Chapter 12, Problem 13CR

To determine

To find: the mean, median, mode and range of the given data set.

Answer to Problem 13CR

The mean here is $12.06$ , median is $12$ ,mode is $12$ and range here is $2$ .

Explanation of Solution

Given:

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  $11,12,12,12,12,12,12,12,12,12,12,12,12,12,12,13,13$ .

Concept used:

Arithmetic mean:

Mean: $\stackrel{-}{x}=\frac{{\displaystyle \sum xi}}{N}$ .

Median $=\left\{\begin{array}{c}\frac{N}{2}\text{th term; }N\text{ is odd}\\ \frac{N}{2}\text{th term; }N\text{ is even}\end{array}\right\}$ .

Mode= the value in the data set that occurs most frequently.

Calculation:

Mean is the average of the data:

Here from the given data:

There is $11$ one times, $12$ fourteen times, $13$ two time.

  $\begin{array}{l}\stackrel{-}{x}=\frac{11+12\left(14\right)+13\left(2\right)}{17}.\\ \Rightarrow \frac{205}{17}\approx \mathrm{12.06.}\\ \text{the mean }\stackrel{-}{x}=\mathrm{12.06.}\end{array}$

Median is calculated by making the order in increasing order and then choose the middle value.

If the number of the data is even then there will have two middle value so, by taking the mean of two middle value get the median as:

The data in increasing order is:

  $11,12,12,12,12,12,12,12,12,12,12,12,12,12,12,13,13$ .

Total number of data is $17$ which is odd so, there will be one middle value which is $12$ .

The median is $12$ .

Mode is the value that occurs maximum or most often.

Taking the increasing order of the data:

  $11,12,12,12,12,12,12,12,12,12,12,12,12,12,12,13,13$ .

Her the maximum value that occur is $12$ .

The mode is $12$ .

Range is the difference between largest and smallest value.

Taking the increasing order of the data:

  $11,12,12,12,12,12,12,12,12,12,12,12,12,12,12,13,13$ .

Here largest value is $13$ and smallest value is $11$ .

  $13-11=2$ .

The range here is $2$ .

Since, the data set has no outlies, then the mean best describes the data set.

Hence, the mean here is $12.06$ , median is $12$ ,mode is $12$ and range here is $2$ .