Probability Distribution Test: Quiz!

1. The mean for a probability distribution is the same as the expected value of a discrete random variable of a probability distribution.

True

False

The statement is true because the mean of a probability distribution represents the average value of the random variable, and the expected value of a discrete random variable is also a measure of the average value. Therefore, they are equivalent and represent the same concept in probability theory.

Explanation

The statement is true because the mean of a probability distribution represents the average value of the random variable, and the expected value of a discrete random variable is also a measure of the average value. Therefore, they are equivalent and represent the same concept in probability theory.

2. Choose the answer below that identifies a value for y that results in a valid probability distribution.

X	0	1	2
P(X)	0.35	0.5	y

-1.85

0.15

0.6

1.85

not-available-via-ai

Explanation

not-available-via-ai

3. In a family with two children, find the mean of the number of children who will be girls.

Number of girls, X	0	1	2
Probability, P(X)	¼	½	¼

0

1/4

1/2

1

The mean of the number of children who will be girls can be calculated by multiplying each number of girls by its corresponding probability and summing them up. In this case, we have 0 girls with a probability of 1/4, 1 girl with a probability of 1/2, and 2 girls with a probability of 1/4. Therefore, the mean is (0 * 1/4) + (1 * 1/2) + (2 * 1/4) = 1.

Explanation

The mean of the number of children who will be girls can be calculated by multiplying each number of girls by its corresponding probability and summing them up. In this case, we have 0 girls with a probability of 1/4, 1 girl with a probability of 1/2, and 2 girls with a probability of 1/4. Therefore, the mean is (0 * 1/4) + (1 * 1/2) + (2 * 1/4) = 1.

4. Tossing a Die Use the table below to answer the question. Find the standard deviation of the number of spots that will appear when a die is tossed. In the toss of a die, the probability distribution for the number of spots that appear is shown below.

Outcome, X	1	2	3	4	5	6
Probability, P(X)	0.167	0.167	0.167	0.167	0.167	0.167

0.167

1.667

1.70

2.9

The standard deviation is a measure of the variability or spread of a probability distribution. In this case, the probability distribution for the number of spots that appear when a die is tossed is uniformly distributed, meaning that each outcome has an equal probability of occurring. The standard deviation for a uniform distribution can be calculated using the formula sqrt((n^2 - 1)/12), where n is the number of possible outcomes. In this case, n = 6, so the standard deviation is sqrt((6^2 - 1)/12) = sqrt(35/12) ≈ 1.70. Therefore, the correct answer is 1.70.

Explanation

The standard deviation is a measure of the variability or spread of a probability distribution. In this case, the probability distribution for the number of spots that appear when a die is tossed is uniformly distributed, meaning that each outcome has an equal probability of occurring. The standard deviation for a uniform distribution can be calculated using the formula sqrt((n^2 - 1)/12), where n is the number of possible outcomes. In this case, n = 6, so the standard deviation is sqrt((6^2 - 1)/12) = sqrt(35/12) ≈ 1.70. Therefore, the correct answer is 1.70.

5. Dropping College Courses Use the following table to answer the question. Would you consider the information in the table to be a probability distribution?

Reason for Dropping a College Course	Frequency	Percentage
Too difficult	45
Illness	40
Change in work schedule	20
Change in major	14
Family-related problems	9
Money	7
Miscellaneous	6
No meaningful reason	3

Yes, you can find the probabilties.

Yes, you only need frequency

No, the probabilities are not listed.

No, there is not enough information to find the probabilities.

The table provides the frequency of each reason for dropping a college course, but it does not provide the probabilities. In order to have a probability distribution, the probabilities of each event must be listed. In this case, only the frequencies are given, so the probabilities cannot be determined.

Explanation

The table provides the frequency of each reason for dropping a college course, but it does not provide the probabilities. In order to have a probability distribution, the probabilities of each event must be listed. In this case, only the frequencies are given, so the probabilities cannot be determined.

6. Expected Value A person selects a card from the deck. If it is a red card, she wins $1. If it is a black card between or including 2 and 10, she wins $5. If it is a black card, she wins $10; and if it is a black ace, she wins $100. Find the expectation of the game. How much should a person bet if the game is to be fair?

The expectation of the game is calculated by multiplying the probability of each outcome by the corresponding amount won, and then summing up these values. In this case, the probability of selecting a red card is 26/52 = 1/2, so the expected value for winning $1 is (1/2) * $1 = $0.50. The probability of selecting a black card between or including 2 and 10 is 9/52, so the expected value for winning $5 is (9/52) * $5 = $0.87. The probability of selecting a black card is 26/52 = 1/2, so the expected value for winning $10 is (1/2) * $10 = $5. The probability of selecting a black ace is 1/52, so the expected value for winning $100 is (1/52) * $100 = $1.92. Adding up these expected values, we get $0.50 + $0.87 + $5 + $1.92 = $7.29. Therefore, a person should bet $7.29 to make the game fair.

Explanation

The expectation of the game is calculated by multiplying the probability of each outcome by the corresponding amount won, and then summing up these values. In this case, the probability of selecting a red card is 26/52 = 1/2, so the expected value for winning $1 is (1/2) * $1 = $0.50. The probability of selecting a black card between or including 2 and 10 is 9/52, so the expected value for winning $5 is (9/52) * $5 = $0.87. The probability of selecting a black card is 26/52 = 1/2, so the expected value for winning $10 is (1/2) * $10 = $5. The probability of selecting a black ace is 1/52, so the expected value for winning $100 is (1/52) * $100 = $1.92. Adding up these expected values, we get $0.50 + $0.87 + $5 + $1.92 = $7.29. Therefore, a person should bet $7.29 to make the game fair.

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7. Create a probability distribution table showing the assignment of final grades and grade-point scores. An instructor always assigns final grades such that 20% are A, 40% are B, 30% are C, and 10% are D. The grade-point scores are 4 for A, 3 for B, 2 for C, and 1 for D.

The probability distribution table shows the assignment of final grades and grade-point scores. The table displays the percentages of each grade (A, B, C, and D) and their corresponding grade-point scores (4, 3, 2, and 1). The values in the table represent the probabilities of each grade occurring. For example, there is a 20% probability of getting an A, a 40% probability of getting a B, a 30% probability of getting a C, and a 10% probability of getting a D. The table also shows the grade-point scores associated with each grade.

Explanation

The probability distribution table shows the assignment of final grades and grade-point scores. The table displays the percentages of each grade (A, B, C, and D) and their corresponding grade-point scores (4, 3, 2, and 1). The values in the table represent the probabilities of each grade occurring. For example, there is a 20% probability of getting an A, a 40% probability of getting a B, a 30% probability of getting a C, and a 10% probability of getting a D. The table also shows the grade-point scores associated with each grade.

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