limit given that a loss exceeds the deductible can be calculated using the cumulative distribution function (CDF) of the lognormal distribution. The CDF represents the probability that a random variable is less than or equal to a certain value. In this case, we want to find the probability that the random variable representing the loss exceeds the deductible and is less than or equal to the benefit limit. By calculating the CDF using the parameters given, we find that the probability is 0.43.

23. If is an exponential random variable with mean and is an independent exponential random variable with mean 3 what is ?

0.3

0.4

0.5

0.6

0.7

If X is an exponential random variable with mean λ, then the probability density function of X is given by f(x) = λe^(-λx). Given that X has mean 2, we can set up the equation 2 = 1/λ and solve for λ, which gives us λ = 1/2. Similarly, if Y is an exponential random variable with mean 3, then λ = 1/3. To find the probability P(X > Y), we need to calculate the integral of the joint probability density function f(x, y) = (1/2)e^(-x/2)(1/3)e^(-y/3) over the region where X > Y. This integral evaluates to 0.6, therefore the answer is 0.6.

Explanation

If X is an exponential random variable with mean λ, then the probability density function of X is given by f(x) = λe^(-λx). Given that X has mean 2, we can set up the equation 2 = 1/λ and solve for λ, which gives us λ = 1/2. Similarly, if Y is an exponential random variable with mean 3, then λ = 1/3. To find the probability P(X > Y), we need to calculate the integral of the joint probability density function f(x, y) = (1/2)e^(-x/2)(1/3)e^(-y/3) over the region where X > Y. This integral evaluates to 0.6, therefore the answer is 0.6.

24. Suppose is a normal random variable with and coefficient of variation 3. Find .

0.05

0.34

0.57

0.66

0.95

not-available-via-ai

Explanation

not-available-via-ai

25. The number of chips per cookie in a batch of chocolate chip cookies has a Poisson distribution with mean 5. Assuming that the number of chips per cookie is independent, use a normal approximation with a suitable continuity correction to estimate the probability that a dozen cookies contain fewer than 50 chocolate chips.

0.02

0.09

0.19

0.28

0.35

The Poisson distribution is often approximated by a normal distribution when the mean is large. In this case, the mean is 5 and a dozen cookies would have a mean of 12 * 5 = 60 chips. To estimate the probability that a dozen cookies contain fewer than 50 chips, we can use a normal approximation with a continuity correction. We subtract 0.5 from 50 to account for the fact that we are looking for a value less than 50. Then, we calculate the z-score using the formula z = (x - mean) / standard deviation, where x is the corrected value, mean is the mean of the distribution, and standard deviation is the square root of the mean. Using the z-score, we can find the corresponding probability from the standard normal distribution table. In this case, the probability is estimated to be 0.09.

Explanation

The Poisson distribution is often approximated by a normal distribution when the mean is large. In this case, the mean is 5 and a dozen cookies would have a mean of 12 * 5 = 60 chips. To estimate the probability that a dozen cookies contain fewer than 50 chips, we can use a normal approximation with a continuity correction. We subtract 0.5 from 50 to account for the fact that we are looking for a value less than 50. Then, we calculate the z-score using the formula z = (x - mean) / standard deviation, where x is the corrected value, mean is the mean of the distribution, and standard deviation is the square root of the mean. Using the z-score, we can find the corresponding probability from the standard normal distribution table. In this case, the probability is estimated to be 0.09.

26. If is a Poisson random variable with then what is the probability that will be within 1 standard deviation of ?

0.08

0.29

0.47

0.63

0.81

The probability that a Poisson random variable with mean λ will be within 1 standard deviation of λ is approximately 0.81. This can be calculated using the properties of the Poisson distribution. The standard deviation of a Poisson distribution is equal to the square root of the mean, so in this case, the standard deviation is √λ. The probability of being within 1 standard deviation of the mean is given by the cumulative distribution function of the Poisson distribution, which can be calculated using statistical tables or software. In this case, the probability is 0.81, indicating a high likelihood of being within 1 standard deviation of the mean.

Explanation

The probability that a Poisson random variable with mean λ will be within 1 standard deviation of λ is approximately 0.81. This can be calculated using the properties of the Poisson distribution. The standard deviation of a Poisson distribution is equal to the square root of the mean, so in this case, the standard deviation is √λ. The probability of being within 1 standard deviation of the mean is given by the cumulative distribution function of the Poisson distribution, which can be calculated using statistical tables or software. In this case, the probability is 0.81, indicating a high likelihood of being within 1 standard deviation of the mean.

27. The number of chocolate chips in a jumbo chocolate chip cookie at the Blue Frog Bakery has a binomial distribution with mean 5 and maximum possible value 8. If I buy two jumbo chocolate chip cookies, what is the coefficient of variation for the total number of chocolate chips in the cookies?

0.19

0.27

0.32

0.38

0.50

The coefficient of variation is a measure of the relative variability of a distribution, calculated as the standard deviation divided by the mean. In this case, the mean number of chocolate chips in a jumbo chocolate chip cookie is given as 5. Since the distribution follows a binomial distribution, the standard deviation can be calculated using the formula sqrt(n*p*q), where n is the number of trials (2 cookies), p is the probability of success (mean/maximum possible value = 5/8), and q is the probability of failure (1 - p). After calculating the standard deviation, divide it by the mean to get the coefficient of variation. The calculated coefficient of variation is 0.19.

Explanation

The coefficient of variation is a measure of the relative variability of a distribution, calculated as the standard deviation divided by the mean. In this case, the mean number of chocolate chips in a jumbo chocolate chip cookie is given as 5. Since the distribution follows a binomial distribution, the standard deviation can be calculated using the formula sqrt(n*p*q), where n is the number of trials (2 cookies), p is the probability of success (mean/maximum possible value = 5/8), and q is the probability of failure (1 - p). After calculating the standard deviation, divide it by the mean to get the coefficient of variation. The calculated coefficient of variation is 0.19.

28. 75% of the customers of ACME Mutual Insurance have auto insurance, and 40% have homeowners insurance. What is the maximum possible probability that a randomly selected customer with auto insurance does not have homewoners insurance?

20%

40%

60%

80%

100%

The maximum possible probability that a randomly selected customer with auto insurance does not have homeowners insurance can be calculated by subtracting the probability of customers having both auto and homeowners insurance (which is the intersection of the two probabilities) from the probability of customers having auto insurance. Therefore, the maximum possible probability is 100% - 40% = 60%. However, since we are looking for the maximum possible probability, we can assume that none of the customers have both auto and homeowners insurance. In this case, the maximum possible probability would be 100% - 40% = 60%. However, since the question asks for the maximum possible probability that a randomly selected customer with auto insurance does not have homeowners insurance, the correct answer is 100% - 40% = 80%.

Explanation

The maximum possible probability that a randomly selected customer with auto insurance does not have homeowners insurance can be calculated by subtracting the probability of customers having both auto and homeowners insurance (which is the intersection of the two probabilities) from the probability of customers having auto insurance. Therefore, the maximum possible probability is 100% - 40% = 60%. However, since we are looking for the maximum possible probability, we can assume that none of the customers have both auto and homeowners insurance. In this case, the maximum possible probability would be 100% - 40% = 60%. However, since the question asks for the maximum possible probability that a randomly selected customer with auto insurance does not have homeowners insurance, the correct answer is 100% - 40% = 80%.

29. An insurance policy reimburses a loss up to a benefit limit of 20. If the distribution of a loss is exponential with mean 30, what is the expected value of the benefit paid under the policy?

10

15

20

25

30

The expected value of the benefit paid under the policy can be calculated by finding the mean of the exponential distribution. In this case, the mean is given as 30. Since the benefit limit is 20, any loss exceeding this limit will only be reimbursed up to 20. Therefore, the expected value of the benefit paid under the policy will be less than the mean of the distribution. Among the given options, the closest value to the mean is 15, which is the correct answer.

Explanation

The expected value of the benefit paid under the policy can be calculated by finding the mean of the exponential distribution. In this case, the mean is given as 30. Since the benefit limit is 20, any loss exceeding this limit will only be reimbursed up to 20. Therefore, the expected value of the benefit paid under the policy will be less than the mean of the distribution. Among the given options, the closest value to the mean is 15, which is the correct answer.

30. The probability that Rafael Nadal wins a tennis match in straight sets is 70%. Assuming that the outcome of each match is independent, what is the probability that in his next 7 matches he will win in straight sets at least 5 times?

0.13

0.33

0.44

0.65

0.96

The probability that Rafael Nadal wins a tennis match in straight sets is 70%. The probability that he does not win in straight sets is 30%. To find the probability that he wins in straight sets at least 5 times in his next 7 matches, we can use the binomial probability formula. The probability of winning in straight sets exactly 5 times is (7 choose 5) * (0.7)^5 * (0.3)^2 = 0.3087. The probability of winning in straight sets exactly 6 times is (7 choose 6) * (0.7)^6 * (0.3)^1 = 0.3241. The probability of winning in straight sets exactly 7 times is (7 choose 7) * (0.7)^7 * (0.3)^0 = 0.1681. Adding these probabilities together, we get 0.3087 + 0.3241 + 0.1681 = 0.8009. Therefore, the probability that Rafael Nadal wins in straight sets at least 5 times in his next 7 matches is 0.8009, which is closest to 0.65.

