If you randomly purchase 18 items, what is the probability that their mean life will be longer than 12 years?

Question

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 10.7 years, and standard deviation of 2.9 years.

If you randomly purchase 18 items, what is the probability that their mean life will be longer than 12 years?

Answer 1

To answer this question, you need to evaluate probabilities using the sampling distribution for n=18

The mean of this distribution is the same as the population mean for individual items. µ_x-bar= 10.7

The standard deviation is adjusted for the sampling distribution. It is the Sd for the individual observations divided by the square root of the sample size. σ_xbar = 2.9/(√18) = .684

You will then find the Z value for 12 based on this distribution: z = (12-10.7)/.684 = 1.9

To find the probability of greater than 12 years, we need area to the right of z=1.9. We find this by using complement of table value for 1.9

P (z > 1.9) = 1- P (z<1.9) = 1- table value = 1- .9713 = .0287