Question

he 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 more than $4000 annually on food and drink

Answer 1

Answer:

87.08% percentage of families spend more than $4000 annually on food and drink

Step-by-step explanation:

Given -

average annual expenditure on food and drink for all families is $5700

Mean  $(\nu)$ = 5700

standard deviation $(\sigma )$ = 1500

Let X be the no of families spend annually on food and drink

percentage of families spend more than $4000 annually on food and drink =

$P(X > 4000)$  = $P(\frac{X - \nu }{\sigma }> \frac{ 4000 - 5700}{1500})$

                      =  $P(Z > - 1.13)$     Using  $(Z = \frac{X - \nu }{\sigma })$

                      =  1 - $P(Z$

                      = 1 - .1292

                      = .8708

                       = 87.08%