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Answer on Statistics and Probability Question for John Dewer

Question #4251
If a shipment of 30,000 batteries is assumed to have a lifetimes which are normally distributed with mean 360 days and standard deviation 25 days what percentage could be expected to last more than 365 days?

b) what percentage could be expected to last more than 180 days?
Expert's answer
Let X be the random variable equal to the lifetime of a battery. Then by
assumption it has normal distribution with
mean 360 days and standard
deviation 25 days.
We have to find the probabilities P(X>365) and
P(X>180)

Notice that the random variable Y=(X-360)/25 has normal
distribution with mean 0 days and standard deviation 1, and the
values of its
probabilities is known from tables.
Therefore
P(X>365) = P( (X-360)/25
> (365-360)/25 )&
= P( Y > 5/25 )
=
P(Y>0.2)
= 1 - P(Y<0.2) =
= 1 - F^{-1}(0.2)
=
= 1 - 0,5792597094 =
= 0,4207402906

Similarly,

P(X>180) = P( (X-360)/25 > (180-360)/25 )&
= P( Y >
-180/25 )
= P(Y>-7.2)
= 1 - P(Y<-7.2) =

= 1 - F^{-1}(-7.2) =
= 1 - 3,01062798111744E-013
~
= 0.9999999999997

Thus the probability is very close to 1,
so it is expected that all 30,000 batteries will have lifetime >
180

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