The average number of push-ups a united states marine does daily is 300, with a standard deviationof 50. a random sample of 36 marines is selected. what is the probability that the sample mean is atmost 320 push-ups?

To solve we need to calculate the z-score, but since we are given the sample size then we shall begin as follows:

μ=300

σ=50

n=36

x=320

thus

σ/√n

=50/√36

=8.33333

thus the z-score will be:

z = (320-300) / 8.33333

z=2.4

Thus

P (x>320) = P (z<2.4) = 0.9918