Question

Find mean, median, and mode for the following data

Classes	0−10	10−20	20−30	30−40	40−50	50−60
Frequencies	4	15	10	7	3	1

Answer 1

Class Interval	Mid Values (m)	Frequency (f)	Cumulative Frequency (c.f.)	fm
0 − 10	5	4(f0)	4	20
10 − 20	15	15(f1)	19	225	→ Modal Class
20 − 30	25	10(f2)	29	250
30 − 40 40 − 50 50 − 60	35 45 55	7 3 1	36 39 40	245 135 55
		N = ∑f=40		∑fm =930

$$\begin{gathered} \underline{\mathbf{Mean}} \\ \mathrm{Mean}\left(\overline{X}\right)=\frac{\Sigma fm}{\Sigma f}=\frac{930}{40}=23.25 \end{gathered}$$

$\underline{\mathbf{Median}}$
Median class is given by the size of ${\left(\frac{N}{2}\right)}^{th}$ item, i.e.${\left(\frac{40}{2}\right)}^{\mathrm{th}}$ item, which is 20^th item.
This corresponds to the class interval of (20 − 30), so this is the median class.
$$\begin{gathered} \mathrm{Median}=l_1+\frac{{\displaystyle \frac{N}{2}}-c.f.}{f}\times i \\ \mathrm{so},\mathrm{Median}=20+\frac{{\displaystyle \frac{40}{2}}-c.f.}{f}\times i \\ \mathrm{or},\mathrm{Median}=20+\frac{20-19}{10}\times 10 \\ \mathrm{or},\mathrm{Median}=20+1=21 \\ \mathrm{Thus},\mathrm{median}=21 \end{gathered}$$

Mode

By inspection, we can say that the modal class is (10 – 20) as it has the highest frequency of 15.
$$\begin{gathered} \mathrm{Mode}\left(Z\right)=l_1+\frac{f_1-f_0}{2f_1-f_0-f_2}\times i \\ \mathrm{or},Z=10+\frac{15-4}{2\left(15\right)-4-10}\times 10 \\ \mathrm{or},Z=10+\frac{11}{30-14}\times 10 \\ \mathrm{or},Z=10+\frac{110}{16} \\ \mathrm{or},Z=10+6.875=16.875 \\ \mathrm{Hence},\mathrm{mode}\left(Z\right)\mathrm{is}16.875. \end{gathered}$$