An automobile manufacturer has offered a 6 yr / 60,000 mi. power train warranty. If the...

An automobile manufacturer has offered a 6 yr / 60,000 mi. power train warranty. If the power train has a life expectancy of 61,000 mi. with a standard deviation of 8,000 mi., what is the probability that the average life of powertrains from 625 randomly selected automobiles will be fewer than 60,497.6 mi.?

Solutions

Expert Solution

The values provided in the question are as below

Mean = = 61000

standard deviation = = 8000

sample size = = 625

We have to find the probability that the average life of powertrains from 625 randomly selected automobiles will be fewer than 60,497.6

------------(1)

We convert this into Z using following way

------------------(2)

Using equation (2) in equation (1) we get

We have to find the probability of Z fewer than -1.57 using the following command in Excel.

The command is =NORMSDIST(Z) then press Enter

here, Z = -1.57

=NORMSDIST(-1.57) then press Enter we get the value probability is 0.0582

The probability that the average life of powertrains from 625 randomly selected automobiles will be fewer than 60,497.6 is 0.0582

orchestra answered 1 year ago

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