Question 1

Use the given values of n and p to find the minimum usual value µ - 2? and the maximum usual value µ + 2?
-$$n = 2277 , p = \frac { 3 } { 4 }$$

Accepted Answer

To find the minimum and maximum usual values, we first calculate the mean (µ) and standard deviation (σ) of the binomial distribution. The mean of a binomial distribution is given by µ = n * p, and the standard deviation is given by σ = sqrt(n * p * (1 - p)). For n = 2277 and p = 3/4, µ = 2277 * (3/4) = 1707.75, and σ = sqrt(2277 * (3/4) * (1/4)) ≈ 20.66. The minimum usual value is µ - 2σ = 1707.75 - 2(20.66) = 1666.43, and the maximum usual value is µ + 2σ = 1707.75 + 2(20.66) = 1749.07.

Question 2

Use the given values of n and p to find the minimum usual value µ - 2? and the maximum usual value µ + 2?
-n = 94, p = 0.20

Accepted Answer

The mean µ is calculated as n * p = 94 * 0.20 = 18.8. The standard deviation ? is sqrt(n * p * (1-p)) = sqrt(94 * 0.20 * 0.80) = 3.88. Therefore, the minimum usual value is µ - 2? = 18.8 - 2(3.88) = 11.04, and the maximum usual value is µ + 2? = 18.8 + 2(3.88) = 26.56.

Question 3

Find the standard deviation, ?, for the binomial distribution which has the stated values of n and p. Round your answer to the nearest hundredth.
-n = 2661; p = .63

Accepted Answer

The answer of Find the standard deviation, ?, for the...

Question 4

Use the given values of n and p to find the minimum usual value µ - 2? and the maximum usual value µ + 2?
-n = 189, p = 0.13

Accepted Answer

For a binomial distribution, the mean (µ) is given by n*p and the standard deviation (σ) is given by sqrt(n*p*(1-p)). Here, n = 189 and p = 0.13, so µ = 189*0.13 = 24.57 and σ = sqrt(189*0.13*(1-0.13)) ≈ 4.625. The minimum usual value is µ - 2σ = 24.57 - 2*4.625 = 15.32, and the maximum usual value is µ + 2σ = 24.57 + 2*4.625 = 33.82.

Question 5

Use the given values of n and p to find the minimum usual value µ - 2? and the maximum usual value µ + 2?
-n = 1647, p = 0.493

Accepted Answer

The minimum usual value is µ - 2? = n * p - 2? = 1647 * 0.493 - 2? = 771.392.The maximum usual value is µ + 2? = n * p + 2? = 1647 * 0.493 + 2? = 852.55.

Question 6

Solve the problem.
-The probability that a radish seed will germinate is 0.7. A gardener plants seeds in batches of 15. Find the mean for the number of seeds germinating in each batch.

Accepted Answer

The mean for a binomial distribution is given by $n \times p$, where $n$ is the number of trials (in this case, seeds planted, which is 15) and $p$ is the probability of success (in this case, the probability a seed will germinate, which is 0.7). Therefore, the mean is $15 \times 0.7 = 10.5$.

Question 7

Use the given values of n and p to find the minimum usual value µ - 2? and the maximum usual value µ + 2?
-$$n = 604 , p = \frac { 4 } { 7 }$$

Accepted Answer

The minimum usual value (µ - 2σ) and the maximum usual value (µ + 2σ) for a binomial distribution can be found using the formulas for the mean (µ = np) and standard deviation (σ = sqrt(np(1-p))). Here, n = 604 and p = 4/7, so µ = 604 * (4/7) = 345.14 and σ = sqrt(604 * (4/7) * (3/7)) ≈ 24.32. Therefore, µ - 2σ ≈ 345.14 - 2*24.32 ≈ 296.5 and µ + 2σ ≈ 345.14 + 2*24.32 ≈ 393.78. However, none of the options exactly match these calculations, suggesting a possible miscalculation or typo in the question or answer choices. Based on the closest values provided in the options, B) Minimum: 320.82; maximum: 369.47 seems to be the intended correct choice, assuming there might have been a calculation or typographical error in the presentation of the question or answers.

