Chapter 10.5, Problem 13PPS

To determine

To calculate mean, median, mode, rangeand standard deviation after multiplying the given constant.

Answer to Problem 13PPS

Mean is 28.6.

Median is 27.2.

Mode is 29.6.

Standard deviation is 3.47

Range is10.4

Explanation of Solution

Given information:

Frequency: 33,37,38,29,35,37,27,40,28,31

Constant to be multiplied= 0.8

Formula used:Mean formula is,

  $\overline{X}=\frac{{\displaystyle \sum fX}}{N}$

Where, f is frequency

N is total number of items

X is item.

Median formula is,

  $median=\frac{{\left(\frac{n}{2}\right)}^{th}term+{\left(\frac{n}{2}+1\right)}^{th}term}{2}$

Where, N cumulative frequency

Mode is the item with the maximum frequency.

Standard deviation is,

  $SD=\sqrt{\frac{{\displaystyle \sum }{(x-\overline{x})}^2}{n}}$

Where, SD= standard deviation,

x = value

The Range is the difference between the lowest and highest values.

Calculation:

According to question,

Multiplying 0.8 to all the frequency values,

New frequency:

26.4,29.6,30.4,23.2,28,29.6,21.6,32,22.4,24.8

Sorting the frequency

21.6, 22.4, 23.2, 24.8, 26.4, 28, 29.6, 29.6, 30.4,32

For Mean,

Substituting the values in mean formula,

  $=\frac{26.4+29.6+30.4+23.2+28+29.6+21.6+32+22.4+24.8}{10}$

So,

  $=\frac{286}{10}$

On solving,

  $=28.6$

Hence, Mean is 28.6

For median,

Substituting values,

  $=\frac{{\left(\frac{10}{2}\right)}^{th}term+{\left(\frac{10}{2}+1\right)}^{th}term}{2}$

On solving,

  $=\frac{{\left(5\right)}^{th}term+{\left(6\right)}^{th}term}{2}$

Substituting the values,

  $=\frac{26.4+28}{2}$

On solving,

  $=27.2$

Hence, Median is 27.2.

For mode,

Mode is the item with the maximum frequency which is 29.6.

For standard deviation,

Substituting values,

  $SD=\sqrt{\frac{\begin{array}{}$ {(21.6-28.6)}^2+{(22.4-28.6)}^2+{(23.2-28.6)}^2+{(24.8-28.6)}^2+{(26.4-28.6)}^2$$ +{(28-28.6)}^2+{(29.6-28.6)}^2+{(29.6-28.6)}^2+{(30.4-28.6)}^2+{(32-28.6)}^2$\end{array}}{10}}$

So,

  $SD=\sqrt{\frac{\begin{array}{}$ {(-5.2)}^2+{(-4.4)}^2+{(-3.6)}^2+{(-2)}^2+{(-0.4)}^2$$ +{(1.2)}^2+{(2.8)}^2+{(2.8)}^2+{(-3.6)}^2+{(5.2)}^2$\end{array}}{10}}$

On solving,

  $SD=3.473$

Hence, Standard deviation is 3.473

For Range,

Range is the difference between the lowest and highest values.

  $range=\max -\min$

Substituting values,

  $=32-21.6$

On solving,

  $=10.4$

Hence, range is 10.4