Question

The times, in seconds, taken by 150 athletes to run a 110m hurdle race are tabulated below:

Class	Frequency
$13.8 - 14$	$2$
$14 - 14.2$	$4$
$14.2 - 14.4$	$5$
$14.4 - 14.6$	$71$
$14.6 - 14.8$	$48$
$14.8 - 15$	$20$

How many athletes completed the race in less than $14.6$ seconds?
A) $11$
B) $71$
C) $82$
D) $130$

Answer 1

In this question, we are given the time taken by 150 athletes to run a 110m hurdle race. We have been asked the number of athletes who completed the race before a certain time period. For this, we will calculate cumulative frequency. Once it has been calculated, we will observe the number of athletes who completed the race before 14.6 seconds, it will be the cumulative frequency written in front of $14.4 - 14.6$ interval. That frequency will be our answer.

Complete step-by-step solution:
We are given the time taken by 150 athletes to run a 110m hurdle race. We have been asked the number of athletes who completed the race before $14.6$ seconds.
To answer this question, let us make a column of cumulative frequency.

Class	Frequency	Cumulative frequency
$13.8 - 14$	$2$	$2$
$14 - 14.2$	$4$	$2 + 4 = 6$
$14.2 - 14.4$	$5$	$6 + 5 = 11$
$14.4 - 14.6$	$71$	$11 + 71 = 82$
$14.6 - 14.8$	$48$	$82 + 48 = 130$
$14.8 - 15$	$20$	$130 + 20 = 150$

As we can see that the cumulative frequency written beside the interval $14.4 - 14.6$ is 82. We have chosen the interval $14.4 - 14.6$ because it shows the number of athletes who completed the race before that interval and during that interval, combined.

Therefore, our answer is option C.

Note: If you are having difficulty in understanding the concept of cumulative frequency, you can also use the following method-
The number of athletes who finished in 14.6 seconds will be the sum of all those who finished at $14.6^{th}$ second and before that. Therefore,
$n( \leqslant 14.6\sec ) = n(13.8 - 14) + n(14 - 14.2) + n(14.2 - 14.4) + n(14.4 - 14.6)$
$\Rightarrow n( \leqslant 14.6\sec ) = 2 + 4 + 5 + 71 = 82$
Hence, 82 athletes completed the race before 14.6 seconds.