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

Answer:

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

Step-by-step 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 $\frac{\sigma}{\sqrt{n}}$

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:

$Z = \frac{X - \mu}{\sigma}$

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:

$s = \frac{\sigma}{\sqrt{n}} = \frac{3.1}{\sqrt{12}} = 0.89$

$Z = \frac{X - \mu}{s}$

$Z = \frac{12-13.2}{0.89}$

$Z = -1.35$

$Z = -1.35$ 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.