Question

Suppose the number of residents within five miles of each of your stores is asymmetrically distributed with a mean of 25 thousand and a standard deviation of 10.6 thousand. What is the probability that the average number of residents within five miles of each store in a sample of 50 stores will be more than 27.8 thousand?

Answer 1

Answer:

$P(\bar X\:>27.8\:thousand)=3.07\%$

Step-by-step explanation:

Since the number of residents within five miles of each of your stores is asymmetrically distributed, the distribution of the sample means will be approximately normal with a mean of 25 thousand.

The standard deviation of the sample means is:

$\sigma_X=\frac{\sigma}{\sqrt{n} }$

$\implies \sigma_X=\frac{10.6}{\sqrt{50} }=1.4991$

The z value is $z=\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n}} }$

We plug in the values to get:

$z=\frac{27.8-25}{1.4991}=\frac{2.8}{1.4991}=1.87$

The area to the right of 1.87 is $1-0.96926=0.03074$.

The probability that the average number of residents within five miles of each store in a sample of 50 stores will be more than 27.8 thousand is 3.07%

See attachment.