Question

A sample of 25 undergraduates reported the following dollar amounts of entertainment expenses last year:

Answer 1

Answer:

$Range = 85$

$\sigma = 28.71$

$Interval = [666.78, 781.62]$

Step-by-step explanation:

Given

The data for 25 undergraduates

Solving (a): Range and Standard deviation

The range is:

$Range = Highest - Least$

From the dataset:

$Highest = 772$

$Least = 687$

So:

$Range = Highest - Least$

$Range = 772-687$

$Range = 85$

The standard deviation is:

$\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n}}$

First, calculate the mean

$\bar x = \frac{769 +691 +............+715}{25}$

$\bar x = \frac{18105}{25}$

$\bar x = 724.2$

So, the standard deviation is:

$\sigma = \sqrt{\frac{(769-724.2)^2 +(691-724.2)^2 +(699-724.2)^2 +(730-724.2)^2 +............+(715-724.2)^2}{25}}$

$\sigma = \sqrt{\frac{20604}{25}}$

$\sigma = \sqrt{824.16}$

$\sigma = 28.71$

Solving (b): The interval of the 95% of the observation.

Using the emperical rule, we have:

$Interval = [\bar x - 2*\sigma, \bar x+ 2*\sigma]$

$Interval = [724.2 - 2*28.71, 724.2 + 2*28.71]$

$Interval = [666.78, 781.62]$