Chapter 10.5, Problem 1CYU

To determine

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

Answer to Problem 1CYU

Mean is 3.5, Median is 3.5, Mode is 1, Standard deviation is 2.43 and Range is 8

Explanation of Solution

Given information:

Frequency: 10,13,9,8,15,8,13,12,7,8,11,12

Constant to be added: -7

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 is,

  $range=\max -\min$

Calculation:

According to question,

Add +8 to all the frequency values,

New frequency:

3,6,2,1,8,1,6,5,0,1,4,5

For Mean substituting the values in mean formula,

  $=\frac{3+6+2+1+8+1+6+5+0+1+4+5}{12}$

So,

  $=\frac{42}{12}$

On solving,

  $=3.5$

Hence, Mean is 3.5

For median sorting the frequency

0,1,1,1,2,3,4,5,5,6,6,8

Substituting values in the median formula,

  $=\frac{{\left(\frac{12}{2}\right)}^{th}term+{\left(\frac{12}{2}+1\right)}^{th}term}{2}$

On solving,

  $=\frac{{\left(6\right)}^{th}term+{\left(7\right)}^{th}term}{2}$

Substituting the values,

  $=\frac{3+4}{2}$

On solving,

  $=3.5$

Hence, Median is 3.5.

For mode,

Mode is the item with the maximum frequency which is 1.

For standard deviation substituting the values in standard deviation formula,

  $SD=\sqrt{\frac{\begin{array}{}$ {(3-3.5)}^2+{(6-3.5)}^2+{(2-3.5)}^2+{(1-3.5)}^2+{(8-3.5)}^2+{(1-3.5)}^2$$ +{(6-3.5)}^2+{(5-3.5)}^2+{(0-3.5)}^2+{(1-3.5)}^2+{(4-3.5)}^2+{(5-3.5)}^2$\end{array}}{12}}$

So,

  $SD=\sqrt{\frac{\begin{array}{}$ {(-0.5)}^2+{(2.5)}^2+{(-1.5)}^2+{(-2.5)}^2+{(4.5)}^2+{(-2.5)}^2$$ +{(2.5)}^2+{(1.5)}^2+{(-3.5)}^2+{(-2.5)}^2+{(0.5)}^2+{(1.5)}^2$\end{array}}{12}}$

On solving,

  $SD=\sqrt{5.91}$

On solving,

  $SD=2.43$

Hence, Standard deviation is 2.43

For Range substituting the values,

  $range=8-0$

Solving,

  $range=8$

Hence, range is 8