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?
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Expert's answer
2011-10-07T09:01:18-0400
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
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