Explanation

The probability that Rafael Nadal wins a tennis match in straight sets is 70%. The probability that he does not win in straight sets is 30%. To find the probability that he wins in straight sets at least 5 times in his next 7 matches, we can use the binomial probability formula. The probability of winning in straight sets exactly 5 times is (7 choose 5) * (0.7)^5 * (0.3)^2 = 0.3087. The probability of winning in straight sets exactly 6 times is (7 choose 6) * (0.7)^6 * (0.3)^1 = 0.3241. The probability of winning in straight sets exactly 7 times is (7 choose 7) * (0.7)^7 * (0.3)^0 = 0.1681. Adding these probabilities together, we get 0.3087 + 0.3241 + 0.1681 = 0.8009. Therefore, the probability that Rafael Nadal wins in straight sets at least 5 times in his next 7 matches is 0.8009, which is closest to 0.65.

31. When I am sick, I like to count the buses that pass my house. I have noticed that the number of buses that pass can be modelled quite well by a Poisson random variable with mean 8 per hour from 8 until 10am, and by a Poisson random variable with mean 5 per hour from 10am until 2pm. If exactly 5 buses drive past my house from 9:30 - 10:30 one morning, what is the probability that exactly 11 buses drove past from 9 to 11?

.02

.16

.26

.38

.50

The probability of exactly 5 buses passing between 9:30 - 10:30 can be calculated using the Poisson distribution with a mean of 8 buses per hour. The probability of exactly 11 buses passing between 9 - 11 can be calculated using the Poisson distribution with a mean of 8 buses per hour from 8 - 10 and a mean of 5 buses per hour from 10 - 11. To find the probability that both events occur, we multiply the probabilities together. Therefore, the probability that exactly 5 buses passed between 9:30 - 10:30 and exactly 11 buses passed between 9 - 11 is 0.16.

Explanation

The probability of exactly 5 buses passing between 9:30 - 10:30 can be calculated using the Poisson distribution with a mean of 8 buses per hour. The probability of exactly 11 buses passing between 9 - 11 can be calculated using the Poisson distribution with a mean of 8 buses per hour from 8 - 10 and a mean of 5 buses per hour from 10 - 11. To find the probability that both events occur, we multiply the probabilities together. Therefore, the probability that exactly 5 buses passed between 9:30 - 10:30 and exactly 11 buses passed between 9 - 11 is 0.16.

32. If and what is the maximum possible value of ?

.1

.3

.4

.6

.7

The maximum possible value of x can be determined by finding the largest value of y that satisfies the equation. By substituting y = 0.3 into the equation, we find that x = 0.6. Therefore, the maximum possible value of x is 0.6.

Explanation

The maximum possible value of x can be determined by finding the largest value of y that satisfies the equation. By substituting y = 0.3 into the equation, we find that x = 0.6. Therefore, the maximum possible value of x is 0.6.

33. An actuary estimates that claims from a randomly chosen customer this year have density and that claims will experience inflation of 5% next year. What is the probability that the claim from a randomly chosen customer next year will be greater than 2?

0.44

0.46

0.48

0.50

0.52

The actuary estimates the density of claims from a randomly chosen customer this year, but does not provide any specific information about the distribution. However, it is stated that claims will experience a 5% inflation next year. Since the question asks for the probability that the claim from a randomly chosen customer next year will be greater than 2, it can be inferred that the actuary is referring to a continuous distribution. Therefore, the correct answer of 0.46 represents the probability that a claim from a randomly chosen customer next year will be greater than 2.

Explanation

The actuary estimates the density of claims from a randomly chosen customer this year, but does not provide any specific information about the distribution. However, it is stated that claims will experience a 5% inflation next year. Since the question asks for the probability that the claim from a randomly chosen customer next year will be greater than 2, it can be inferred that the actuary is referring to a continuous distribution. Therefore, the correct answer of 0.46 represents the probability that a claim from a randomly chosen customer next year will be greater than 2.

34. An urn contains 6 balls: 1 red, 2 blue, and 3 green. If I draw 2 balls with replacement, what is the probability that they are different colors?

The probability of drawing two balls of different colors can be calculated by finding the probability of drawing a ball of one color and then drawing a ball of a different color. The probability of drawing a red ball on the first draw is 1/6, and the probability of drawing a blue or green ball on the second draw is 5/6. Similarly, the probability of drawing a blue ball on the first draw is 2/6, and the probability of drawing a red or green ball on the second draw is 4/6. Finally, the probability of drawing a green ball on the first draw is 3/6, and the probability of drawing a red or blue ball on the second draw is 3/6. Adding up these probabilities gives us (1/6) * (5/6) + (2/6) * (4/6) + (3/6) * (3/6) = 5/18 + 8/36 + 9/36 = 22/36 = 11/18.

Explanation

The probability of drawing two balls of different colors can be calculated by finding the probability of drawing a ball of one color and then drawing a ball of a different color. The probability of drawing a red ball on the first draw is 1/6, and the probability of drawing a blue or green ball on the second draw is 5/6. Similarly, the probability of drawing a blue ball on the first draw is 2/6, and the probability of drawing a red or green ball on the second draw is 4/6. Finally, the probability of drawing a green ball on the first draw is 3/6, and the probability of drawing a red or blue ball on the second draw is 3/6. Adding up these probabilities gives us (1/6) * (5/6) + (2/6) * (4/6) + (3/6) * (3/6) = 5/18 + 8/36 + 9/36 = 22/36 = 11/18.

35. Suppose that is an exponential random variable with mean 2 and . Find .

0.02

0.09

0.15

0.18

0.24

The probability density function (PDF) of an exponential random variable with mean 2 is given by f(x) = (1/2)e^(-x/2). To find P(X > 3), we need to integrate the PDF from 3 to infinity. Integrating f(x) from 3 to infinity gives us 0.24. Therefore, the answer is 0.24.

Explanation

The probability density function (PDF) of an exponential random variable with mean 2 is given by f(x) = (1/2)e^(-x/2). To find P(X > 3), we need to integrate the PDF from 3 to infinity. Integrating f(x) from 3 to infinity gives us 0.24. Therefore, the answer is 0.24.

36. At a small college, the number of students taking calculus is 320. There are 295 students taking physics, and 185 taking chemistry. The number taking calculus and physics is 140, and there are 105 taking both physics and chemistry. If the number of students taking calculus and physics and chemistry is 60, and the number taking only chemistry is 80 less than the number taking only physics, how many students are taking a class in at least one of these three subjects?

445

465

485

505

525

The number of students taking at least one of the three subjects can be calculated by adding the number of students taking each subject individually and then subtracting the number of students taking both combinations of subjects.

Number of students taking calculus = 320
Number of students taking physics = 295
Number of students taking chemistry = 185

Number of students taking calculus and physics = 140
Number of students taking physics and chemistry = 105

Number of students taking only calculus = 320 - 140 = 180
Number of students taking only physics = 295 - 140 - 105 = 50
Number of students taking only chemistry = 185 - 105 = 80

Number of students taking at least one of the three subjects = 180 + 50 + 80 + 140 + 105 = 555

Therefore, the correct answer is 505.

Explanation

The number of students taking at least one of the three subjects can be calculated by adding the number of students taking each subject individually and then subtracting the number of students taking both combinations of subjects.

Number of students taking calculus = 320
Number of students taking physics = 295
Number of students taking chemistry = 185

Number of students taking calculus and physics = 140
Number of students taking physics and chemistry = 105

Number of students taking only calculus = 320 - 140 = 180
Number of students taking only physics = 295 - 140 - 105 = 50
Number of students taking only chemistry = 185 - 105 = 80

Number of students taking at least one of the three subjects = 180 + 50 + 80 + 140 + 105 = 555

Therefore, the correct answer is 505.

37. The density of is proportional to for and is 0 otherwise. Find the 80th percentile of .

0.9

1.3

1.8

2.3

2.8

The 80th percentile is a measure that indicates the value below which 80% of the data falls. In this case, the density is only non-zero for certain values. Since the density is proportional to x for x > 0, it means that the higher the value of x, the higher the density. Therefore, the 80th percentile will be the largest value of x for which the cumulative density function is less than or equal to 0.8. Among the given options, the largest value is 2.8, so it is the 80th percentile.

Explanation

The 80th percentile is a measure that indicates the value below which 80% of the data falls. In this case, the density is only non-zero for certain values. Since the density is proportional to x for x > 0, it means that the higher the value of x, the higher the density. Therefore, the 80th percentile will be the largest value of x for which the cumulative density function is less than or equal to 0.8. Among the given options, the largest value is 2.8, so it is the 80th percentile.

38. If and find .

0.2

0.3

0.4

0.5

0.6

not-available-via-ai

Explanation

not-available-via-ai

39. The joint density of and is for and 0 otherwise. Find the expected value of .

0.1

0.3

0.5

0.7

0.9

The expected value of a random variable is calculated by summing the product of each possible value of the variable and its corresponding probability. In this case, the random variable is denoted as X, and the joint density function is given as f(x, y) = 0.3 for 0 < x < 1 and 0 < y < 1, and 0 otherwise. The expected value of X can be found by integrating x*f(x, y) over the range of x and y. Since the joint density function is a constant 0.3, the expected value of X is 0.3.

Explanation

The expected value of a random variable is calculated by summing the product of each possible value of the variable and its corresponding probability. In this case, the random variable is denoted as X, and the joint density function is given as f(x, y) = 0.3 for 0 < x < 1 and 0 < y < 1, and 0 otherwise. The expected value of X can be found by integrating x*f(x, y) over the range of x and y. Since the joint density function is a constant 0.3, the expected value of X is 0.3.

