Question

For the data given in problem 12, obtain modal distance travelled per litre by a moped.

Answer 1

Distance travelled (in km)	No. of mopeds (frequency)
62-65	5
65-68	8
68-71	12
71-74	28
74-77	35
77-80	10
80-83	2

Here, the maximum frequency f_m is 35. So, the corresponding modal class is 74-77.
Mode = L+$\left[\frac{f_m-f_1}{2f_m-f_1-f_2}\right]h$
Here, L = lower limit of the modal class = 74
$f_m=$maximum frequency = 35
$f_1=$frequency of pre-modal class = 28
$f_2=$frequency of post-tmodal class = 10
h = width of modal class = 3
$$\begin{gathered} \therefore \mathrm{Mode}=74+\left[\frac{35-28}{2\left(35\right)-28-10}\right]\times 3 \\ =74+\left[\frac{7}{32}\right]\times 3 \\ =74+0.65 \\ =74.65 \\ \therefore \mathrm{The}\mathrm{modal}\mathrm{distance}\mathrm{travelled}\mathrm{per}\mathrm{litre}\mathrm{by}\mathrm{a}\mathrm{moped}\mathrm{is}74.65\mathrm{km}. \end{gathered}$$