Question 8

Use the given values of n and p to find the minimum usual value µ - 2? and the maximum usual value µ + 2?
-n = 252, p = 0.261

Accepted Answer

The mean (µ) of a binomial distribution is found using µ = np, and the standard deviation (σ) is found using the formula σ = sqrt(np(1-p)). For n = 252 and p = 0.261, µ = 252 * 0.261 = 65.772 and σ = sqrt(252 * 0.261 * (1 - 0.261)) = 6.972. The minimum usual value is µ - 2σ = 65.772 - 2(6.972) = 51.828, and the maximum usual value is µ + 2σ = 65.772 + 2(6.972) = 79.716.

Question 9

Solve the problem.
-A company manufactures batteries in batches of 29 and there is a 3% rate of defects. Find the mean number of defects per batch.

Accepted Answer

The mean number of defects per batch can be found by multiplying the total number of items in a batch (29) by the defect rate (3%). So, $29 \times 0.03 = 0.87$.

Question 10

Solve the problem.
-The probability that a person has immunity to a particular disease is 0.8. Find the mean number who have immunity in samples of size 25.

Accepted Answer

The mean number of people who have immunity in samples of size 25 is calculated by multiplying the probability of immunity (0.8) by the sample size (25), which equals 20.0.

Question 11

Use the given values of n and p to find the minimum usual value µ - 2? and the maximum usual value µ + 2?
-$$n = 270 , p = \frac { 1 } { 4 }$$

Accepted Answer

The answer of Use the given values of n and...

Question 12

Solve the problem.
-The probability is 0.6 that a person shopping at a certain store will spend less than $20. For groups of size 19, find the mean number who spend less than $20.

Accepted Answer

The answer of Solve the problem.
-The probability is 0.6 that...

Question 13

Use the given values of n and p to find the minimum usual value µ - 2? and the maximum usual value µ + 2?
-n = 1100, p = 0.84

Accepted Answer

The minimum usual value is µ - 2? = 899.68 and the maximum usual value is µ + 2? = 948.32.

Question 14

Use the given values of n and p to find the minimum usual value µ - 2? and the maximum usual value µ + 2?
-$$p = \frac { 2 } { 3 }$$ n = 377

Accepted Answer

The answer of Use the given values of n and...

Question 15

Solve the problem.
-A die is rolled 7 times and the number of times that two shows on the upper face is counted. If this experiment is repeated many times, find the mean for the number of twos.

Accepted Answer

The answer of Solve the problem.
-A die is rolled 7...

Question 16

Solve the problem.
-On a multiple choice test with 11 questions, each question has four possible answers, one of which is correct. For students who guess at all answers, find the mean for the number of correct answers.

Accepted Answer

The mean number of correct answers can be found by multiplying the probability of getting a question correct (1/4) by the total number of questions (11). Thus, the mean is 11 * (1/4) = 2.75, which rounds to 2.8.

Question 17

Solve the problem.
-According to a college survey, 22% of all students work full time. Find the mean for the number of students who work full time in samples of size 16.

Accepted Answer

The mean for the number of students who work full time in samples of size 16 can be found by multiplying the sample size (16) by the probability that a student works full time (0.22). Thus, the mean is 16 * 0.22 = 3.52.

Question 18

Use the given values of n and p to find the minimum usual value µ - 2? and the maximum usual value µ + 2?
-n = 48; p = .4

Accepted Answer

The standard deviation $\sigma$ for a binomial distribution is calculated using the formula $\sigma = \sqrt{n \cdot p \cdot (1 - p)}$, where $n$ is the number of trials and $p$ is the probability of success. Substituting $n = 48$ and $p = 0.4$ gives $\sigma = \sqrt{48 \cdot 0.4 \cdot (1 - 0.4)} = \sqrt{48 \cdot 0.4 \cdot 0.6} = \sqrt{11.52} \approx 3.39$.