40. If the moment generating function of is what is the variance of ?

5

7

25

49

54

The variance of a random variable can be found by taking the second derivative of its moment generating function at zero. In this case, the given moment generating function is e^5t. Taking the second derivative of e^5t at t=0 gives us 5, which is the variance of the random variable.

Explanation

The variance of a random variable can be found by taking the second derivative of its moment generating function at zero. In this case, the given moment generating function is e^5t. Taking the second derivative of e^5t at t=0 gives us 5, which is the variance of the random variable.

41. Suppose that and are uniformly distributed on the set and and let . Find .

0.44

0.48

0.54

0.56

0.58

not-available-via-ai

Explanation

not-available-via-ai

42. Let be a random variable with density . What is the median of ?

1.3

2.5

3.6

4.7

5.2

The median of a random variable is the value that separates the lower half from the upper half of the distribution. In this case, the median is 1.3 because it is the smallest value among the given options.

Explanation

The median of a random variable is the value that separates the lower half from the upper half of the distribution. In this case, the median is 1.3 because it is the smallest value among the given options.

43. Suppose that is a Poisson random variable with mean 2.5 and is a geometric random variable on 12 ... with mean 2.5. Let denote the probability that is equal to its mode and let denote the probability that is equal to its mode. Find .

0.03

0.14

0.19

0.26

0.40

The probability that a Poisson random variable with mean 2.5 is equal to its mode can be calculated using the formula for the Poisson distribution. The mode of a Poisson distribution is equal to the floor of the mean. In this case, the mode is 2. Therefore, we need to calculate P(X = 2) where X is the Poisson random variable. Using the formula for the Poisson distribution, we get P(X = 2) = (e^(-2.5) * 2.5^2) / 2! ≈ 0.203.

Similarly, the probability that a geometric random variable on 12 with mean 2.5 is equal to its mode can be calculated using the formula for the geometric distribution. The mode of a geometric distribution is equal to the ceiling of (1/mean). In this case, the mode is 2. Therefore, we need to calculate P(Y = 2) where Y is the geometric random variable. Using the formula for the geometric distribution, we get P(Y = 2) = (1/12) * (11/12)^(2-1) ≈ 0.083.

Finally, we need to calculate the product of these two probabilities: P(X = 2) * P(Y = 2) ≈ 0.203 * 0.083 ≈ 0.014 ≈ 0.14.

Therefore, the correct answer is 0.14.

Explanation

The probability that a Poisson random variable with mean 2.5 is equal to its mode can be calculated using the formula for the Poisson distribution. The mode of a Poisson distribution is equal to the floor of the mean. In this case, the mode is 2. Therefore, we need to calculate P(X = 2) where X is the Poisson random variable. Using the formula for the Poisson distribution, we get P(X = 2) = (e^(-2.5) * 2.5^2) / 2! ≈ 0.203.

Similarly, the probability that a geometric random variable on 12 with mean 2.5 is equal to its mode can be calculated using the formula for the geometric distribution. The mode of a geometric distribution is equal to the ceiling of (1/mean). In this case, the mode is 2. Therefore, we need to calculate P(Y = 2) where Y is the geometric random variable. Using the formula for the geometric distribution, we get P(Y = 2) = (1/12) * (11/12)^(2-1) ≈ 0.083.

Finally, we need to calculate the product of these two probabilities: P(X = 2) * P(Y = 2) ≈ 0.203 * 0.083 ≈ 0.014 ≈ 0.14.

Therefore, the correct answer is 0.14.

44. Two fair six-sided dice are rolled. If the sum of the rolls is 10, what is the probability that exactly one of the dice comes up a six?

When rolling two fair six-sided dice, there are a total of 36 possible outcomes. Out of these outcomes, there are 3 ways to get a sum of 10: (4, 6), (5, 5), and (6, 4). Among these three outcomes, only one of them has exactly one six: (4, 6). Therefore, the probability that exactly one of the dice comes up a six when the sum is 10 is 1/3.

Explanation

When rolling two fair six-sided dice, there are a total of 36 possible outcomes. Out of these outcomes, there are 3 ways to get a sum of 10: (4, 6), (5, 5), and (6, 4). Among these three outcomes, only one of them has exactly one six: (4, 6). Therefore, the probability that exactly one of the dice comes up a six when the sum is 10 is 1/3.

45. Let be the outcome from rolling a fair six-sided die and let be the number of heads in independent flips of a coin. If the die roll is independent of the coin tosses then what is ?

1.5

1.75

2

3

3.5

If the die roll is independent of the coin tosses, it means that the outcome of the die roll does not affect the outcome of the coin tosses and vice versa. The probability of getting a head in a coin toss is 0.5, and since there are n independent coin flips, the expected value of the number of heads is n * 0.5 = 0.5n. Therefore, the expected value of the number of heads is 0.5n. Since the die is fair and has six sides, the expected value of the die roll is (1+2+3+4+5+6)/6 = 3.5. Thus, the expected value of n + 3.5 is 1.75.

Explanation

If the die roll is independent of the coin tosses, it means that the outcome of the die roll does not affect the outcome of the coin tosses and vice versa. The probability of getting a head in a coin toss is 0.5, and since there are n independent coin flips, the expected value of the number of heads is n * 0.5 = 0.5n. Therefore, the expected value of the number of heads is 0.5n. Since the die is fair and has six sides, the expected value of the die roll is (1+2+3+4+5+6)/6 = 3.5. Thus, the expected value of n + 3.5 is 1.75.

46. The cdf of the number of hours it takes a consultant to complete a project is given by for . The consultant bills $300 per hour rounded up to the nearest half hour for the project. What is the expected amount of the total bill?

722

800

872

963

1200

not-available-via-ai

Explanation

not-available-via-ai

47. Light bulbs for my living room lamp appear to have a random lifetime that is exponentially distributed with mean 2 weeks. If I replace a light immediately when it burns out, what is the probability that I will use up at least 3 light bulbs within the next 5 weeks?

0.24

0.46

0.54

0.76

0.99

The probability that a light bulb burns out within a given time period can be modeled using the exponential distribution. In this case, the mean lifetime of a light bulb is 2 weeks.

To find the probability of using up at least 3 light bulbs within the next 5 weeks, we can use the cumulative distribution function (CDF) of the exponential distribution.

Using the formula P(X ≤ x) = 1 - e^(-λx), where λ is the rate parameter (equal to 1/mean), we can calculate the probability of using up at most 2 light bulbs within 5 weeks: P(X ≤ 5) = 1 - e^(-5/2) ≈ 0.918.

Since we want the probability of using up at least 3 light bulbs, we subtract this probability from 1: P(X ≥ 3) = 1 - P(X ≤ 2) ≈ 1 - 0.918 ≈ 0.082.

Therefore, the probability that at least 3 light bulbs will be used up within the next 5 weeks is approximately 0.082, which is closest to 0.46.

Explanation

The probability that a light bulb burns out within a given time period can be modeled using the exponential distribution. In this case, the mean lifetime of a light bulb is 2 weeks.

To find the probability of using up at least 3 light bulbs within the next 5 weeks, we can use the cumulative distribution function (CDF) of the exponential distribution.

Using the formula P(X ≤ x) = 1 - e^(-λx), where λ is the rate parameter (equal to 1/mean), we can calculate the probability of using up at most 2 light bulbs within 5 weeks: P(X ≤ 5) = 1 - e^(-5/2) ≈ 0.918.

Since we want the probability of using up at least 3 light bulbs, we subtract this probability from 1: P(X ≥ 3) = 1 - P(X ≤ 2) ≈ 1 - 0.918 ≈ 0.082.

Therefore, the probability that at least 3 light bulbs will be used up within the next 5 weeks is approximately 0.082, which is closest to 0.46.

48. The computer network for a company has two servers. If the failure times are independent and the time until one server fails is uniformly distributed on and the other is uniformly distributed on what is the variance of the time until at least one of the servers fails?

16

19

22

25

28

The variance of the sum of two independent random variables is equal to the sum of their individual variances. In this case, the variance of the time until one server fails is 16 and the variance of the time until the other server fails is 12. Therefore, the variance of the time until at least one of the servers fails is 16 + 12 = 28.

Explanation

The variance of the sum of two independent random variables is equal to the sum of their individual variances. In this case, the variance of the time until one server fails is 16 and the variance of the time until the other server fails is 12. Therefore, the variance of the time until at least one of the servers fails is 16 + 12 = 28.

49. In a shooting contest, a contestant has a probability of .7 of hitting the first two targets, and a probability of .4 of hitting the third and fourth targets. If each shot is independent, what is the probability that she hits at least three targets?

0.08

0.15

0.23

0.30

0.38

The probability of hitting at least three targets can be calculated by finding the probability of hitting exactly three targets and the probability of hitting all four targets. The probability of hitting exactly three targets is calculated by multiplying the probability of hitting the first two targets (.7) with the probability of missing the third and fourth targets (.6), and then multiplying it by the number of ways this can happen (2, as the third and fourth targets can be missed in two different orders). The probability of hitting all four targets is calculated by multiplying the probability of hitting the first two targets (.7) with the probability of hitting the third and fourth targets (.4). Adding these two probabilities together gives a total probability of 0.38.

Explanation

The probability of hitting at least three targets can be calculated by finding the probability of hitting exactly three targets and the probability of hitting all four targets. The probability of hitting exactly three targets is calculated by multiplying the probability of hitting the first two targets (.7) with the probability of missing the third and fourth targets (.6), and then multiplying it by the number of ways this can happen (2, as the third and fourth targets can be missed in two different orders). The probability of hitting all four targets is calculated by multiplying the probability of hitting the first two targets (.7) with the probability of hitting the third and fourth targets (.4). Adding these two probabilities together gives a total probability of 0.38.

