Question

Find out the Mode from any of the following two distributions:
(a)

X	:	30-40	40-50	50-60	60-70	70-80	80-90	90-100
f	:	6	10	16	14	10	5	2

(b)

Marks	:	0-9	10-19	20-29	30-39	40-49
No of candidates	:	6	29	87	181	247
Marks	:	50-59	60-69	70-79	80-89	90-99
No of candidates	:	263	113	49	9	2

Answer 1

(a)

X	f
30 – 40	6
40 – 50	10
50 – 60	16	$\underset{}{\to }$Modal class
60 – 70	14
70 – 80	10
80 – 90	5
90 – 100	2

By inspection, we can spot that (50–60) has the highest frequency, i.e. 16. So, this is the modal class.

$$\begin{gathered} \mathrm{Mode}=l_1+\frac{f_1-f_0}{2f_1-f_0-f_2}\times i \\ =50+\frac{16-10}{2\times 16-10-14}\times 10 \\ =50+\frac{60}{8} \\ =50+7.5 \\ =57.5 \end{gathered}$$

Thus, the mode is 57.5.

(b)
Since this is an inclusive class interval series, to calculate mode, first we need to convert the inclusive class intervals into the exclusive class intervals by using the following formula.

$\mathrm{Value}\mathrm{of}\mathrm{Adjustment}=\frac{\mathrm{Lower}\mathrm{limit}\mathrm{of}\mathrm{ine}\mathrm{class}-\mathrm{Upper}\mathrm{limit}\mathrm{of}\mathrm{the}\mathrm{preceding}\mathrm{class}}{2}$Valueofadjustment=Lowerlimitofoneclass−Upperlimitofthepreceedingclass2

The value of adjustment (as calculated) is then added to the upper limit of each class and subtracted from the lower limit of each class. In this manner, we get the following distribution.

Marks	f
–0.5 – 9.5	6
9.5 – 19.5	29
19.5 – 29.5	87
29.5 – 39.5	181
39.5 – 49.5	247
49.5 – 59.5	263	$\underset{}{\to }$Modal class
59.5 – 69.5	113
69.5 – 79.5	49
79.5 – 89.5	9
89.5 – 99.5	2

By inspection, we can spot that (49.5 – 59.5) has the highest frequency, i.e.263. So, this is the modal class.
$$\begin{gathered} \mathrm{Mode}=l_1+\frac{f_1-f_0}{2f_1-f_0-f_2}\times i \\ =49.5+\frac{263-247}{2\times 263-247-113}\times 10 \\ =49.5+\frac{160}{166} \\ =49.5+0.963 \\ =50.46 \end{gathered}$$

Thus, the mode is 50.46.