Let x denotes the score of a man.
As per given , we have
[tex]\mu = 512\\\\ \sigma=106[/tex]
If 1 of the men is randomly selected, find the probability that his score is at least 559.5 will be :-
[tex]P(X > 559.5) =1- P(X\leq X)\\\\=1-P(\dfrac{x-\mu}{\sigma}\leq\dfrac{559.5-512}{106})\\\\=1-P(z\leq 0.448\ \ [\because\ z=\dfrac{x-\mu}{\sigma}]\\\\ =1 - 0.6729= 0.3271[/tex]
Hence, the probability that his score is at least 559.5 is 0.3271 .
Let M be the sample mean.
[tex]P(M > 559.5) =1-P(M\leq 559.5)\\\ =1- P(\dfrac{M-\mu}{\dfrac{\sigma}{\sqrt{18}}}\leq\dfrac{559.5-512}{\dfrac{106}{\sqrt{18}}})\\\\ =1-P(z\leq1.90 )\ \ [ \because\ z=\dfrac{M-\mu}{\dfrac{\sigma}{\sqrt{18}}}]\\\\ =1-0.9713\\\\=0.0287[/tex]
Hence, if 18 of the men are randomly selected, find the probability that their mean score is at least 559.5 is 0.0287.
