need some help
a. p(x>570)
z509 = 570-509/112 = .54
p(z>(570-509)/112) = p(z>0.545) = 1 - p(z<0.545) ~= 1-0.707 = 0.293
The sample mean is normaly distributed with mean 509 and sd 112/sqrt(100)=11.2b. If 100 men are randomly selected, find the probability that their mean SAT verbal score is more than 570.
So we want:
p(z>(570-509)/11.2) = p(z>5.45) ~= 0.
RonL