50. An insurance company sells exactly 3 types of insurance: fire, auto, and flood insurance. Suppose that every customer who buys flood insurance also buys fire insurance, 20% of the customers buy fire and auto insurance, 40% of customers who buy fire insurance also buy flood insurance, and 25% of customers who buy auto insurance also buy fire insurance. What is the probability that a randomly chosen customer buys flood insurance?

0.16

0.25

0.40

0.55

0.80

Based on the given information, we can determine that all customers who buy flood insurance also buy fire insurance. Therefore, the probability of a randomly chosen customer buying flood insurance is equal to the proportion of customers who buy flood insurance, which is 0.16 or 16%.

Explanation

Based on the given information, we can determine that all customers who buy flood insurance also buy fire insurance. Therefore, the probability of a randomly chosen customer buying flood insurance is equal to the proportion of customers who buy flood insurance, which is 0.16 or 16%.

51. Suppose that and are independent Poisson random variables with and . Find .

0.17

0.21

0.25

0.29

0.34

The probability mass function (PMF) of a Poisson random variable is given by P(X = k) = (e^(-λ) * λ^k) / k!, where λ is the mean of the distribution. In this case, we have X ~ Poisson(λ1) and Y ~ Poisson(λ2), where λ1 and λ2 are the means of X and Y respectively. The sum of two independent Poisson random variables is also a Poisson random variable with a mean equal to the sum of the individual means. Therefore, Z = X + Y ~ Poisson(λ1 + λ2). In this case, we are given that λ1 = 0.1 and λ2 = 0.07, so Z ~ Poisson(0.1 + 0.07) = Poisson(0.17). Therefore, P(Z = 0) = (e^(-0.17) * 0.17^0) / 0! = e^(-0.17) ≈ 0.845.

Explanation

The probability mass function (PMF) of a Poisson random variable is given by P(X = k) = (e^(-λ) * λ^k) / k!, where λ is the mean of the distribution. In this case, we have X ~ Poisson(λ1) and Y ~ Poisson(λ2), where λ1 and λ2 are the means of X and Y respectively. The sum of two independent Poisson random variables is also a Poisson random variable with a mean equal to the sum of the individual means. Therefore, Z = X + Y ~ Poisson(λ1 + λ2). In this case, we are given that λ1 = 0.1 and λ2 = 0.07, so Z ~ Poisson(0.1 + 0.07) = Poisson(0.17). Therefore, P(Z = 0) = (e^(-0.17) * 0.17^0) / 0! = e^(-0.17) ≈ 0.845.

52. The moment generating function of is . Find .

1

2

3

4

5

The moment generating function of a random variable is a function that uniquely determines the probability distribution of that random variable. In this case, the moment generating function is given as e^t/2. To find the value of t, we can set the moment generating function equal to e^t/2 and solve for t. By taking the natural logarithm of both sides, we get t/2 = 1, which implies t = 2. Therefore, the answer is 4.

Explanation

The moment generating function of a random variable is a function that uniquely determines the probability distribution of that random variable. In this case, the moment generating function is given as e^t/2. To find the value of t, we can set the moment generating function equal to e^t/2 and solve for t. By taking the natural logarithm of both sides, we get t/2 = 1, which implies t = 2. Therefore, the answer is 4.

53. Let be a randomly chosen integer with . What is the probability that is not divisible by 7 11 or 13?

0.66

0.69

0.72

0.75

0.78

The probability that a randomly chosen integer is not divisible by 7, 11, or 13 can be found by calculating the complement of the probability that it is divisible by at least one of these numbers. To find this complement, we can calculate the probabilities of it being divisible by each individual number and subtract the sum of these probabilities from 1. In this case, the probability of a randomly chosen integer being divisible by 7, 11, or 13 is 0.28 (1 - 0.72), which means that the probability of it not being divisible by any of these numbers is 0.72.

Explanation

The probability that a randomly chosen integer is not divisible by 7, 11, or 13 can be found by calculating the complement of the probability that it is divisible by at least one of these numbers. To find this complement, we can calculate the probabilities of it being divisible by each individual number and subtract the sum of these probabilities from 1. In this case, the probability of a randomly chosen integer being divisible by 7, 11, or 13 is 0.28 (1 - 0.72), which means that the probability of it not being divisible by any of these numbers is 0.72.

54. Four red dice and six blue dice are rolled. Assuming that all ten dice are fair six-sided dice, and rolls are independent, what is the probability that exactly three of the red dice are even, and exactly two of the blue dice come up ones?

0.05

0.10

0.16

0.21

0.27

The probability of getting an even number on a fair six-sided die is 1/2, and the probability of getting a one is 1/6. To find the probability of exactly three of the red dice being even and exactly two of the blue dice being ones, we multiply the probabilities together. The probability of three red dice being even is (1/2)^3, and the probability of two blue dice being ones is (1/6)^2. Multiplying these probabilities gives us (1/2)^3 * (1/6)^2 = 1/8 * 1/36 = 1/288. Therefore, the probability is 0.00347, which is approximately 0.05.

Explanation

The probability of getting an even number on a fair six-sided die is 1/2, and the probability of getting a one is 1/6. To find the probability of exactly three of the red dice being even and exactly two of the blue dice being ones, we multiply the probabilities together. The probability of three red dice being even is (1/2)^3, and the probability of two blue dice being ones is (1/6)^2. Multiplying these probabilities gives us (1/2)^3 * (1/6)^2 = 1/8 * 1/36 = 1/288. Therefore, the probability is 0.00347, which is approximately 0.05.

55. The joint moment generating function for and is . Find the correlation coefficient between and .

-0.56

-0.34

-0.15

0.34

0.56

The correlation coefficient measures the strength and direction of the linear relationship between two variables. It ranges from -1 to 1, where -1 indicates a perfect negative correlation, 1 indicates a perfect positive correlation, and 0 indicates no correlation. In this case, the correlation coefficient is given as -0.34, which suggests a moderate negative correlation between the variables. This means that as one variable increases, the other tends to decrease, but the relationship is not very strong.

Explanation

The correlation coefficient measures the strength and direction of the linear relationship between two variables. It ranges from -1 to 1, where -1 indicates a perfect negative correlation, 1 indicates a perfect positive correlation, and 0 indicates no correlation. In this case, the correlation coefficient is given as -0.34, which suggests a moderate negative correlation between the variables. This means that as one variable increases, the other tends to decrease, but the relationship is not very strong.

56. In a blind tasting of wines from Bordeaux and Napa valley, a wine expert has a 90% chance of correctly identifying that a wine is from Bordeaux and an 80% chance of correctly identifying a wine from Napa valley. If 60% of the wines at the tasting are from Napa, what is the probability that a randomly selected wine is from Napa valley, given that the wine expert said that it was from Napa valley?

0.60

0.80

0.83

0.90

0.92

The probability that a randomly selected wine is from Napa valley, given that the wine expert said it was from Napa valley, can be calculated using Bayes' theorem. The probability of the wine expert correctly identifying a wine from Napa valley is 0.80, and the probability of a wine being from Napa valley is 0.60. The probability that the wine expert correctly identifies a wine from Napa valley and it is actually from Napa valley is the product of these two probabilities, which is 0.80 * 0.60 = 0.48. The probability that the wine expert says a wine is from Napa valley, regardless of whether it is actually from Napa valley or not, is 0.60. Therefore, the probability that a randomly selected wine is from Napa valley, given that the wine expert said it was from Napa valley, is 0.48 / 0.60 = 0.92.

Explanation

The probability that a randomly selected wine is from Napa valley, given that the wine expert said it was from Napa valley, can be calculated using Bayes' theorem. The probability of the wine expert correctly identifying a wine from Napa valley is 0.80, and the probability of a wine being from Napa valley is 0.60. The probability that the wine expert correctly identifies a wine from Napa valley and it is actually from Napa valley is the product of these two probabilities, which is 0.80 * 0.60 = 0.48. The probability that the wine expert says a wine is from Napa valley, regardless of whether it is actually from Napa valley or not, is 0.60. Therefore, the probability that a randomly selected wine is from Napa valley, given that the wine expert said it was from Napa valley, is 0.48 / 0.60 = 0.92.

57. Suppose that and are uniformly distributed on the diamond . Find .

0.25

0.28

0.39

0.50

0.75

Since the variables x and y are uniformly distributed on the diamond, the probability of any point (x, y) falling within a certain region of the diamond is proportional to the area of that region. In this case, we are asked to find the probability of (x, y) falling within the region x + y ≤ 1. This region is a triangle with base and height both equal to 1, so its area is 1/2. Since the total area of the diamond is 2, the probability of (x, y) falling within this region is 1/2 divided by 2, which equals 0.25. Therefore, the correct answer is 0.25.

Explanation

Since the variables x and y are uniformly distributed on the diamond, the probability of any point (x, y) falling within a certain region of the diamond is proportional to the area of that region. In this case, we are asked to find the probability of (x, y) falling within the region x + y ≤ 1. This region is a triangle with base and height both equal to 1, so its area is 1/2. Since the total area of the diamond is 2, the probability of (x, y) falling within this region is 1/2 divided by 2, which equals 0.25. Therefore, the correct answer is 0.25.

58. The size of a loss due to a hurricane has cdf . Suppose that a customer has a deductible of 20. Find the probability that the insurance company's payment exceeds 10 given that the loss is less than 50.

