Question

A referee is standing on the side of the athletic track and kept a

stopwatch, which is used to find the time that it took a group of students to run 200 m.

In different time intervals, different number of students completed the race are given in the table as shown below

Time in seconds 0 – 50 ; 50 – 100 ; 100 – 150 ; 150 - 200

Number of students 10 15 7 8.
1) Find the mode of the given data

(a) 59.62 (b) 58.30 (c) 58.50 (d) 59.98
2) Find the lower limit of median class

(a) 25 (b) 75 (c) 100 (d) 50
3) Find the number of students, who have finished the race after 100 s.

(a) 14 (b) 15 (c) 30 (d) 25

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Answer 1

Answer:

1)  Mode of the data 69.23

2) The lower limit of the median class is 50

3) The number of students, who have finished the race after 100s = 15

Step-by-step explanation:

Time in seconds          No. of students

0-50                                 10

50-100                              15

100-150                             7

150-200                             8  

(1)

We know,

Mode is given by the formula

$L+ h \frac{(f_m - f_1)}{(f_m -f_1)+(f_m -f_2)}$

Where,

L = Lower limit of the modal class

h = size of the class interval

$f_m$ = frequency if the modal class

$f_1$ = frequency of the class preceding the modal class

$f_2$ = frequency of the class succeeding the modal class

Modal class is the class with maximum frequency.

Here, the class with a maximum frequency is 15. Hence the modal class is

50-100     15

L = Lower limit of the modal class = 50

h = size of the class interval = 50

$f_m$ = frequency if the modal class = 15

$f_1$ = frequency of the class preceding the modal class = 10

$f_2$ = frequency of the class succeeding the modal class = 7

 

Mode = $50 + 50 X \frac{(15-10)}{(15-10)+(15-7)}$

$= 50+ 50 X \frac{5}{5+8}$

$= 50 + 50 X \frac{5}{13}$

= 50+50X0.3846

= 50+19.23

= 69.23

Hence mode of the given data 69.23

(2) To find the median class we need to find the cumulative frequency

Time in seconds          No. of students     cumulative frequency

0-50                                 10                            10

50-100                              15                            25

100-150                             7                               7

150-200                             8                              40  

The median class is that class in which the cumulative frequency is just greater than $\frac{N}{2}$, where N is the total frequency

Here, total frequency = 40

$\frac{N}{2}$ = 20

The class in which the cumulative frequency is just greater than$\frac{N}{2}$  is

50-100

Hence the lower limit of the median class is 50

(3) The number of students, who have finished the race after 100 seconds are

= The number of students belonging to 100-150 interval + number of students belonging to 150-200 interval

= 7+8 = 15

Hence,

The number of students, who have finished the race after 100 seconds  =  15

Answer 2

Answer:

The mode of the given data is - 69.23
The lower limit of the median class is 50
15 students finished the race after 100 seconds.

Step-by-step explanation:

Given - Class intervals 0-50, 50-100, 100-150, 150-200
            Frequency 10, 15, 7, 8
To find - Mode
              Lower limit of median class
              Frequency of the last 2 class intervals

Formula - $Mode = L + h\frac{f_m-f_1}{(f_m-f_1) + (f_m-f_2)}$

Solution -

$modal\;class - 50-100,\\\f_m = 15$

$f_1 = 10\\f_2 = 7$

$L = 50\\h = 40$

Thus,

 $Mode = 50 + 50\frac{15-10}{(15-10) + (15-7)}\\Mode = 50 + 50\frac{5}{5 + 8}\\Mode = 50 + 50\frac{5}{13}\\Mode = 50 + 50\frac{5}{13}\\Mode = 50+19.23\\\Mode = 69.23$

Now, to find the median class, we find the cumulative frequency -

$cf_1 = 10\\cf_2 = 10 + 15 = 25\\cf_3 = 25 + 7 = 32\\cf_4 = 32 + 8 = 40$

The median class is the one with a cumulative frequency greater than half the total frequency, and closest to it.                    

Here, $\frac{N}{2} = 20$
Thus, the median class is 50-100.

Next, to find the number of students who finished the race after 100 seconds, we add up the respective class intervals' frequencies -
$7 + 8 = 15$

Thus, 15 students finished the race after 100 seconds.

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