Question

SeeClear Windows makes windows for use in homes and commercial buildings. The standards for glass thickness call for the glass to average 0.375 inches with a standard deviation equal to 0.050 inch. Suppose a random sample of nequals50 windows yields a sample mean of 0.392 inches. What is the probability if the windows meet the​ standards?

Answer 1

Answer:  0.0164

Step-by-step explanation:

Given : The standards for glass thickness call for the glass to average $\mu=\text{ 0.375 inches}$ with a standard deviation 0$\sigma=\text{ 0.050 inch}$.

$H_0:\mu=0.375\\\\H_a:\mu\neq0.375$

Since the alternative hypothesis is two-tailed , so the test is two-tailed test.

Sample size : $n=50$ , a large sample ( n>30) so we use z-test.

Sample mean :-$\overline{x}=0.392$

We assume its a normal distribution.

$z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

$z=\dfrac{0.392-0.375}{\dfrac{0.050}{\sqrt{50}}}\approx2.40$

By using standard normal distribution table , the p-value =

$0.0163951\approx0.0164$

Thus, the probability if the windows meet the​ standards = 0.0164