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:[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.