Question

In a recent survey of 655 working americans ages 25-34, the average weekly amount spent on lunch was $43.29 with standard deviation $2.75. The weekly amounts are approximately bell-shaped

Answer 1

99.7%

Explanation:

Given that:

sample size (n) = 655

sample mean (
`\overline x`) = 43.29

Standard deviation
`\sigma` = 2.75

Let's assume we are to estimate the percentage between $33.72 and $51.54

Then;

The test statistics at 33.72 is :

`Z = (X - \mu)/(\sigma)`

`Z = (33.72 -43.29)/(2.75)`

`Z = -3.48`

Z at 51.54

`Z = (X - \mu)/(\sigma)`

`Z = (51.54 -43.29)/(2.75)`

`Z = 3.00`

Using the empirical standard rule, the percentage will be 99.7%