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

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

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=(\sigma)/(√(n) )`

`\implies \sigma_X=(10.6)/(√(50) )=1.4991`

The z value is
`z=(\bar X-\mu)/((\sigma)/(√(n)) )`

We plug in the values to get:

`z=(27.8-25)/(1.4991)=(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.