Question

Calculate the Mean and Standard Deviation from the following distribution.

Age (years)	15−19	20−24	25−29	30−34	35−39	40−44
No. of Persons	4	20	38	24	10	4

Answer 1

In order to calculate the mean and standard deviation, we first need to convert the inclusive series into exclusive series as given below:

Age (X)	Frequency (f)	Mid-Values (m)	fm	m2	fm2
14.5−19.5	4	17	68	289	1156
19.5−24.5	20	22	440	484	9680
24.5−29.5	38	27	1026	729	27702
29.5−34.5	24	32	768	1024	24576
34.5−39.5	10	37	370	1369	13690
39.5−44.5	4	42	168	1764	7056
	Σf=100		Σfm=2840		Σfm2=83860

$$\begin{gathered} Mean,\overline{X}=\frac{\Sigma fm}{\Sigma f}=\frac{2840}{100}=28.4years \\ \mathrm{Standard}\mathrm{deviation},\left(\sigma \right)=\sqrt{\frac{\Sigma f{m}^2}{\Sigma f}-{\left(\overline{X}\right)}^2} \\ or,\left(\sigma \right)=\sqrt{\frac{83860}{100}-{\left(28.4\right)}^2} \\ or,\left(\sigma \right)=\sqrt{838.6-806.56} \\ or,\left(\sigma \right)=\sqrt{32.04} \\ \Rightarrow \left(\sigma \right)=5.66years \end{gathered}$$

Hence, mean age of the persons is 28.4 years and standard deviation is 5.66 years.