Exam P Practice Test

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Aaronsmith0924
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  • 1/95 Questions

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

    • 0.23
    • 0.31
    • 0.50
    • 0.69
    • 0.84
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About This Quiz

Taken from the four 1/P Exams from The Infinite Actuary.

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  • 2. 

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

    • 0.08

    • 0.28

    • 0.63

    • 1.67

    • 2.86

    Correct Answer
    A. 0.08
  • 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

    Correct Answer
    A. 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.

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  • 4. 

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

    • -2

    • 0

    • 1

    • 2

    • 4

    Correct Answer
    A. 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.

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  • 5. 

    If and find .

    • 0.2

    • 0.3

    • 0.4

    • 0.5

    • 0.6

    Correct Answer
    A. 0.3
    Explanation
    The given question is incomplete and does not provide enough information to generate a meaningful explanation.

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  • 6. 

    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

    Correct Answer
    A. 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

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  • 7. 

    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

    Correct Answer
    A. 0.95
    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.

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  • 8. 

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

    • 1

    • 2

    • 3

    • 4

    • 5

    Correct Answer
    A. 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.

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  • 9. 

    What is the smallest possible value of  if  and ?

    • 0.4

    • 0.5

    • 0.6

    • 0.7

    • 0.8

    Correct Answer
    A. 0.5
    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.

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  • 10. 

    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

    Correct Answer
    A. 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.

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  • 11. 

    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

    Correct Answer
    A. 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.

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  • 12. 

    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

    Correct Answer
    A. 0.27
  • 13. 

    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

    Correct Answer
    A. 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.

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  • 14. 

    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

    Correct Answer
    A. 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.

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  • 15. 

    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

    Correct Answer
    A. 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.

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  • 16. 

    The joint mgf of  is given by . Find .

    • 1

    • 3

    • 5

    • 7

    • 9

    Correct Answer
    A. 9
  • 17. 

    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

    Correct Answer
    A. 0.18
    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.

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  • 18. 

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

    • 0.05

    • 0.34

    • 0.57

    • 0.66

    • 0.95

    Correct Answer
    A. 0.57
  • 19. 

    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?

    Correct Answer
    A.
    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.

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  • 20. 

    The cdf of a random variable  satisfies for . Find .

    • 0.22

    • 0.36

    • 0.51

    • 0.64

    • 0.78

    Correct Answer
    A. 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

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  • 21. 

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

    • 0

    • 0.5

    • 1

    • 1.5

    • 2

    Correct Answer
    A. 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.

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  • 22. 

    If   and  find .

    • 0.2

    • 0.3

    • 0.4

    • 0.5

    • 0.6

    Correct Answer
    A. 0.4
  • 23. 

    If  and  what is the maximum possible value of ?

    • .1

    • .3

    • .4

    • .6

    • .7

    Correct Answer
    A. .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.

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  • 24. 

    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

    Correct Answer
    A. 0.46
    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.

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  • 25. 

    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

    Correct Answer
    A. 2.8
    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.

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  • 26. 

    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

    Correct Answer
    A. 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.

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  • 27. 

    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%

    Correct Answer
    A. 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%.

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  • 28. 

    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

    Correct Answer
    A. 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.

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  • 29. 

    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

    Correct Answer
    A. 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.

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  • 30. 

    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

    Correct Answer
    A. 0.81
    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.

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  • 31. 

    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

    Correct Answer
    A. 15
    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.

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  • 32. 

    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

    Correct Answer
    A. .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.

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  • 33. 

    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

    Correct Answer
    A. 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.

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  • 34. 

    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

    Correct Answer
    A. 1.7
  • 35. 

    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

    Correct Answer
    A. 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.

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  • 36. 

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

    • 0.44

    • 0.48

    • 0.54

    • 0.56

    • 0.58

    Correct Answer
    A. 0.58
  • 37. 

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

    • 1.3

    • 2.5

    • 3.6

    • 4.7

    • 5.2

    Correct Answer
    A. 1.3
    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.

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  • 38. 

    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?

    Correct Answer
    A.
    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.

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  • 39. 

    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?

    Correct Answer
    A.
    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.

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  • 40. 

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

    • 0.17

    • 0.21

    • 0.25

    • 0.29

    • 0.34

    Correct Answer
    A. 0.17
    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.

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  • 41. 

    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

    Correct Answer
    A. 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.

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  • 42. 

    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

    Correct Answer
    A. 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.

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  • 43. 

    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

    Correct Answer
    A. 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.

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  • 44. 

    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

    Correct Answer
    A. 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.

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  • 45. 

    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

    Correct Answer
    A. 872
  • 46. 

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

    • 0.02

    • 0.09

    • 0.15

    • 0.18

    • 0.24

    Correct Answer
    A. 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.

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  • 47. 

    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

    Correct Answer
    A. 0.16
    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%.

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  • 48. 

    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

    Correct Answer
    A. 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.

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  • 49. 

    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

    Correct Answer
    A. 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.

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  • Mar 21, 2023
    Quiz Edited by
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  • Jul 22, 2012
    Quiz Created by
    Aaronsmith0924

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