Chapter 14.2, Problem 25E

(a)

To determine

To find: The mean of the verbal scores at Kennedy High School.

Answer to Problem 25E

The mean is $425.6$ .

Explanation of Solution

Given:

Scores	Number of students
$\begin{array}{l}200\text{-250}\\ 250\text{-300}\\ \text{300-350}\\ \text{350-400}\\ \text{400-450}\\ \text{450-500}\\ \text{500-550}\\ \text{550-600}\\ \text{600-650}\\ 650\text{-700}\\ 700\text{-750}\\ 750\text{-800}\end{array}$	$\begin{array}{l}9\\ 14\\ 23\\ 30\\ 33\\ 28\\ 18\\ 12\\ 7\\ 3\\ 1\\ 1\end{array}$

Arrange the data in ascending order:

  $\left\{3.6,3.6,3.7,3.9,4.8\right\}$

The mean is calculated as:

  $\begin{array}{c}\overline{x}=\frac{{\displaystyle \sum x_i{f}_i}}{{\displaystyle \sum f_i}}\\ =\frac{\begin{array}{l}9\left(245\right)+14\left(275\right)+30\left(375\right)+33\left(425\right)+28\left(475\right)+18\left(525\right)+12\left(575\right)+\\ 7\left(625\right)+3\left(675\right)+1\left(725\right)+1\left(775\right)\end{array}}{9+14+23+30+33+28+12+7+3+1+1}\\ \approx 425.6\end{array}$

The meanis $425.6$ .

(b)

To determine

To find: The median class of the frequency distribution.

Answer to Problem 25E

The median class is $400\text{-450}$ .

Explanation of Solution

The median is calculated as:

Scores	Number of students	Number of Students
$\begin{array}{l}200\text{-250}\\ 250\text{-300}\\ \text{300-350}\\ \text{350-400}\\ \text{400-450}\\ \text{450-500}\\ \text{500-550}\\ \text{550-600}\\ \text{600-650}\\ 650\text{-700}\\ 700\text{-750}\\ 750\text{-800}\end{array}$	$\begin{array}{l}9\\ 14\\ 23\\ 30\\ 33\\ 28\\ 18\\ 12\\ 7\\ 3\\ 1\\ 1\end{array}$	$\begin{array}{l}9\\ 23\\ 46\\ 76\\ 109\\ 137\\ 155\\ 167\\ 174\\ 177\\ 178\\ 179\end{array}$

Since, $\frac{179}{2}=89.5$ , half of the data has been gathered in the $400\text{-450}$ class.

The median class is $400\text{-450}$ .

(c)

To determine

To find: The median of the verbal scores at Kennedy High School.

Answer to Problem 25E

The median is $420.45$ .

Explanation of Solution

The median is calculated as:

  $\begin{array}{l}\text{Median}=l+\left(\frac{\frac{n}{2}-c.f.}{f}\right)\cdot h\\ =400+\left(\frac{\frac{179}{2}-76}{33}\right)\cdot 50\\ \approx 420.45\end{array}$

The median is $420.45$ .