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

Answer:

99.7%

Step-by-step 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 :

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

$Z = \dfrac{33.72 -43.29}{2.75}$

$Z = -3.48$

Z at 51.54

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

$Z = \dfrac{51.54 -43.29}{2.75}$

$Z = 3.00$

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