Chapter 14.2, Problem 32E

(a)

To determine

To find: The mean of the data.

Answer to Problem 32E

The mean is $30.4$ .

Explanation of Solution

Given:

The frequency table is given below:

Grade Point Averages	Frequency
$\begin{array}{l}\text{1}\text{.75-2}\text{.25}\\ \text{2}\text{.25-2}\text{.75}\\ \text{2}\text{.75-3}\text{.25}\\ \text{3}\text{.25-3}\text{.75}\\ \text{3}\text{.75-4}\text{.25}\end{array}$	$\begin{array}{l}12\\ 15\\ 31\\ 37\\ 5\end{array}$

The mean is calculated as:

  $\begin{array}{c}\overline{x}=\frac{{\displaystyle \sum x_i{f}_i}}{{\displaystyle \sum f_i}}\\ =\frac{12\left(2.00\right)+15\left(2.50\right)+31\left(3.00\right)+37\left(3.50\right)+5\left(4.00\right)}{100}\\ =30.4\end{array}$

The mean is $30.4$ .

(b)

To determine

To find: The median of the data.

Answer to Problem 32E

The median is approximately is $3.12$ .

Explanation of Solution

The median is calculated as:

Grade Point Averages	Frequency	Cumulative frequency
$\begin{array}{l}\text{1}\text{.75-2}\text{.25}\\ \text{2}\text{.25-2}\text{.75}\\ \text{2}\text{.75-3}\text{.25}\\ \text{3}\text{.25-3}\text{.75}\\ \text{3}\text{.75-4}\text{.25}\end{array}$	$\begin{array}{l}12\\ 15\\ 31\\ 37\\ 5\end{array}$	$\begin{array}{l}12\\ 27\\ 58\\ 95\\ 100\end{array}$

Since, $\frac{100}{2}=50$ , half of the data has been gathered in the $\text{2}\text{.75-3}\text{.25}$ class.

The median class is $\text{2}\text{.75-3}\text{.25}$ .

The median is calculated as:

  $\begin{array}{c}\text{Median}=l+\left(\frac{\frac{n}{2}-c\cdot f}{f}\right)\cdot h\\ =2.75+\left(\frac{50-27}{31}\right)\cdot 0.50\\ \approx 3.12\end{array}$

The median is approximately is $3.12$ .