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
