Chapter B.2, Problem 6E

To determine

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

Answer to Problem 6E

Mean is $33.85$ ,median is $33$ and mode is $32$ .

Explanation of Solution

Given information: Given data is $30,37,32,39,33,34,32$ .

Calculation: Given data is $30,37,32,39,33,34,32$ .

Mean of the data :

  $\begin{array}{l}=\frac{\mathrm{s}\mathrm{u}\mathrm{m}\mathrm{o}\mathrm{f}\mathrm{o}\mathrm{b}\mathrm{s}\mathrm{e}\mathrm{r}\mathrm{v}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{o}\mathrm{f}\mathrm{o}\mathrm{b}\mathrm{s}\mathrm{e}\mathrm{r}\mathrm{v}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}\\ =\frac{30+37+32+39+33+34+32}{7}\\ =\frac{237}{7}=33.85\end{array}$

Median :

Data in ascending order $30,32,32,33,34,37,39$ .

  $\begin{array}{l}\mathrm{Median}=\frac{\mathrm{n}+1}{2}\mathrm{t}\mathrm{h}\\ \mathrm{Median}=\frac{7+1}{2}\mathrm{t}\mathrm{h}=\frac{8\mathrm{t}\mathrm{h}}{2}=4\mathrm{t}\mathrm{h}\\ \mathrm{Median}=33\end{array}$

Mode :

  $32$ comes highest time .

So mode is $32$ .