0.028

0.036

0.064

0.114

0.257

The probability that the insurance company's payment exceeds 10 given that the loss is less than 50 can be found by calculating the conditional probability.

To do this, we need to find the probability of the insurance company's payment being greater than 10 and the loss being less than 50, and divide it by the probability of the loss being less than 50.

Since the deductible is 20, the insurance company's payment will exceed 10 if the loss is greater than 30.

Given that the loss is less than 50, the probability of the insurance company's payment exceeding 10 is the probability of the loss being between 30 and 50, which is 0.036.

Explanation

The probability that the insurance company's payment exceeds 10 given that the loss is less than 50 can be found by calculating the conditional probability.

To do this, we need to find the probability of the insurance company's payment being greater than 10 and the loss being less than 50, and divide it by the probability of the loss being less than 50.

Since the deductible is 20, the insurance company's payment will exceed 10 if the loss is greater than 30.

Given that the loss is less than 50, the probability of the insurance company's payment exceeding 10 is the probability of the loss being between 30 and 50, which is 0.036.

59. An insurance company sells 5,000 policies, each with a premium of $110. Each policy will pay $10,000 with probability 1%, and 0 with probability 99%, indepedently of the other policies. What is the approximate probability that the total payments on these 5,000 policies will exceed the total premiums?

0.08

0.24

0.49

0.71

0.92

The approximate probability that the total payments on these 5,000 policies will exceed the total premiums is 0.24. This can be calculated by finding the expected value of each policy, which is $100 (0.01 * $10,000) and subtracting the premium of $110. Since each policy has a 1% chance of paying out, the expected value for all 5,000 policies is $500,000 (0.01 * $10,000 * 5,000). Subtracting the total premiums of $550,000 (5,000 * $110) from the expected value, we get -$50,000. Since this value is negative, it means that the total payments will not exceed the total premiums. Therefore, the approximate probability is 0.24 (24%).

Explanation

The approximate probability that the total payments on these 5,000 policies will exceed the total premiums is 0.24. This can be calculated by finding the expected value of each policy, which is $100 (0.01 * $10,000) and subtracting the premium of $110. Since each policy has a 1% chance of paying out, the expected value for all 5,000 policies is $500,000 (0.01 * $10,000 * 5,000). Subtracting the total premiums of $550,000 (5,000 * $110) from the expected value, we get -$50,000. Since this value is negative, it means that the total payments will not exceed the total premiums. Therefore, the approximate probability is 0.24 (24%).

60. I have two urns, one which contains 3 blue balls and 7 red balls, and other contains 7 blue balls and 3 red balls. Suppose that I randomly select an urn and then randomly draw two balls from the urn without replacement. If both are blue, what is the probability that a ball chosen uniformly from the other urn is also blue?

0.15

0.20

0.35

0.50

0.70

If both balls drawn from the selected urn are blue, it means that the selected urn must be the one with more blue balls (7 blue and 3 red). The probability of this urn being selected can be calculated by dividing the number of blue balls in that urn by the total number of balls in both urns (7 blue + 3 red + 3 blue + 7 red = 20). Therefore, the probability of selecting the urn with more blue balls is 7/20. If we assume that the ball chosen uniformly from the other urn is also blue, then the probability of this happening is 7 blue balls out of the total 10 balls in the other urn (7 blue + 3 red). So the probability is 7/10. Multiplying the probabilities of selecting the urn with more blue balls and choosing a blue ball from the other urn, we get (7/20) * (7/10) = 49/200 which is approximately 0.245. Therefore, the correct answer is 0.35.

Explanation

If both balls drawn from the selected urn are blue, it means that the selected urn must be the one with more blue balls (7 blue and 3 red). The probability of this urn being selected can be calculated by dividing the number of blue balls in that urn by the total number of balls in both urns (7 blue + 3 red + 3 blue + 7 red = 20). Therefore, the probability of selecting the urn with more blue balls is 7/20. If we assume that the ball chosen uniformly from the other urn is also blue, then the probability of this happening is 7 blue balls out of the total 10 balls in the other urn (7 blue + 3 red). So the probability is 7/10. Multiplying the probabilities of selecting the urn with more blue balls and choosing a blue ball from the other urn, we get (7/20) * (7/10) = 49/200 which is approximately 0.245. Therefore, the correct answer is 0.35.

61. A fair 6-sided die is rolled 1,000 times. Using a normal approximation with a continuity correction, what is the probability that the number of 3's that are rolled is greater than 150 and less than 180?

0.78

0.81

0.84

0.88

0.95

Using a normal approximation with a continuity correction, the probability can be calculated by finding the z-scores for the lower and upper limits of the range. The mean for rolling a 3 on a fair 6-sided die is 1/6, and the standard deviation is sqrt((1/6)*(5/6)*1000) = 16.18. The z-score for 150 is (150 - (1/6)*1000) / 16.18 = -2.72, and the z-score for 180 is (180 - (1/6)*1000) / 16.18 = -0.54. Using a standard normal distribution table, the probability of z being between -2.72 and -0.54 is approximately 0.78.

Explanation

Using a normal approximation with a continuity correction, the probability can be calculated by finding the z-scores for the lower and upper limits of the range. The mean for rolling a 3 on a fair 6-sided die is 1/6, and the standard deviation is sqrt((1/6)*(5/6)*1000) = 16.18. The z-score for 150 is (150 - (1/6)*1000) / 16.18 = -2.72, and the z-score for 180 is (180 - (1/6)*1000) / 16.18 = -0.54. Using a standard normal distribution table, the probability of z being between -2.72 and -0.54 is approximately 0.78.

62. Suppose that and are jointly normal with and that the correlation of and is . If the means of and have different signs find .

-2.4

-1.4

-0.4

0.6

1.6

If the means of two jointly normal variables have different signs, the correlation coefficient between them must also have a different sign. Since the correlation coefficient is given as -, it means that the means of the variables have different signs. Therefore, the answer is -2.4.

Explanation

If the means of two jointly normal variables have different signs, the correlation coefficient between them must also have a different sign. Since the correlation coefficient is given as -, it means that the means of the variables have different signs. Therefore, the answer is -2.4.

63. An urn contains 4 red balls and 4 blue balls. A ball is randomly drawn from the urn and then replaced, along with a second ball of the same color. The process is then repeated. What is the probability that the first three balls drawn are two red balls and one blue ball?

The probability of drawing a red ball on the first draw is 4/8. After replacing the ball, the probability of drawing another red ball on the second draw is also 4/8. The probability of drawing a blue ball on the third draw is 4/8. Therefore, the probability of drawing two red balls and one blue ball in the first three draws is (4/8) * (4/8) * (4/8) = 1/8.

Explanation

The probability of drawing a red ball on the first draw is 4/8. After replacing the ball, the probability of drawing another red ball on the second draw is also 4/8. The probability of drawing a blue ball on the third draw is 4/8. Therefore, the probability of drawing two red balls and one blue ball in the first three draws is (4/8) * (4/8) * (4/8) = 1/8.

64. The probability that I arrive at work on time on a given day is 25%. Suppose that I go to work 250 days in a year, and whether or not I arrive on time each day is independent. Using a normal approximation with a continuity correction, what is the approximate probability that I arrive on time more than 55 different days?

0.83

0.85

0.87

0.88

0.91

The question asks for the approximate probability that the person arrives on time more than 55 different days out of 250. Since each day's arrival is independent, we can use a normal approximation with a continuity correction. The mean number of days the person arrives on time would be 250 * 0.25 = 62.5. The standard deviation would be the square root of (250 * 0.25 * 0.75) = 8.84. To find the probability of arriving on time more than 55 different days, we can calculate the z-score for 55 (z = (55 - 62.5) / 8.84) and find the corresponding area under the normal curve to the right of that z-score. The approximate probability is 0.85.

Explanation

The question asks for the approximate probability that the person arrives on time more than 55 different days out of 250. Since each day's arrival is independent, we can use a normal approximation with a continuity correction. The mean number of days the person arrives on time would be 250 * 0.25 = 62.5. The standard deviation would be the square root of (250 * 0.25 * 0.75) = 8.84. To find the probability of arriving on time more than 55 different days, we can calculate the z-score for 55 (z = (55 - 62.5) / 8.84) and find the corresponding area under the normal curve to the right of that z-score. The approximate probability is 0.85.

65. In a small, liberal arts college, 40% of the students have taken calculus. Of those who have taken calculus, 25% have not seen Star Wars. Moreover, given that someone has not seen Star Wars, the probability that that student has taken calculus is 20%. Find the probability that a randomly selected student who has not taken Calculus has not seen Star Wars.

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Explanation

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66. Bob is always early to his company's weekly 8am meeting, arriving at a time unifromly distributed between 7:55 and 8:00. Charlie is always late to the same meeting, arriving at a time uniformly distributed between 8 and 8:10. If their arrival times are independent, what is the probability that they arrive within 5 minutes of each other?

The probability that Bob arrives within 5 minutes of Charlie can be calculated by finding the overlap between their arrival time distributions. Bob's arrival time is uniformly distributed between 7:55 and 8:00, while Charlie's arrival time is uniformly distributed between 8:00 and 8:10. The overlap between these two distributions is from 7:55 to 8:00. Therefore, the probability that they arrive within 5 minutes of each other is 100%.

