Question

The U.S. Bureau of Labor Statistics reports that the average annual expenditure on food and drink for all families is $5700. Assume that the annual expenditure on food and drink is normally distributed and that the standard deviation is $1500. What percentage of families spend less than $7000 annually on food and drink?

Answer 1

Answer:

80.78% of families spend less than $7000 annually on food and drink

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:

$\mu = 5700, \sigma = 1500$

What percentage of families spend less than $7000 annually on food and drink?

This is the pvalue of Z when X = 7000. So

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

$Z = \frac{7000 - 5700}{1500}$

$Z = 0.87$

$Z = 0.87$ has a pvalue of 0.8078

80.78% of families spend less than $7000 annually on food and drink