Question

Forty persons were examined for their haemoglobin % in blood (in mg per 100 ml) and the results were grouped as shown.

Haemoglobin %(mg/100 ml)	13.1-14	14.1-15	15.1-16	16.1-17	17.1-18
No. of persons	8	12	10	6	4

Determine the modal value of haemoglobin% in the blood of the persons.

Answer 1

Here, the class intervals are inclusive type. For calculating the mode, we have to convert the class
intervals into continuous type by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of
each interval. Then, we have to prepare the frequency distribution table as shown:

Original class	Class Boundaries	Frequency
13.1-14	13.05-14.05	$8\to f_1$
14.1-15	14.05-15.05	$12\to f_m$
15.1-16	15.05-16.05	$10\to f_2$
16.1-17	16.05-17.05	6
17.1-18	17.05-18.05	4

Here, the maximum frequency f_m is 12. So, the corresponding modal class is 14.05-15.05.
$$\begin{gathered} \mathrm{Mode}=L+\left[\frac{f_m-f_1}{2f_m-f_1-f_2}\right]h \\ \text{H}\mathrm{ere},\mathrm{L}=\mathrm{lower}\mathrm{limit}\mathrm{of}\mathrm{the}\mathrm{modal}\mathrm{class}=14.05 \\ {f}_m=\mathrm{maximum}\mathrm{frequency}=12 \\ {f}_1=\mathrm{frequency}\mathrm{of}\text{the}\mathrm{premodal}\mathrm{class}=8 \\ {f}_2=\mathrm{frequency}\mathrm{of}\text{the}\mathrm{postmodal}\mathrm{class}=10 \\ h=\mathrm{width}\mathrm{of}\mathrm{modal}\mathrm{class}=1 \\ \therefore \mathrm{Mode}=14.05+\left[\frac{12-8}{2\left(12\right)-8-10}\right]\times 1 \\ =14.05+\left[\frac{4}{24-18}\right] \\ =14.05+\frac{4}{6} \\ =14.71 \\ \therefore \mathrm{The}\mathrm{modal}\mathrm{value}\mathrm{of}\mathrm{heamoglobin}\%\mathrm{in}\text{the}\mathrm{blood}\mathrm{of}\text{the}\mathrm{person}s\mathrm{is}14.71\mathrm{mg}/100\mathrm{ml}. \end{gathered}$$