Answer: 0.5467
Step-by-step explanation:
Let X be the random variable that represents the income (in dollars) of a randomly selected person.
Given : [tex]\mu=15451[/tex]
[tex]\sigma^2=298116\\\\\Rightarrow\ \sigma=\sqrt{298116}=546[/tex]
Sample size : n=350
z-score : [tex]\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
To find the probability that the sample mean would differ from the true mean by less than 22 dollars, the interval will be
[tex]\mu-22,\ \mu+22\\\\=15,451 -22,\ 15,451 +22\\\\=15429,\ 15473[/tex]
For x=15429
[tex]z=\dfrac{15429-15451}{\dfrac{546}{\sqrt{350}}}\approx-0.75[/tex]
For x=15473
[tex]z=\dfrac{15473-15451}{\dfrac{546}{\sqrt{350}}}\approx0.75[/tex]
The required probability :-
[tex]P(15429<X<15473)=P(-0.75<z<0.75)\\\\=1-2(P(z<0.75))=1-2(0.2266274)=0.5467452\approx0.5467[/tex]
Hence, the required probability is 0.5467.
