Question 1186222: It has been estimated that 40% of the televisions that the manufacturer makes need repairs in the first three years of operation. A new hotel buys 90 televisions from the company. Find the probability that in the first three years operation:
a. Less than 32 of the television needs repairs
b. Between 38 and 42 of the televisions, needs repairs
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! this is a binomial (n, p) with n=90 and p=0.4
with normal approximation with mean 36 and sd (see below) of 4.65
z <(31.5-36)/4.65=-0.968
that probability is 0.1665
from calculator (binomcdf(90,0.4,31)) which gives everything 31 and fewer is 0.1666
second can be done using formula where
px=38) is 90C38*0.4^38*0.6^52=0.0774
px=39) is 0.0688 (note the mean is 90*0.4=36, so after that, the probability should fall slightly, which it is.
p(x=40) is 0.0585
p(x=41)= is 0.0476
p(x=42) is 0.0370
That sum 0.2893, which is exact.
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check with normal approximation
mean=np=36
variance is np(1-p)=36*0.6=21.6
sort (V)=sd=4.65
so between 38 and 42
z=(42.5-36)/4.65, using continuity correction factor
=1.40
and
z=(37.5-36)/4.65
=0.323
probability is in that z interval which is 0.2925, which is a decent approximation.
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