Chapter 10, Problem 9PT

To determine

To calculate mean, median, mode, range and standard deviation

Answer to Problem 9PT

Range is 15, mean is 5.75, standard deviation is 4.044, median is 6 and mode is 6

Explanation of Solution

Given information:

The given data is,

6,9,0,15,9,14,11,13,9,5,8,6

The constant to be added is “-3”

Formula used:

Mean is,

  $\overline{x}=\frac{{\displaystyle \sum x}}{n}$

Median is,

  $M=\frac{{(\frac{n}{2})}^{th}+{(\frac{n}{2}+1)}^{th}}{2}$

Mode is,

  $z=Most\mathrm{Re}peatedValue$

Range is,

  $range=\max -\min$

Standard deviation is,

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

Calculation:

Re-arranging the data in ascending order,

0,5,6,6,8,9,9,9,11,13,14,15

The required data is,

-3,2,3,3,5,6,6,6,8,10,11,12

Minimum value is -3

Maximum value is 12.

For range, substituting the value,

  $range=12-(-3)$

On solving,

  $range=15$

Hence, range is 15.

For mean, substituting the values,

  $\overline{x}=\frac{-3+2+3+3+5+6+6+6+8+10+11+12}{12}$

So,

  $\overline{x}=\frac{69}{12}$

On solving,

  $\overline{x}=5.75$

Hence, mean is 5.75

For standard deviation, substituting the values,

  $sd=\sqrt{\frac{\begin{array}{l}{(}^{-}+{(}^2+{(}^3+{(}^3+{(}^5+{(}^6+{(}^6\\ +{(}^6+{(}^8+{(}^{10}+{(}^{11}+{(}^{12}\end{array}}{12}}$

So,

  $sd=\sqrt{\frac{\begin{array}{l}\text{76}\text{.5625+14}\text{.0625+7}\text{.5625+7}\text{.5625+0}\text{.5625+0}\text{.0625+0}\text{.0625}\\ \text{+0}\text{.0625+5}\text{.0625+18}\text{.0625+27}\text{.5625+39}\text{.0625}\end{array}}{12}}$

On solving,

  $sd=\sqrt{\frac{196.25}{12}}$

On solving,

  $sd=4.044$

Hence, standard deviation is 4.044

For median, substituting the values,

  $M=\frac{{(\frac{12}{2})}^{th}+{(\frac{12}{2}+1)}^{th}}{2}$

On solving,

  $M=\frac{6^{th}+7^{th}}{2}$

Substituting the values,

  $M=\frac{6+6}{2}$

On solving,

  $M=6$

Hence, median is 9.

For mode, substituting the value,

  $z=6$

Hence, mode is 6