Chapter 10.5, Problem 7PPS

To determine

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

Answer to Problem 7PPS

Mean is 60.9.

Median is 60.

Mode is 60.

Standard deviation is 4.65

Range is 14

Explanation of Solution

Given information:

Frequency: 52,53,49,61,57,52,48,60,50,47

Constant to be added: +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

  $range=\max -\min$

Calculation:

According to question,

Add +8 to all the frequency values,

New frequency:

60,61,57,69,65,60,56,68,58,55

For Mean,

Substituting the values in mean formula,

  $\begin{array}{l}=\frac{60+61+57+69+65+60+56+68+58+55}{10}\\ solving,\\ =\frac{609}{10}\\ =60.9\end{array}$

Mean is 60.9

For median,

Sorting the frequency

55,56,57,58,60,60,61,65,68,69

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

Where n=10(total frequency)

  $\begin{array}{l}=\frac{{\left(\frac{10}{2}\right)}^{th}term+{\left(\frac{10}{2}+1\right)}^{th}term}{2}\\ solving,\\ =\frac{{\left(5\right)}^{th}term+{\left(6\right)}^{th}term}{2}\\ =\frac{60+60}{2}\\ =60\end{array}$

Median is 60.

For mode,

Mode is the item with the maximum frequency which is 60.

For standard deviation,

Substituting the values in standard deviation formula,

  

Standard deviation is 4.65

For Range,

Range is the difference between the lowest and highest values.

  $range=\max -\min$

  =69-55

  =14