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?

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 and standard deviation , the zscore of a measure X is given by:

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:

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

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

has a pvalue of 0.8078

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

x = 25

step-by-step explanation:

a^2+b^2=c^2

7^2+24^2=c^2

49+576=625

sqrt625=25

x=25

21

step-by-step explanation: