Question

The amount spent on textbooks for the fall term was recorded for a sample of five university students. The following data were observed: \(400, \)350, \(600, \)525, and $450. Calculate each of the following values for this data:

a.The sample mean

b.The sample median

c.The sample range

d.The standard deviation

Short answer

a. Mean = $465

b. Median = $450

c. Range = $250

d. Standard Deviation = $99.37

Step-by-step solution

Calculating the mean

x =400, 350, 600, 525, 450

$\begin{array}{c}\text{Mean=}\frac{{\displaystyle \sum \text{χ}}}{\text{n}}\\ =\frac{400+350+600+525+450}{5}\\ =\frac{2325}{5}\\ =\$465\end{array}$

Therefore, the mean value of books is $465.

Computing the median

Arranging the data in ascending order,

350, 400, 450, 525, 600

As the sample size is odd, the central value is 450. Therefore, $450 is the median.

Finding the range

Maximum value = $600

Minimum value = $350

$\begin{array}{c}\text{Range}=\text{Max}–\text{Min}\\ =600–350\\ =\$250\end{array}$

Therefore, the range is $250.

Estimating the standard deviation

$\begin{array}{c}\text{Variance=}\frac{{\displaystyle \sum {\left(\text{χ}-\right)}^2}}{\text{n}-\text{1}}\\ \text{=}\frac{{\left(\text{400}-\text{465}\right)}^{\text{2}}\text{+}{\left(\text{350}-\text{465}\right)}^{\text{2}}\text{+}{\left(\text{600}-\text{465}\right)}^{\text{2}}\text{+}{\left(\text{525}-\text{465}\right)}^{\text{2}}\text{+}{\left(\text{450}-\text{465}\right)}^{\text{2}}}{\text{5}-\text{1}}\\ \text{=}\frac{\text{4225+13225+18225+3600+225}}{\text{4}}\\ \text{=}\frac{\text{39500}}{\text{4}}\end{array}$

$\text{=9875}$

$\begin{array}{c}\text{StandardDeviation=}\sqrt{\text{Variance}}\\ =\sqrt{9875}\\ =99.37\end{array}$

Therefore, the standard deviation is $99.37.