Question

An appliance dealer's website features the mean life expectancy values for major household appliances. A standard refrigerator's mean life expectancy is 18 years with a standard deviation of 2 years. What is the probability that your standard refrigerator will last between 12 and 20 years? Express the probability as a percentage rounded to the nearest hundredth.

Answer 1

Answer:

There is an 84% probability that your standard refrigerator will last between 12 and 20 years.

Step-by-step explanation:

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 standard refrigerator's mean life expectancy is 18 years with a standard deviation of 2 years. This means that $\mu = 18, \sigma = 2$.

What is the probability that your standard refrigerator will last between 12 and 20 years?

This probability is the pvalue of Z when $X = 20$ subtracted by the pvalue of Z when $X = 12$. So

X = 20

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

$Z = \frac{20 - 18}{2}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

X = 12

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

$Z = \frac{12- 18}{2}$

$Z = -3$

$Z = -3$ has a pvalue of 0.0014

This means that there is a 0.8413-0.0014 = 0.8399 = 0.84 = 84% probability that your standard refrigerator will last between 12 and 20 years.