Question

A recent study showed that 23% of cars, of a particular model, required warrantycovered repair work during the first year after purchase. A random sample of 36 cars was chosen.

A) The mean number of cars requiring warranty-covered renair work is ____
B) The standard deviation of the number of cars requiring warranty-covered repair work is ____
C) The probability that you find exactly 10 cars requiring warranty-covered repair work is ____

Answer 1

The mean number of cars requiring warranty-covered repair work is 8.28. The standard deviation of the number of cars requiring warranty-covered repair work is 2.58. The probability that you find exactly 10 cars requiring warranty-covered repair work is 0.080.

Step-by-step explanation:

A) To find the mean number of cars requiring warranty-covered repair work, we multiply the percentage by the sample size: 0.23 x 36 = 8.28. So, the mean number of cars requiring warranty-covered repair work is 8.28.

B) To find the standard deviation of the number of cars requiring warranty-covered repair work, we use the formula:

Standard Deviation = √ ( p * (1 - p) * n ), where p is the percentage and n is the sample size.

Standard Deviation = √ (0.23 * (1 - 0.23) * 36) = √(0.23 * 0.77 * 36) = √(6.6468) ≈ 2.58. So, the standard deviation of the number of cars requiring warranty-covered repair work is 2.58.

C) To find the probability of exactly 10 cars requiring warranty-covered repair work, we use the binomial probability formula:

Probability = ( nCk ) * ( p^k ) * (1 - p)^(n-k), where n is the sample size, k is the number of cars requiring warranty-covered repair work, and p is the percentage.

Probability = ( 36C10 ) * ( 0.23^10 ) * ( 1 - 0.23 )^(36-10) ≈ 0.080. So, the probability that you find exactly 10 cars requiring warranty-covered repair work is 0.080.