Question

Find lower quartile, median and upper quartile from the data given below

Class-interval (More than)	10	20	30	40	50	60	70
Frequency	100	99	96	85	64	31	9

Answer 1

Class Interval	Frequency (f)	Cumulative Frequency (c.f.)
10 − 20 20 − 30 30 − 40 40 − 50	1 3 11 21	1 4 15 36
50 − 60	33 (f)	69
60 − 70 70 − 80	22 9	91 100
	$\underset{}{N=\sum }f=100$

Lower Quartile

$Q_1=\mathrm{Size}\mathrm{of}{\left(\frac{N}{4}\right)}^{\mathrm{th}}\mathrm{item}={\left(\frac{100}{4}\right)}^{\mathrm{th}}={25}^{\mathrm{th}}\mathrm{item}$
This lies in the class interval (40-50).
Now,
$$\begin{gathered} Q_1=l_1=\frac{{\displaystyle \frac{N}{4}}-c.f.}{f}\times i \\ \mathrm{or},Q_1=40+\frac{25-15}{21}\times 10 \\ \Rightarrow Q_1=44.76 \end{gathered}$$

Thus, the value of lower quartile is 44.76.

Median
Median class is given by the size of ${\left(\frac{N}{2}\right)}^{th}$ item, i.e.${\left(\frac{100}{2}\right)}^{\mathrm{th}}$ item, which is 50^th item.
This corresponds to the class interval of 50 − 60, so this is the median class.Uppe
$$\begin{gathered} \mathrm{Median}=l_1+\frac{{\displaystyle \frac{N}{2}}-c.f.}{f}\times i \\ \mathrm{so},\mathrm{Median}=l_1+\frac{{\displaystyle \frac{100}{2}}-c.f.}{f}\times i \\ \mathrm{or},\mathrm{Median}=50+\frac{50-36}{33}\times 10 \\ \mathrm{or},\mathrm{Median}=50+4.24=54.24 \\ \mathrm{Thus},\mathrm{Median}=54.24 \end{gathered}$$

Upper Quartile

$Q_3=\mathrm{Size}\mathrm{of}3{\left(\frac{100}{4}\right)}^{\mathrm{th}}={75}^{\mathrm{th}}\mathrm{item}$
This lies in the class interval (60-70).

$$\begin{gathered} Q_3=l_1=\frac{3({\displaystyle \frac{N}{4})}-c.f.}{f}\times i \\ \mathrm{or},Q_3=60+\frac{75-69}{22}\times 10 \\ \Rightarrow Q_3=62.72 \end{gathered}$$

Thus, the value of upper quartile is 62.72.