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

99.7%

Step-by-step explanation:

Given that:

sample size (n) = 655

sample mean () = 43.29

Standard deviation = 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 at 51.54

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

one set of data has more variation than the other. hope this !