# Answer on Statistics and Probability Question for Jorda Goll

Question #4252

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?

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 probability P(X>365).

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

Thus about 42% of all batteries, that is

& 30000 * 0.42 = 12622 batteries

have lifetime > 365 days.

assumption it has normal distribution with

mean 360 days and standard

deviation 25 days.

We have to find the probability P(X>365).

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

Thus about 42% of all batteries, that is

& 30000 * 0.42 = 12622 batteries

have lifetime > 365 days.

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