Question

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 13.2 years, and standard deviation of 3.1 years.If you randomly purchase 12 items, what is the probability that their mean life will be longer than 12 years?

Answer 1

There is a 91.15% probability that their mean life will be longer than 12 years.

Explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean
`\mu` and standard deviation
`\sigma`, a large sample size can be approximated to a normal distribution with mean
`\mu` and standard deviation
`(\sigma)/(√(n))`

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 13.2 years, and standard deviation of 3.1 years. This means that
`\mu = 13.2, \sigma = 3.1`.

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

There are 12 items, so
`n = 12`.

This is 1 subtracted by the pvalue of Z when
`X = 12`.

By the Central Limit Theorem, we use the standard deviation of the sample in the Z score formula. That is:

`s = (\sigma)/(√(n)) = (3.1)/(√(12)) = 0.89`

`Z = (X - \mu)/(s)`

`Z = (12-13.2)/(0.89)`

`Z = -1.35`

`Z = -1.35` has a pvalue of 0.0885

This means that there is a 1-0.0885 = 0.9115 = 91.15% probability that their mean life will be longer than 12 years.