Question

A paint company claims their paint will be completely dry within 45 minutes after application. Recently, customers have complained drying times are longer than the claimed 45 minutes. A consumer advocate group takes a random sample of 25 paint specimens and records their drying times. The average drying time x is 52. Consider dryng time, for all test specimens, to be normally distributed with ? = 6.
Suppose the claimed drying time is true, that is ? = 45 minutes, what is the probability of observing a sample mean of x = 52 or greater from a sample size of 25? (Round your answer to four decimal places.)

Answer 1

Answer:

The probability of observing a sample mean of x = 52 or greater from a sample size of 25 is 0.0000026

Step-by-step explanation:

Mean = $\mu = 45$

Population standard deviation =$\sigma = 6$

Sample size = n =25

Sample mean = $\bar{x} = 52$

We are supposed to find  the probability of observing a sample mean of x = 52 or greater from a sample size of 25 i.e.$P(x\geq 52)$

$Z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}\\Z=\frac{52-45}{\frac{6}{\sqrt{25}}}$

Z=5.83

P(Z<52)=0.9999974

$P(Z\geq 52)=1-P(z$

Hence  the probability of observing a sample mean of x = 52 or greater from a sample size of 25 is 0.0000026