Question

) furnace repair bills are normally distributed with a mean of 272 dollars and a standard deviation of 15 dollars. if 36 of these repair bills are randomly selected, find the probability that they have a mean cost between 272 dollars and 274 dollars.

Answer 1

To evaluate the given probability we proceed as follows:
The z-score is given by:
z=(x-μ)/σ
where:
μ-mean
σ-standard deviation
from the question:
mean=272, sig=15
but the sample is n=36, thus we shall have:
√σ/n
=15/√36
=2.5
thus the z-score will be as follows:
x=272
z=(272-272)/2.5=0
x=274
z=(274-272)/2.5==0.8
thus
P(272<x<274)
=P(z<0.8)-P(z<0)
=0.7881-0.5
=0.2881
Hence the answer is 0.2881