Question

Lunch break in a recent survey of 624 working Americans ages 25:34, the average weekly amount spent on lunch was $45.03 with a standard deviation of $2.90. The weekly amounts are approximately bell-shaped.

(a) Estimate the percentage of amounts that were less than $42.13. Round the answer to one decimal place.

Approximately _____ % of the amounts were less than $42.13.

Answer 1

To estimate the percentage of working Americans aged 25 to 34 who spend less than $42.13 weekly on lunch, calculate the z-score and refer to the z-table. The z-score for $42.13 is -1.0, which corresponds to approximately 15.9%.

Step-by-step explanation:

The question asks to estimate the percentage of working Americans aged 25 to 34 who spend less than $42.13 weekly on lunch. This involves a statistical concept called the z-score, which represents the number of standard deviations an element is from the mean. In this case, the average weekly amount spent on lunch is $45.03, with a standard deviation of $2.90.

First, calculate the z-score for $42.13 using the formula z = (X - μ) / σ, where X is the value in question, μ (mu) is the mean, and σ (sigma) is the standard deviation.

z = ($42.13 - $45.03) / $2.90 = -1.0

Next, we refer to the standard normal distribution (z) table to find the percentage of values that lie below a z-score of -1.0. This percentage corresponds to approximately 15.9%.

Therefore, approximately 15.9% of the amounts were less than $42.13.