Chapter 7.1, Problem 39E

(a)

To determine

The mean, median and mode of the given data.

Answer to Problem 39E

  $7.7$

Explanation of Solution

Given:

The given data set is:

  

Calculation:

The frequency of the data set will be:

  $\begin{array}{l}\frac{200\times 30}{100}=60\\ \frac{200\times 14}{100}=28\\ \frac{200\times 20}{100}=40\\ \frac{200\times 35}{100}=70\\ \frac{200\times 1}{100}=2\end{array}$

Hence, the data set is:

  $2,8,40,60,70$

The value of mean, median mode will be:

  $\begin{array}{l}\text{Mean}=\frac{2+28+40+60+70}{5}=\frac{200}{5}=40\\ \text{Median}=40\\ \text{Mode = None}\end{array}$

(b)

To determine

The outliers and the effect of outliers on mean, median and mode.

Answer to Problem 39E

Outlier = 2.

Mean and median will decrease and mode remains the same.

Explanation of Solution

Given:

The given data set is:

  

Calculation:

Hence, the data set is:

  $2,8,40,60,70$

Outlier = 2.

The value of new mean, median mode will be:

  $\begin{array}{l}\text{Mean}=\frac{28+40+60+70}{5}=\frac{198}{5}=39.6\\ \text{Median}=\frac{28+40}{2}=34\\ \text{Mode = None}\end{array}$

Mean and median will decrease and mode remains the same.

(c)

To determine

The circle graph with the mentioned changes.

Answer to Problem 39E

The mean, median and mode will be the Psychology I class.

Explanation of Solution

Given:

The given data set is:

  

Calculation:

All students take the Psychology II class 1 year later.

The new circle graph will be:

  

The mean, median and mode will be the Psychology I class.