Question

A survey regarding the heights (in cm) of 50 girls of a class was conducted and the following data was obtained:

Height in cm	120-130	130-140	140-150	150-160	160-170	Total
Number of girls	2	8	12	20	8	50

Find the mean, median and mode of the above data.

Answer 1

We have the following:

Height in cm	Mid value $\left(x_i\right)$	Frequency $\left(f_i\right)$	Cumulative frequency	$f_i\times x_i$
120-130	125	2	2	250
130-140	135	8	10	1080
140-150	145	12	22	1740
150-160	155	20	42	3100
160-170	165	8	50	1320
		∑$f_i=50$		∑$\left(f_i\times x_i\right)=$7490

Mean, $\overline{x}$=$\frac{\sum \left(f_i\times x_i\right)}{\sum f_i}$
=$\frac{7490}{50}$
=149.8

$$\begin{gathered} \mathrm{Now},N=50 \\ \Rightarrow \frac{N}{2}=25 \end{gathered}$$
The cumulative frequency just greater than 25 is 42 and the corresponding class is 150-160.
Thus, the median class is 150-160.
$\mathrm{Now},l=150,h=10,f=20,c=cf$of preceding class = 22 and $\frac{N}{2}=25$
$$\begin{gathered} \mathrm{Median},M_e=l+\left\{h\times \frac{\left(\frac{N}{2}-c\right)}{f}\right\} \\ =150+\left\{10\times \frac{\left(25-22\right)}{20}\right\} \\ =\left(150+10\times \frac{3}{20}\right) \end{gathered}$$
=151.5
∴ Mode = 3(Median) $-2$(Mean)
= 3$\times 151.5-2\times 149.8$
= 154.9