Answer:
0.4998
Explanation:
Given that:
Mean life of computer μ = 80
standard deviation σ = 8
Mean of sample n = 73
x = 83.28
We have to find P(Mean of sample is less than 83.38 months) =
P(X<x=83.38), to do this we have to first determine the z score.
Since we are dealing with multiple samples, the z score will :
z = (x – μ) / (σ / √n)
z = (83.28 - 80)/ (8/√73)
z = 3.28/0.936
z = 3.5
P(X<x=83.38) = P ( Z < z = 3.5)
Use the standard normal table to find P ( Z < 3.5 )
We will have P (Z < 3.5 ) = 0.4998
The probability that the mean of a sample of 73 computers would be less than 83.28 months is 0.4998
