Chapter B.2, Problem 5E

To determine

To Find: The mean,median and mode of the data.

Answer to Problem 5E

Mean is $8$ ,median is $8.5$ and mode is $7\mathrm{and}12$ .

Explanation of Solution

Given information: Given data is $5,12,7,14,8,9,7,12$ .

Calculation: Given data is $5,12,7,14,8,9,7,12$ .

Mean of the data :

  $\begin{array}{l}=\frac{\mathrm{s}\mathrm{u}\mathrm{m}\mathrm{o}\mathrm{f}\mathrm{o}\mathrm{b}\mathrm{s}\mathrm{e}\mathrm{r}\mathrm{v}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{o}\mathrm{f}\mathrm{o}\mathrm{b}\mathrm{s}\mathrm{e}\mathrm{r}\mathrm{v}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}\\ =\frac{5+12+7+14+8+9+7+12}{8}\\ =8\end{array}$

Median :

Data in ascending order $5,7,7,8,9,12,12,14$ .

  $\begin{array}{l}\mathrm{Median}=\frac{\frac{\mathrm{n}}{2}\mathrm{t}\mathrm{h}+(\frac{\mathrm{n}}{2}+1)\mathrm{t}\mathrm{h}}{2}\\ \mathrm{Median}=\frac{\frac{8}{2}\mathrm{t}\mathrm{h}+(\frac{8}{2}+1)\mathrm{t}\mathrm{h}}{2}\\ \mathrm{Median}=\frac{4\mathrm{t}\mathrm{h}+5\mathrm{t}\mathrm{h}}{2}\\ \mathrm{Median}=\frac{8+9}{2}=\frac{17}{2}=8.5\end{array}$

Mode :

  $7\mathrm{and}12$ comes highest time .

So mode is $7\mathrm{and}12$ .