Question

A manufacturer produces a commodity where the length of the commodity has approximately normaldistribution with a mean of 15.7 inches and standard deviation of 1.4 inches. If a sample of 45 items are chosenat random, what is the probability the sample's mean length is greater than 15.7 inches? Round answer to fourdecimal places

Answer 1

Given:

- The Mean:

$\mu=15.7in$

- The Standard Deviation:

$\sigma=1.4in$

- And the sample:

$n=45$

You need to find:

$P(\mu^{\prime}>15.7)$

Where this the Sample Mean:

$\mu^{\prime}$

Therefore, you need to use the formula to find the z-statistics:

$z=\frac{\mu^{\prime}-\mu}{\frac{\sigma}{\sqrt[]{n}}}$

Knowing that:

$\mu^{\prime}=15.7in$

You can substitute values into the formula and evaluate:

$z=\frac{15.7-15.7}{\frac{1.4}{\sqrt[]{45}}}=\frac{0}{\frac{1.4}{\sqrt[]{45}}}=0$

Now you need to look for the probability of that value of "z" in the table of Right-Tail Normal Standard Deviation. This is:

$P(\mu^{\prime}>15.7)=0.5$

Hence, the answer is:

$P(\mu^{\prime}>15.7)=0.5$