Question

Calculate the median and mode from the following data:

Marks	0−10	10−20	20−30	30−40	40−50	50−60	60−70	70−80
Number of Students	2	18	30	45	35	20	6	3

Answer 1

Marks	No. of Workers (f)	Cumulative Frequency (c.f.)
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80	2 18 30 = f0 45 = f1 35 = f2 20 6 3	2 20 50 95 130 150 156 159
	N = ∑f = 159

$\underline{\mathbf{Median}}$
Median class is given by the size of ${\left(\frac{N}{2}\right)}^{th}$ item, i.e.${\left(\frac{159}{2}\right)}^{th}$ item, which is 79.5^th item.
This corresponds to the class interval of (30 − 40), so this is the median class.
$$\begin{gathered} \mathrm{Median}=l_1+\frac{{\displaystyle \frac{N}{2}}-c.f.}{f}\times i \\ \mathrm{so},\mathrm{Median}=30+\frac{{\displaystyle \frac{159}{2}}-50}{45}\times i \\ \mathrm{or},\mathrm{Median}=30+\frac{29.5}{45}\times 10 \\ \mathrm{or},\mathrm{Median}=30+6.55 \\ \mathrm{Thus},\mathrm{Median}=36.55 \end{gathered}$$

$\underline{\mathbf{Mode}}$
By inspection, we can say that the modal class is (30 – 40) as it has the highest frequency of 45.
$$\begin{gathered} \mathrm{Mode}\left(Z\right)=l_1+\frac{f_1-f_0}{2f_1-f_0-f_2}\times i \\ \mathrm{or},Z=30+\frac{45-30}{2\left(45\right)-30-35}\times 10 \\ \mathrm{or},Z=30+\frac{15}{90-65}\times 10 \\ \mathrm{or},Z=30+\frac{150}{25}=30+6=36 \\ \mathrm{Hence},\mathrm{mode}\left(Z\right)\mathrm{is}36 \end{gathered}$$