Chapter 10.3, Problem 36E

Solution

To calculate : The standard deviation and variance for the given student debt data.

The standard deviation and variance of the given data is $\$637.11\text{ billions and \$}4462.24\text{ billions}$ respectively.

Given information :

The data is:

Year	Total Student Loan Debt in billions of dollars
$\begin{array}{l}2004\\ 2005\\ 2006\\ 2007\\ 2008\\ 2009\\ 2010\\ 2011\\ 2012\end{array}$	$\begin{array}{l}346\\ 392\\ 477\\ 538\\ 631\\ 715\\ 804\\ 865\\ 966\end{array}$

Formula used :

The standard deviation of a population $\left\{x_1,x_2,x_3,\mathrm{......},x_n\right\}$ is given by,

  $s=\sqrt{\frac{1}{n-1}{\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}}$ .

Variance $={\left(s\right)}^2$ .

Calculation :

Consider the given data and present it in tabular form as below:

$x$	$x-\overline{x}$	${\left(x-\overline{x}\right)}^2$
$\begin{array}{l}346\\ 392\\ 477\\ 538\\ 631\\ 715\\ 804\\ 865\\ 966\end{array}$	$\begin{array}{l}-291.11\\ -245.11\\ -160.11\\ -99.11\\ -6.11\\ 77.89\\ 166.89\\ 227.89\\ 328.89\end{array}$	$\begin{array}{l}84745.0321\\ 60078.9121\\ 25635.2121\\ 9822.7921\\ 37.3321\\ 6066.8521\\ 27852.2721\\ 51933.8521\\ 108168.6321\end{array}$
$\sum x=5734$		$\sum {\left(x-\overline{x}\right)}^2=374340.8889$

Find the mean of the given data.

  $\begin{array}{c}\overline{x}=\frac{5734}{9}\\ =637.11\end{array}$

Find the standard deviation $\left(s\right)$ of the given data.

  $\begin{array}{c}s=\sqrt{\frac{1}{n-1}{\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}}\\ =\sqrt{\frac{1}{9-1}\left(374340.8889\right)}\\ \approx 216.32\end{array}$

Find the variance for the given data.

  $s^2\approx 46794.34$

Therefore, the standard deviation and variance of the given data is $\$637.11\text{ billions and \$}4462.24\text{ billions}$ respectively.