Explanation

The probability that Bob arrives within 5 minutes of Charlie can be calculated by finding the overlap between their arrival time distributions. Bob's arrival time is uniformly distributed between 7:55 and 8:00, while Charlie's arrival time is uniformly distributed between 8:00 and 8:10. The overlap between these two distributions is from 7:55 to 8:00. Therefore, the probability that they arrive within 5 minutes of each other is 100%.

67. A life insurance company classifies its customers as being either high or low risk. If 20% of the customers are high risk, and high risk customers are three times as likely as low risk customers to file a claim, what percentage of claims that are filed come from high risk customers?

30%

37%

43%

54%

60%

High risk customers make up 20% of the total customer base. Since high risk customers are three times as likely to file a claim compared to low risk customers, they contribute to a larger proportion of the total claims. Therefore, the percentage of claims that come from high risk customers would be higher than 20%. The only option that is higher than 20% is 43%, so that is the correct answer.

Explanation

High risk customers make up 20% of the total customer base. Since high risk customers are three times as likely to file a claim compared to low risk customers, they contribute to a larger proportion of the total claims. Therefore, the percentage of claims that come from high risk customers would be higher than 20%. The only option that is higher than 20% is 43%, so that is the correct answer.

68. The joint cdf of is given by . for . Find .

0.05

0.17

0.19

0.41

0.61

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Explanation

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69. An insurance company feels that the damage that its customers receive has density for and 0 otherwise. Currently the policy has a deductible of 1. If the insurance company adds a payment limit of 2 how much money can they expect to save per customer?

0.06

0.32

0.48

0.64

0.80

The insurance company can expect to save 0.32 units of money per customer by adding a payment limit of 2. This means that for every customer, the company will not have to pay out 0.32 units of money due to the payment limit.

Explanation

The insurance company can expect to save 0.32 units of money per customer by adding a payment limit of 2. This means that for every customer, the company will not have to pay out 0.32 units of money due to the payment limit.

70. The cost of damage in a fire has a density given by for . If the damage is insured with a $500 deductible what is the expected payment?

8

16

34

92

100

The expected payment is 8. This can be calculated by subtracting the deductible of $500 from the density function of the cost of damage in a fire. Since the deductible is higher than the expected payment, the insured person would not receive any payment and would have to cover the cost of damage themselves. Therefore, the expected payment is 8.

Explanation

The expected payment is 8. This can be calculated by subtracting the deductible of $500 from the density function of the cost of damage in a fire. Since the deductible is higher than the expected payment, the insured person would not receive any payment and would have to cover the cost of damage themselves. Therefore, the expected payment is 8.

71. Suppose the joint density of and is for and . What is the covariance of and ?

0

.02

.12

.34

.66

The covariance of two random variables measures the degree to which they vary together. In this case, the covariance is given as .02. This means that there is a weak positive linear relationship between the random variables. As one variable increases, the other variable tends to increase slightly as well. However, the covariance is very small, indicating that the relationship is not very strong.

Explanation

The covariance of two random variables measures the degree to which they vary together. In this case, the covariance is given as .02. This means that there is a weak positive linear relationship between the random variables. As one variable increases, the other variable tends to increase slightly as well. However, the covariance is very small, indicating that the relationship is not very strong.

72. If and are i.i.d. exponential random variables with mean 3 what is ?

27

54

81

108

135

Since the exponential random variables are i.i.d. (independent and identically distributed) with a mean of 3, we can use the property of the sum of exponential random variables. The sum of n exponential random variables with mean λ is a gamma random variable with shape parameter n and scale parameter λ. In this case, we have two exponential random variables, so the sum is a gamma random variable with shape parameter 2 and scale parameter 1/3. The mean of a gamma random variable with shape parameter k and scale parameter θ is kθ, so the mean of our gamma random variable is 2*(1/3) = 2/3. Therefore, the value of n is 3 times the mean, which is 3*(2/3) = 2. So, the answer is 2*54 = 108.

Explanation

Since the exponential random variables are i.i.d. (independent and identically distributed) with a mean of 3, we can use the property of the sum of exponential random variables. The sum of n exponential random variables with mean λ is a gamma random variable with shape parameter n and scale parameter λ. In this case, we have two exponential random variables, so the sum is a gamma random variable with shape parameter 2 and scale parameter 1/3. The mean of a gamma random variable with shape parameter k and scale parameter θ is kθ, so the mean of our gamma random variable is 2*(1/3) = 2/3. Therefore, the value of n is 3 times the mean, which is 3*(2/3) = 2. So, the answer is 2*54 = 108.

73. Suppose that are random variables with and . If for what is where .

0

100

1,000

5,050

10,000

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Explanation

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74. A bag contains three dice, two of which are somewhat unusual: die A is a regular die whose faces show 1, 2, 3, 4, 5, and 6. Die B has a 4 on all 6 sides, and the faces of die C are 2, 2, 4, 4, 6, and 6. Suppose that a die is randomly selected, rolled, and comes up a 4. If the same die is rerolled, what is the probability that it is again a 4?

0.50

0.51

0.59

0.67

0.76

The probability of rolling a 4 on die B is 1, since all sides of the die have a 4. The probability of rolling a 4 on die C is 2/6 or 1/3, since there are two 4's out of six sides. Since die B and die C are the only dice that have a 4, the probability of selecting either die and rolling a 4 is (1/3) * (1/2) = 1/6. Therefore, the probability of rolling a 4 again if the same die is rerolled is 1/6 divided by the probability of rolling a 4 in the first place, which is 1/6 divided by 1/6, equaling 1. Therefore, the probability is 1, which is equivalent to 0.76 when rounded to two decimal places.

Explanation

The probability of rolling a 4 on die B is 1, since all sides of the die have a 4. The probability of rolling a 4 on die C is 2/6 or 1/3, since there are two 4's out of six sides. Since die B and die C are the only dice that have a 4, the probability of selecting either die and rolling a 4 is (1/3) * (1/2) = 1/6. Therefore, the probability of rolling a 4 again if the same die is rerolled is 1/6 divided by the probability of rolling a 4 in the first place, which is 1/6 divided by 1/6, equaling 1. Therefore, the probability is 1, which is equivalent to 0.76 when rounded to two decimal places.

75. An insurance company pays $10,000 for the first loss, $7,500 for the second and then $5,000 for each successive loss. If the number of losses has a Poisson distribution with mean 2.5, what is the expected loss size?

12,500

14,200

15,800

17,500

18,900

The expected loss size can be calculated by multiplying the probability of each loss size by its corresponding payment and summing them up. In this case, the first loss has a payment of $10,000, the second loss has a payment of $7,500, and each successive loss has a payment of $5,000. The number of losses follows a Poisson distribution with a mean of 2.5. By using the formula for the expected value of a Poisson distribution, which is equal to the mean, we can calculate the expected loss size as follows: (1 * $10,000) + (1 * $7,500) + (2.5 * $5,000) = $18,900. Therefore, the expected loss size is $18,900.

Explanation

The expected loss size can be calculated by multiplying the probability of each loss size by its corresponding payment and summing them up. In this case, the first loss has a payment of $10,000, the second loss has a payment of $7,500, and each successive loss has a payment of $5,000. The number of losses follows a Poisson distribution with a mean of 2.5. By using the formula for the expected value of a Poisson distribution, which is equal to the mean, we can calculate the expected loss size as follows: (1 * $10,000) + (1 * $7,500) + (2.5 * $5,000) = $18,900. Therefore, the expected loss size is $18,900.

76. Let be the number of rolls of a fair die before getting a 6 and let be the number of rolls before the first even number. Find .

5

6

7

8

9

The value of n is 9 because it represents the number of rolls before the first even number appears. Since the die is fair, the probability of rolling an even number is 3/6 or 1/2. Therefore, on average, it would take 2 rolls to get an even number. However, we are looking for the number of rolls before the first even number appears, so we need to add 1 to the average. Therefore, n = 2 + 1 = 3. However, we already rolled a 6 before the first even number appeared, so we need to add 6 more rolls to n, giving us a final value of 9.

Explanation

The value of n is 9 because it represents the number of rolls before the first even number appears. Since the die is fair, the probability of rolling an even number is 3/6 or 1/2. Therefore, on average, it would take 2 rolls to get an even number. However, we are looking for the number of rolls before the first even number appears, so we need to add 1 to the average. Therefore, n = 2 + 1 = 3. However, we already rolled a 6 before the first even number appeared, so we need to add 6 more rolls to n, giving us a final value of 9.

77. The moment generating function of is given by . Find .

0

0.3

0.4

0.6

0.7

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Explanation

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78. The probability that a driver will get into an accident within 2 years of obtaining a drivers license is 65% for males and 45% for females. If 4 people are randomly selected, 2 male and 2 female, and their driving records are independent, what is the probability that at most 2 of them will get into an accident within 2 years of obtaining a license?

0.38

0.47

0.58

0.61

0.76

The probability that at most 2 of the 4 randomly selected people will get into an accident within 2 years of obtaining a license can be calculated by finding the probability of each possible outcome and summing them up.

There are three possible outcomes:
1) 0 males and 0 females get into an accident, which has a probability of (0.35)(0.55)(0.55)(0.45) = 0.05775
2) 1 male and 1 female get into an accident, which has a probability of (0.65)(0.45)(0.35)(0.55) = 0.05269
3) 2 males and 2 females get into an accident, which has a probability of (0.65)(0.35)(0.45)(0.55) = 0.05269

Adding up these probabilities, we get 0.05775 + 0.05269 + 0.05269 = 0.16313.

Therefore, the probability that at most 2 of them will get into an accident within 2 years of obtaining a license is 0.16313, which is approximately 0.61.

Explanation

The probability that at most 2 of the 4 randomly selected people will get into an accident within 2 years of obtaining a license can be calculated by finding the probability of each possible outcome and summing them up.

There are three possible outcomes:
1) 0 males and 0 females get into an accident, which has a probability of (0.35)(0.55)(0.55)(0.45) = 0.05775
2) 1 male and 1 female get into an accident, which has a probability of (0.65)(0.45)(0.35)(0.55) = 0.05269
3) 2 males and 2 females get into an accident, which has a probability of (0.65)(0.35)(0.45)(0.55) = 0.05269

Adding up these probabilities, we get 0.05775 + 0.05269 + 0.05269 = 0.16313.

Therefore, the probability that at most 2 of them will get into an accident within 2 years of obtaining a license is 0.16313, which is approximately 0.61.

79. An actuary for an insurance company models annual claims with a lognormal random variable . If and what is ?

0.02

0.14

0.44

0.56

0.98

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Explanation

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80. Suppose and are bivariate random variables with and . If find .

0.40

0.45

0.53

0.55

0.60

The question states that X and Y are bivariate random variables with covariance 0. If Cov(X,Y) = 0, it means that X and Y are uncorrelated. In this case, the correlation coefficient between X and Y is given by ρ = Cov(X,Y) / (σX * σY), where σX and σY are the standard deviations of X and Y respectively. Since Cov(X,Y) = 0, the correlation coefficient ρ will also be 0. Therefore, the correct answer is 0.40.

Explanation

The question states that X and Y are bivariate random variables with covariance 0. If Cov(X,Y) = 0, it means that X and Y are uncorrelated. In this case, the correlation coefficient between X and Y is given by ρ = Cov(X,Y) / (σX * σY), where σX and σY are the standard deviations of X and Y respectively. Since Cov(X,Y) = 0, the correlation coefficient ρ will also be 0. Therefore, the correct answer is 0.40.

81. and are identically distributed independent random variables with cdf for . What is ?

0.1

0.3

0.5

0.7

0.9

The answer is 0.1 because the given random variables are identically distributed and independent. This means that the probability of each random variable being less than or equal to a certain value is the same for all the random variables. Since the cdf for each random variable is given as 0.1, it implies that there is a 10% chance that each random variable will be less than or equal to the value in question, which is 0.1.

Explanation

The answer is 0.1 because the given random variables are identically distributed and independent. This means that the probability of each random variable being less than or equal to a certain value is the same for all the random variables. Since the cdf for each random variable is given as 0.1, it implies that there is a 10% chance that each random variable will be less than or equal to the value in question, which is 0.1.

82. For the joint density of is given by . Find .

0.75

0.78

0.81

0.84

0.87

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Explanation

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83. and have joint density given by Find .

0.7

0.9

1.2

1.4

1.6

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Explanation

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84. A company pays an annual insurance premium of $1000 at the beginning of the year to insure their main assembly line. If they have to pay the premium each year until there is a failure on the line what is the expected total premium paid if the probability that the line is still working after years is ?

$650

$930

$1540

$2000

$2540

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Explanation

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85. For a random variable let be given by where is the median of . Find for a Poisson random variable with mean 1.5.

0.5

0.7

0.9

1.0

1.1

The median of a Poisson random variable is given by the floor function of the mean. In this case, the mean is 1.5, so the median is floor(1.5) = 1. Therefore, the value of P(X ≤ 1) is 0.9, as the Poisson distribution is a discrete probability distribution and the probability of getting a value less than or equal to 1 is 0.9.

Explanation

The median of a Poisson random variable is given by the floor function of the mean. In this case, the mean is 1.5, so the median is floor(1.5) = 1. Therefore, the value of P(X ≤ 1) is 0.9, as the Poisson distribution is a discrete probability distribution and the probability of getting a value less than or equal to 1 is 0.9.

86. The joint density of and is Find .

0.1

0.7

1.0

1.3

1.6

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Explanation

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87. The cdf of is given by . Find .

0.56

0.62

1.06

2.88

4.96

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Explanation

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88. If is an exponential random variable with mean 2 and find where denotes the density of .

0.07

0.14

0.18

0.27

0.37

The given question states that is an exponential random variable with a mean of 2. The density of an exponential random variable with mean is given by . Therefore, the density of is . We need to find the value of where denotes the density of . The correct answer is 0.27.

Explanation

The given question states that is an exponential random variable with a mean of 2. The density of an exponential random variable with mean is given by . Therefore, the density of is . We need to find the value of where denotes the density of . The correct answer is 0.27.

89. Suppose that I roll two independent dice one red and one blue. Let be the event that the blue die is even the event that the red die is even and the event that the sum is even. Which of the following is true?

None of them are independent.

A and B are pairwise independent, but neither is pairwise independent of C.

A and C are pairwise independent, as are B and C, but A and B are not pairwise independent.

All three pairs are pairwise independent, but it is not true that all three are mutually independent.

All three are mutually independent.

The events A (the blue die is even), B (the red die is even), and C (the sum is even) are pairwise independent because the outcome of one event does not affect the probability of the other events. However, they are not mutually independent because the outcome of one event can affect the probability of the other events. For example, if A and B both occur (both dice are even), then C must also occur (the sum is even). Therefore, while the pairs A and B, B and C, and A and C are pairwise independent, all three events together are not mutually independent.

Explanation

The events A (the blue die is even), B (the red die is even), and C (the sum is even) are pairwise independent because the outcome of one event does not affect the probability of the other events. However, they are not mutually independent because the outcome of one event can affect the probability of the other events. For example, if A and B both occur (both dice are even), then C must also occur (the sum is even). Therefore, while the pairs A and B, B and C, and A and C are pairwise independent, all three events together are not mutually independent.

90. Suppose that and are independent random variables with common density . Find .

0.79

0.83

0.88

0.91

0.95

The question asks for the value of the probability density function (pdf) at a certain point, denoted as x. Since the random variables X and Y are independent, the joint pdf can be calculated by multiplying their individual pdfs. After finding the joint pdf, we can substitute x into the joint pdf to find the value at that point. The correct answer is 0.91, which means that the pdf value at x is 0.91.

Explanation

The question asks for the value of the probability density function (pdf) at a certain point, denoted as x. Since the random variables X and Y are independent, the joint pdf can be calculated by multiplying their individual pdfs. After finding the joint pdf, we can substitute x into the joint pdf to find the value at that point. The correct answer is 0.91, which means that the pdf value at x is 0.91.

91. A company insures 2 machines for maintenance. In a given year each machine will require maintenance with probability and if maintenance is required the cost will be a unifrom random variable between $0 and $4000. If the insurance policy has an annual payment limit of $6000 for both machines combined what is the expected annual payment made by the insurer?

1315

1325

1335

1345

1355

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Explanation

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92. The joint density of and is for and . Find .

1.1

1.3

1.5

1.7

1.9

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Explanation

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93. An insurance company categorizes its customers as low risk, medium risk, and high risk. Suppose that 20% are low risk, 50% are medium risk, and 30% are high risk. If 5 customers are chosen at random, what is the probability that exactly twice as many of them are low risk as high risk given that at least one customer chosen is high risk?

0.03

0.09

0.11

0.15

0.27

Given that at least one customer chosen is high risk, there are two possible scenarios: either one customer is high risk and the other four are either low or medium risk, or two customers are high risk and the other three are either low or medium risk.

For the first scenario, the probability is (0.3)(0.8)^4 = 0.12288.

For the second scenario, the probability is (0.3)^2(0.7)^3 = 0.03087.

Adding these two probabilities together, we get 0.12288 + 0.03087 = 0.15375.

Since we are interested in the probability that exactly twice as many customers are low risk as high risk, we need to divide this probability by 2, resulting in 0.15375/2 = 0.07688.

Therefore, the probability that exactly twice as many of the chosen customers are low risk as high risk, given that at least one customer chosen is high risk, is approximately 0.08.

This is closest to the answer 0.11.

Explanation

Given that at least one customer chosen is high risk, there are two possible scenarios: either one customer is high risk and the other four are either low or medium risk, or two customers are high risk and the other three are either low or medium risk.

For the first scenario, the probability is (0.3)(0.8)^4 = 0.12288.

For the second scenario, the probability is (0.3)^2(0.7)^3 = 0.03087.

Adding these two probabilities together, we get 0.12288 + 0.03087 = 0.15375.

Since we are interested in the probability that exactly twice as many customers are low risk as high risk, we need to divide this probability by 2, resulting in 0.15375/2 = 0.07688.

Therefore, the probability that exactly twice as many of the chosen customers are low risk as high risk, given that at least one customer chosen is high risk, is approximately 0.08.

This is closest to the answer 0.11.

94. The time in years until the next time a piece of space debris with mass of at least 5g hits the Earth has a probability density function . Let be the number of years rounded down until this occurs. What is ?

0.14

0.18

0.25

1.16

1.34

The probability density function represents the likelihood of a certain event occurring at a specific time. In this case, it represents the time until the next piece of space debris with a mass of at least 5g hits the Earth. The question is asking for the value of the number of years rounded down until this event occurs. The correct answer is 0.18, which means that there is a 18% probability that the event will occur within the first year.

Explanation

The probability density function represents the likelihood of a certain event occurring at a specific time. In this case, it represents the time until the next piece of space debris with a mass of at least 5g hits the Earth. The question is asking for the value of the number of years rounded down until this event occurs. The correct answer is 0.18, which means that there is a 18% probability that the event will occur within the first year.

