Question

The marks obtained by 100 students in an examination are given below. Compute the median, 1st and 3rd Quartiles.

Marks	30-35	35-40	40-45	45-50	50-55	55-60	60-65
No of students	14	16	18	23	18	8	3

Answer 1

Marks	f	c.f
30 – 35	14	14
35 – 40	16	30
40 – 45	18	48
45 – 50	23	71
50 – 55	18	89
55 – 60	8	97
60 – 65	3	100

N=100
Median class is given by the size of the ${\left(\frac{N}{2}\right)}^{\mathrm{th}}$ item, i.e. the ${\left(\frac{100}{2}\right)}^{\mathrm{th}}$ item, which is the 50^th item. This corresponds to the class interval 45−50, so this is the median class.
$$\begin{gathered} \mathrm{Median}=l_1+\frac{{\displaystyle \frac{N}{2}-c.f}}{f}\times i \\ =45+\frac{50-48}{23}\times 5 \\ =45+\frac{2}{23}\times 5 \\ =45+0.43 \\ =45.43 \\ \therefore \boxed{\mathrm{Median}=45.43} \end{gathered}$$

First quartile class is given by the size of the ${\left(\frac{N}{4}\right)}^{\mathrm{th}}$ item, i.e. the ${\left(\frac{100}{4}\right)}^{\mathrm{th}}$ item, which is the 25^th item. This corresponds to the class interval 35−40, so this is the first quartile class.

$$\begin{gathered} Q_1=l_1+\frac{\left({\displaystyle \frac{N}{4}}\right)-c.f}{f}\times i \\ =35+\frac{25-14}{16}\times 5 \\ =35+\frac{55}{16} \\ =35+3.43 \\ =38.43 \\ \boxed{\therefore Q_1=38.43} \end{gathered}$$

Third quartile class is given by the size of the $3{\left(\frac{N}{4}\right)}^{\mathrm{th}}$ item, i.e. the $3{\left(\frac{100}{4}\right)}^{\mathrm{th}}$ item, which is the 75^th item. This corresponds to the class interval 50−55, so this is the third quartile class.
$$\begin{gathered} Q_3=l_1+\frac{3\left({\displaystyle \frac{N}{4}}\right)-c.f}{f}\times i \\ =50+\frac{75-71}{18}\times 5 \\ =50+\frac{20}{18} \\ =50+1.11 \\ =51.11 \\ \boxed{\therefore Q_3=51.11} \end{gathered}$$