Statistics question: Women's heights follow a normal distribution with a mean of 64" and standard deviation...?

let X be the woman's height.

X~N(6'4", 9")

first, find P(X >= 6'0") using graphic calculator, or standardizing to standard N(0,1) & read the standard normal table (you should expect it to be greater than 0.5, do this yourself.)

then just multiply the probability to 16053,

we expect 16053*P(X >= 6'0") of the women to be 6'0" or taller. (round off to nearest integer)

A. you're in basic terms a million.2 nicely-known deviations faraway from the conventional, so for a million analyzing, that's available that the declare remains valid. there is extremely no reason to disbelieve at this element. B. while over the years, most of the readings are out of the nicely-known deviation, that's much less in all possibility that the declare is valid. yet at this element, extra information is important to settle on on the declare. If the circumstances have been all clustered around 40 two minutes, there is reliable reason to disbelieve the declare. yet while maximum of them are clustered around 30, and there became right into a million or 2 readings way outdoors of the conventional, you could finally end up with an hassle-free of 40 two ( 19 of the deliveries have been at half-hour and a million at 270 minutes could supply an hassle-free of 40 two minutes). in this 2nd case, the oddball datum could assist you nevertheless have confidence the declare.