Question 19

Find the standard deviation, ?, for the binomial distribution which has the stated values of n and p. Round your answer to the nearest hundredth.
-n = 1507; p = .57

Accepted Answer

The standard deviation for a binomial distribution is given by the formula:$\sigma = \sqrt{np(1-p)}$Plugging in the given values, we get:$\sigma = \sqrt{1507 * 0.57 * (1-0.57)} = 19.22$Therefore, the correct answer is A.

Question 20

Use the given values of n and p to find the minimum usual value µ - 2? and the maximum usual value µ + 2?
-n = 1104, p = 0.93

Accepted Answer

The minimum usual value (µ - 2σ) and the maximum usual value (µ + 2σ) for a binomial distribution can be found using the formulas µ = np and σ = sqrt(np(1-p)). Here, n = 1104 and p = 0.93, so µ = 1104 * 0.93 = 1026.72 and σ = sqrt(1104 * 0.93 * 0.07) ≈ 8.48. Therefore, µ - 2σ ≈ 1026.72 - 2*8.48 ≈ 1009.76 and µ + 2σ ≈ 1026.72 + 2*8.48 ≈ 1043.68.

Quiz 5: Probability Distributions

Use the given values of n and p to find the minimum usual value µ - 2? and the maximum usual value µ + 2? - $n = 2277 , p = \frac { 3 } { 4 }$

Use the given values of n and p to find the minimum usual value µ - 2? and the maximum usual value µ + 2? -n = 94, p = 0.20

Find the standard deviation, ?, for the binomial distribution which has the stated values of n and p. Round your answer to the nearest hundredth. -n = 2661; p = .63

Use the given values of n and p to find the minimum usual value µ - 2? and the maximum usual value µ + 2? -n = 189, p = 0.13

Use the given values of n and p to find the minimum usual value µ - 2? and the maximum usual value µ + 2? -n = 1647, p = 0.493

Solve the problem. -The probability that a radish seed will germinate is 0.7. A gardener plants seeds in batches of 15. Find the mean for the number of seeds germinating in each batch.

Use the given values of n and p to find the minimum usual value µ - 2? and the maximum usual value µ + 2? - $n = 604 , p = \frac { 4 } { 7 }$

Use the given values of n and p to find the minimum usual value µ - 2? and the maximum usual value µ + 2? -n = 252, p = 0.261

Solve the problem. -A company manufactures batteries in batches of 29 and there is a 3% rate of defects. Find the mean number of defects per batch.

Solve the problem. -The probability that a person has immunity to a particular disease is 0.8. Find the mean number who have immunity in samples of size 25.

Use the given values of n and p to find the minimum usual value µ - 2? and the maximum usual value µ + 2? - $n = 270 , p = \frac { 1 } { 4 }$

Solve the problem. -The probability is 0.6 that a person shopping at a certain store will spend less than $20. For groups of size 19, find the mean number who spend less than $20.

Use the given values of n and p to find the minimum usual value µ - 2? and the maximum usual value µ + 2? -n = 1100, p = 0.84

Use the given values of n and p to find the minimum usual value µ - 2? and the maximum usual value µ + 2? - $p = \frac { 2 } { 3 }$ n = 377

Solve the problem. -A die is rolled 7 times and the number of times that two shows on the upper face is counted. If this experiment is repeated many times, find the mean for the number of twos.

Solve the problem. -On a multiple choice test with 11 questions, each question has four possible answers, one of which is correct. For students who guess at all answers, find the mean for the number of correct answers.

Solve the problem. -According to a college survey, 22% of all students work full time. Find the mean for the number of students who work full time in samples of size 16.

Use the given values of n and p to find the minimum usual value µ - 2? and the maximum usual value µ + 2? -n = 48; p = .4

Find the standard deviation, ?, for the binomial distribution which has the stated values of n and p. Round your answer to the nearest hundredth. -n = 1507; p = .57

Use the given values of n and p to find the minimum usual value µ - 2? and the maximum usual value µ + 2? -n = 1104, p = 0.93