Question

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 12.8 years, and standard deviation of 2.9 years. If 5 items are picked at random, 2% of the time their mean life will be less than how many years?

Answer 1

Let X be a mean life span of the items in years.

We want to find the 2% of the time their mean life will be less than C years.

P(X < C) = 0.02

The below 2% of the time their mean life is below C years.

We will solve this problem by standardising.

P(X < C) = P ((X - µ)/σ < (C – 12.8)/2.9) = P(Z < (C – 12.8)/2.9)

P(Z < (C – 12.8)/2.9) = 0.02

The Z score to corresponding area is 0.02 is -2.05(using statistical table).

NORM.S.INV(0.02) = -2.05375 ………….. (Using Excel fuction)

(C – 12.8)/2.9= -2.05

C = -2.05*2.9 + 12.8

C = 6.855

If 5 items are picked at random, 2% of the time their mean life will be less than approximate 7 many years.