The mean output of a certain type of amplifier is 117 watts with a variance of 100. if 42 amplifiers are sampled, what is the probability that the mean of the sample would differ from the population mean by less than 3.3 watts? round your answer to four decimal places.

The probability that mean sample will differ from population mean by 3.3 watts will be given as follows:
P(113.7<x<120.3)
z-score is given by:
z=(x-μ)/σ

but
n=42
hence:
σ/√n
100/√42=15.43
thus
when x=113.7
z=(-3)/15.43
z=-0.1944
P(z<-0.1944)=0.4247

when x=120.3
z=3.3/15.43
z=0.214
P(z<0.214)=0.5832
hence:
P(113.7<x<120.3)=0.5832-0.4247=0.1585

What can we do about it

All you have to do is multiply