Answer:
[tex]P(\bar X\:>27.8\:thousand)=3.07\%[/tex]
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:
[tex]\sigma_X=\frac{\sigma}{\sqrt{n} }[/tex]
[tex]\implies \sigma_X=\frac{10.6}{\sqrt{50} }=1.4991[/tex]
The z value is [tex]z=\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
We plug in the values to get:
[tex]z=\frac{27.8-25}{1.4991}=\frac{2.8}{1.4991}=1.87[/tex]
The area to the right of 1.87 is [tex]1-0.96926=0.03074[/tex].
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.
