The probability that the sample mean would differ from the true mean by greater than 158 dollars if a sample of 226 persons is randomly selected is 0.99
mean per capita income is 19,695 dollars per annum , μ = 19695
a variance of 802,816, σ² = 802816
Standard deviation σ = √802816= 896
Sample size = 226
To find the probability that the sample mean would differ from the true mean by greater than 158 dollars i.e. less than 19537 dollars and more than 19853 dollars.
The formula for z-score :-
[tex]z = \frac{x - mean}{\frac{\alpha }{\sqrt{n} } } \\[/tex]
For x = 19537 dollars
[tex]z = \frac{19537 - 19695}{\frac{\ 896 }{\sqrt{226} } } \\[/tex]
z = -2.65
For x = 19853 dollars
[tex]z = \frac{19853 - 19695}{\frac{\ 896 }{\sqrt{226} } } \\[/tex]
z = 2.65
The P-value=
P(z < -2.65) + P(z > 2.65 = 2P(z > 2.65) = 2x. 0.495975
= 0.99
Therefore, the probability that the sample mean would differ from the true mean by greater than 158 dollars if a sample of 226 persons is randomly selected is 0.99
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