Question

Lunch break: In a recent survey of 647 working Americans ages 25-34,the average weekly amount spent on lunch as $44.01 with standard deviation $2.96 The weekly amounts are approximately bell-shaped. Estimate the percentage of amounts that were less than $38.09 . Round the answer to one decimal place.

Answer 1

Answer :

Given data is :

Mean = 44.01

standard deviation = 2.96

Limit =

= 44.01 - 2(2.96)

= 44.01 - 5.92

= 38.09

Here it is 95% of amount lies between and .

Z value at 95% is 1.96.

It is one tail test and less than ,

So we take required probability as = 1 - 0.975 = 0.025

i.e 0.025 * 100% = 2.5%

Approximately 2.5% of amount were less than $ 38.09.