The mode of the following frequency distribution is

Marks 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100

Frequency 5 10 20 10 15

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Solution

The correct option is B 50

Marks Frequency ( $f_{i}$ )

0 - 20 5

20 - 40 10

40 - 60 20

60 - 80 10

80 - 100 15

Here, the maximum frequency is 20, and the class corresponding to this frequency is 40 - 60.

∴ Modal class = 40 - 60
Here,
Lower limit (l) of modal class = 40
Class size (h) = 60 – 40 = 20
Frequency ( $f_{1}$ ) of the modal class = 20
Frequency ( $f_{0}$ ) of the class preceding to the modal class = 10
Frequency ( $f_{2}$ ) of the class succeeding to the modal class = 10
Now, substituting these values in the formula, we get
$M o d e = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h$
$= 40 + (\frac{20 - 10}{2 (20) - 10 - 10}) \times 20$
$= 40 + (\frac{10}{40 - 20}) \times 20$
$= 40 + (\frac{10}{20}) \times 20$
$= 40 + 10$
$= 50$
Thus, the mode of the given data is 50.
Hence, the correct answer is option b.

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