Answer: 0.0385
Step-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
