The mean per capita income is 24,787 dollars per annum with a variance of 169,744.What is the probability that the sample mean would differ from the true mean by greater than 42 dollars if a sample of 412 persons is randomly selected? Round your answer to four decimal places.

Answer: 0.0385Step-by-step explanation:Given : Mean : [tex]\mu = 24,787 \text{ dollars per annum }[/tex]Variance : [tex]\sigma^2=169,744[/tex]Standard deviation :  [tex]\sigma =\sqrt{169,744}= 412[/tex]Sample size : [tex]n= 412[/tex]To find the probability that the sample mean would differ from the true mean by greater than 42 dollars i.e. less than 24,745 dollars and more than 24,829 dollars.The formula for z-score :-[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]For x = 24,745 dollars[tex]z=\dfrac{24745- 24787}{\dfrac{412}{\sqrt{412}}}=-2.07[/tex]For x = 24,829 dollars[tex]z=\dfrac{24829- 24787}{\dfrac{412}{\sqrt{412}}}=2.07[/tex]The P-value= [tex]P(z<-2.07)+P(z>2.07)=2(P(z>2.07))=2(0.0192262)=0.0384523\approx0.0385[/tex]Hence, the required probability = 0.0385