Question

Calculate the mean, the mode and the median from the following frequency distribution.

Height (in inches)	60−61	61−62	62−63	63−64	64−65	65−66	66−67	67−68	68−69
Frequency	4	16	8	24	35	18	19	16	10

Answer 1

Height	Mid Values (m)	Frequency (f)	Cumulative Frequency (c.f.)	fm
60 − 61 61 − 62 62 − 63 63 − 64	60.5 61.5 62.5 63.5	4 16 8 24 (f0)	4 20 28 52 (c.f.)	242 984 500 1524
64 − 65	64.5	35 (f1)	87	2257.5	Modal Class
65 − 66 66 − 67 67 − 68 68 − 69	65.5 66.5 67.5 68.5	18 (f2) 19 16 10	105 124 140 150	1179 1263.5 1080 685
		N = ∑f=150		∑fm =9715

$$\begin{gathered} \underline{\mathbf{Mean}} \\ \mathrm{Mean}\left(\overline{X}\right)=\frac{\Sigma fm}{\Sigma f}=\frac{9715}{150}=64.76 \end{gathered}$$

$\underline{\mathbf{Median}}$
Median class is given by the size of ${\left(\frac{N}{2}\right)}^{th}$ item, i.e.${\left(\frac{150}{2}\right)}^{\mathrm{th}}$ item, which is 75^th item.
This corresponds to the class interval of (64 − 65), so this is the median class.
$$\begin{gathered} \mathrm{Median}=l_1+\frac{{\displaystyle \frac{N}{2}}-c.f.}{f}\times i \\ \mathrm{so},\mathrm{Median}=64+\frac{{\displaystyle \frac{150}{2}}-c.f.}{f}\times i \\ \mathrm{or},\mathrm{Median}=64+\frac{75-52}{35}\times 1 \\ \mathrm{or},\mathrm{Median}=64+0.65=64.65 \\ \mathrm{Thus},\mathrm{median}=64.65 \end{gathered}$$

Mode

By inspection, we can say that the modal class is (64 – 65) as it has the highest frequency of 35.
$$\begin{gathered} \mathrm{Mode}\left(Z\right)=l_1+\frac{f_1-f_0}{2f_1-f_0-f_2}\times i \\ \mathrm{or},Z=64+\frac{35-24}{2\left(35\right)-24-18}\times 1 \\ \mathrm{or},Z=64+\frac{11}{70-42}\times 1 \\ \mathrm{or},Z=64+\frac{11}{28} \\ \mathrm{or},Z=64+0.39=64.39 \\ \mathrm{Hence},\mathrm{mode}\left(Z\right)\mathrm{is}64.39. \end{gathered}$$