Chapter 10, Problem 11MCQ

To determine

To calculate mean and standard deviation

Answer to Problem 11MCQ

Standard deviation is 2.87 and mean is 7.88

Explanation of Solution

Given information:

The data set is,

9,11,2,6,8,10,6,3,10,11,9,8,3,8,5,11,14,6,8,6,11,5,9,10,8

Formula used:

Mean is,

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

Standard deviations,

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

Calculation:

For mean, substituting the values,

  $\overline{x}=\frac{9+11+2+6+8+10+6+3+10+11+9+8+3+8+5+11+14+6+8+6+11+5+9+10+8}{25}$

So,

  $\overline{x}=\frac{197}{25}$

On solving,

  $\overline{x}=7.88$

Hence, mean is 7.88

For standard deviation, substituting the values,

  $sd=\sqrt{\frac{\begin{array}{}$ {(9-7.88)}^2+{(11-7.88)}^2+{(2-7.88)}^2+{(6-7.88)}^2+{(8-7.88)}^2+{(10-7.88)}^2+{(6-7.88)}^2+$$ {(3-7.88)}^2+{(10-7.88)}^2+{(11-7.88)}^2+{(9-7.88)}^2+{(8-7.88)}^2+{(3-7.88)}^2+{(8-7.88)}^2+$$ {(5-7.88)}^2+{(11-7.88)}^2+{(14-7.88)}^2+{(6-7.88)}^2+{(8-7.88)}^2+{(6-7.88)}^2+{(11-7.88)}^2+$$ {(5-7.88)}^2+{(9-7.88)}^2+{(10-7.88)}^2+{(8-7.88)}^2$\end{array}}{25}}$

So,

  $sd=\sqrt{\frac{\begin{array}{l}\text{1}\text{.2544+9}\text{.7344+34}\text{.5744+3}\text{.5344+0}\text{.0144+4}\text{.4944+3}\text{.5344+23}\text{.8144+4}\text{.4944+9}\text{.7344+1}\text{.2544+ }\\ \text{0}\text{.0144+23}\text{.8144+0}\text{.0144+8}\text{.2944+9}\text{.7344+37}\text{.4544+3}\text{.5344+0}\text{.0144+3}\text{.5344+9}\text{.7344+8}\text{.2944+}\\ \text{1}\text{.2544+4}\text{.4944+0}\text{.0144}\end{array}}{25}}$

On solving,

  $sd=\sqrt{\frac{206.64}{25}}$

On solving,

  $sd=2.87$

Hence, standard deviation 2.87