Question

Obtain the mean, median and mode of the following data:

Marks	0−10	10−20	20−30	30−40	40−50	50−60	60−70	70−80
Number of Students	5	7	15	25	20	15	8	5

Answer 1

Marks	Mid Values (m)	No. of Workers (f)	Cumulative Frequency (c.f.)	fm
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80	5 15 25 35 45 55 65 75	5 7 15 25 20 15 8 5	5 12 27 52 72 87 95 100	25 105 375 875 900 825 520 375
	15	N = ∑f = 100		∑fm = 4000

$$\begin{gathered} \underline{\mathbf{Mean}} \\ \mathrm{Mean}\left(\overline{X}\right)=\frac{\Sigma fm}{\Sigma f}=\frac{4000}{100}=40 \end{gathered}$$

$\underline{\mathbf{Median}}$
Median class is given by the size of ${\left(\frac{N}{2}\right)}^{th}$ item, i.e.${\left(\frac{100}{2}\right)}^{th}$ item, which is 50^th item.
This corresponds to the class interval of (30 − 40), so this is the median class.
$$\begin{gathered} \mathrm{Median}=l_1+\frac{{\displaystyle \frac{N}{2}}-c.f.}{f}\times i \\ \mathrm{so},\mathrm{Median}=l_1+\frac{{\displaystyle \frac{100}{2}}-c.f.}{f}\times i \\ \mathrm{or},\mathrm{Median}=30+\frac{50-27}{25}\times 10 \\ \mathrm{or},\mathrm{Median}=30+\frac{230}{25} \\ \mathrm{Thus},\mathrm{median}=30+9.2=39.2 \end{gathered}$$

Mode

By inspection, we can say that the modal class is (30 – 40) as it has the highest frequency of 25.
$$\begin{gathered} \mathrm{Mode}\left(Z\right)=l_1+\frac{f_1-f_0}{2f_1-f_0-f_2}\times i \\ \mathrm{or},Z=30+\frac{25-15}{2\left(25\right)-15-20}\times 10 \\ \mathrm{or},Z=30+\frac{10}{50-35}\times 10 \\ \mathrm{or},Z=30+\frac{100}{15} \\ \mathrm{or},Z=30+6.6=36.6 \\ \mathrm{Hence},\mathrm{mode}\left(Z\right)\mathrm{is}36.6 \end{gathered}$$