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

The answer is the first one