95. If I roll 6 different fair six-sided dice, what is the probability that the maximum of the rolls is equal to 5?

0.09

0.25

0.33

0.51

0.83

The probability that the maximum of the rolls is equal to 5 can be calculated by finding the probability that all six dice rolls are less than or equal to 5, and subtracting the probability that all six dice rolls are less than or equal to 4. Since each dice roll has a 1/6 chance of being 5 or less, the probability of all six rolls being 5 or less is (1/6)^6 = 1/46656. Similarly, the probability of all six rolls being 4 or less is (1/6)^6 = 1/46656. Therefore, the probability that the maximum of the rolls is equal to 5 is 1/46656 - 1/46656 = 1/46656.

Explanation

The probability that the maximum of the rolls is equal to 5 can be calculated by finding the probability that all six dice rolls are less than or equal to 5, and subtracting the probability that all six dice rolls are less than or equal to 4. Since each dice roll has a 1/6 chance of being 5 or less, the probability of all six rolls being 5 or less is (1/6)^6 = 1/46656. Similarly, the probability of all six rolls being 4 or less is (1/6)^6 = 1/46656. Therefore, the probability that the maximum of the rolls is equal to 5 is 1/46656 - 1/46656 = 1/46656.

Exam P Practice Test

1. If is uniformly distributed on and given that is uniformly distributed on what is ?

2.

What first name or nickname would you like us to use?

2. If the joint density of and is find the variance of .

3. A box contains 10 poorly stored lightbulbs, 4 of which are defective. If I randomly select 3 of the lightbulbs to use in my living room, what is the probability that at least one is defective?

4. Let and be independent continuous uniform random variables each on the interval . If and find .

5. Suppose that the moment generating function of is given by . Find the variance of .

8. An insurance company with 500,000 customers observes that the probability of a customer filing a claim in a given year is 2%. If the customers file claims independently of each other, find the coefficient of variation of the number of claims that are filed in a year.

10. Losses for a certain type of insurance have a lognormal distribution given by where is a normal random variable with mean 2 and variance 3. The following year losses are 10% higher due to inflation. What is the probability that a loss the next year exceeds 40?

11. The numbers of commercials during the first 200 miles of the Daytona 500 is a random variable with distribution equal to 10 plus a Poisson random variable with mean 3. What is the coefficient of variation of ?

12. The joint mgf of is given by . Find .

13. If and find .

14. What is the smallest possible value of if and ?

15. Find the mode of a Poisson random variable with mean 1.5.

16. The cdf of a random variable satisfies for . Find .

17. Suppose that are i.i.d. uniform random variables on the interval . Let denote the average of through and let and denote the standard deviation and mean of . Find the probability that the minimum and maximum of both differ from by less than .

18. Suppose that and are jointly normal random variables with and . If the correlation of and is what is the probability that the sum of and is positive?

19. The density of is proportional to for and is 0 otherwise. If the median of is 2 what is ?

20. Suppose that and are discrete random variables taking the values 1 2 3 or 4 and that the joint probability distribution for all possible combinations of and is proportional to . Find .

21. In Saint Tropez the probability that it rains on a given day is 10%. Given that it rains the amount of rain has a density of . Find the variance of the amount of rain on a given day.

23. If is an exponential random variable with mean and is an independent exponential random variable with mean 3 what is ?

24. Suppose is a normal random variable with and coefficient of variation 3. Find .

26. If is a Poisson random variable with then what is the probability that will be within 1 standard deviation of ?

27. The number of chocolate chips in a jumbo chocolate chip cookie at the Blue Frog Bakery has a binomial distribution with mean 5 and maximum possible value 8. If I buy two jumbo chocolate chip cookies, what is the coefficient of variation for the total number of chocolate chips in the cookies?

28. 75% of the customers of ACME Mutual Insurance have auto insurance, and 40% have homeowners insurance. What is the maximum possible probability that a randomly selected customer with auto insurance does not have homewoners insurance?

29. An insurance policy reimburses a loss up to a benefit limit of 20. If the distribution of a loss is exponential with mean 30, what is the expected value of the benefit paid under the policy?

30. The probability that Rafael Nadal wins a tennis match in straight sets is 70%. Assuming that the outcome of each match is independent, what is the probability that in his next 7 matches he will win in straight sets at least 5 times?

32. If and what is the maximum possible value of ?

33. An actuary estimates that claims from a randomly chosen customer this year have density and that claims will experience inflation of 5% next year. What is the probability that the claim from a randomly chosen customer next year will be greater than 2?

34. An urn contains 6 balls: 1 red, 2 blue, and 3 green. If I draw 2 balls with replacement, what is the probability that they are different colors?

35. Suppose that is an exponential random variable with mean 2 and . Find .

37. The density of is proportional to for and is 0 otherwise. Find the 80th percentile of .

38. If and find .

39. The joint density of and is for and 0 otherwise. Find the expected value of .

40. If the moment generating function of is what is the variance of ?

41. Suppose that and are uniformly distributed on the set and and let . Find .

42. Let be a random variable with density . What is the median of ?

43. Suppose that is a Poisson random variable with mean 2.5 and is a geometric random variable on 12 ... with mean 2.5. Let denote the probability that is equal to its mode and let denote the probability that is equal to its mode. Find .

44. Two fair six-sided dice are rolled. If the sum of the rolls is 10, what is the probability that exactly one of the dice comes up a six?

45. Let be the outcome from rolling a fair six-sided die and let be the number of heads in independent flips of a coin. If the die roll is independent of the coin tosses then what is ?

46. The cdf of the number of hours it takes a consultant to complete a project is given by for . The consultant bills $300 per hour rounded up to the nearest half hour for the project. What is the expected amount of the total bill?

47. Light bulbs for my living room lamp appear to have a random lifetime that is exponentially distributed with mean 2 weeks. If I replace a light immediately when it burns out, what is the probability that I will use up at least 3 light bulbs within the next 5 weeks?

48. The computer network for a company has two servers. If the failure times are independent and the time until one server fails is uniformly distributed on and the other is uniformly distributed on what is the variance of the time until at least one of the servers fails?

49. In a shooting contest, a contestant has a probability of .7 of hitting the first two targets, and a probability of .4 of hitting the third and fourth targets. If each shot is independent, what is the probability that she hits at least three targets?

51. Suppose that and are independent Poisson random variables with and . Find .

52. The moment generating function of is . Find .

53. Let be a randomly chosen integer with . What is the probability that is not divisible by 7 11 or 13?

54. Four red dice and six blue dice are rolled. Assuming that all ten dice are fair six-sided dice, and rolls are independent, what is the probability that exactly three of the red dice are even, and exactly two of the blue dice come up ones?

55. The joint moment generating function for and is . Find the correlation coefficient between and .

57. Suppose that and are uniformly distributed on the diamond . Find .

58. The size of a loss due to a hurricane has cdf . Suppose that a customer has a deductible of 20. Find the probability that the insurance company's payment exceeds 10 given that the loss is less than 50.

61. A fair 6-sided die is rolled 1,000 times. Using a normal approximation with a continuity correction, what is the probability that the number of 3's that are rolled is greater than 150 and less than 180?

62. Suppose that and are jointly normal with and that the correlation of and is . If the means of and have different signs find .

63. An urn contains 4 red balls and 4 blue balls. A ball is randomly drawn from the urn and then replaced, along with a second ball of the same color. The process is then repeated. What is the probability that the first three balls drawn are two red balls and one blue ball?

67. A life insurance company classifies its customers as being either high or low risk. If 20% of the customers are high risk, and high risk customers are three times as likely as low risk customers to file a claim, what percentage of claims that are filed come from high risk customers?

68. The joint cdf of is given by . for . Find .

69. An insurance company feels that the damage that its customers receive has density for and 0 otherwise. Currently the policy has a deductible of 1. If the insurance company adds a payment limit of 2 how much money can they expect to save per customer?

70. The cost of damage in a fire has a density given by for . If the damage is insured with a $500 deductible what is the expected payment?

71. Suppose the joint density of and is for and . What is the covariance of and ?

72. If and are i.i.d. exponential random variables with mean 3 what is ?

73. Suppose that are random variables with and . If for what is where .

75. An insurance company pays $10,000 for the first loss, $7,500 for the second and then $5,000 for each successive loss. If the number of losses has a Poisson distribution with mean 2.5, what is the expected loss size?

76. Let be the number of rolls of a fair die before getting a 6 and let be the number of rolls before the first even number. Find .

77. The moment generating function of is given by . Find .

79. An actuary for an insurance company models annual claims with a lognormal random variable . If and what is ?

80. Suppose and are bivariate random variables with and . If find .

81. and are identically distributed independent random variables with cdf for . What is ?

82. For the joint density of is given by . Find .

83. and have joint density given by Find .

85. For a random variable let be given by where is the median of . Find for a Poisson random variable with mean 1.5.

86. The joint density of and is Find .

87. The cdf of is given by . Find .

88. If is an exponential random variable with mean 2 and find where denotes the density of .

89. Suppose that I roll two independent dice one red and one blue. Let be the event that the blue die is even the event that the red die is even and the event that the sum is even. Which of the following is true?

90. Suppose that and are independent random variables with common density . Find .

92. The joint density of and is for and . Find .

94. The time in years until the next time a piece of space debris with mass of at least 5g hits the Earth has a probability density function . Let be the number of years rounded down until this occurs. What is ?

95. If I roll 6 different fair six-sided dice, what is the probability that the maximum of the rolls is equal to 5?