Thirty students in a class went to cricket match. The results are recording in groups as follows $:$
Calculate the value mode for the foolowing data. (choose your answer to $2$ decimal places)
Over $1 - 5$ $5 - 9$ $9 - 13$ $13 - 17$ $17 - 21$ $21 - 25$ $25 - 29$
Runs $0$ $2$ $4$ $10$ $3$ $11$ $0$

21.84

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20.89

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22.684

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22.68

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Solution

The correct option is A $21.84$

Over

Run

1-5

0

5-9

2

9-13

4

13-17

10

17-21

3

21-25

11

25-29

0

$M o d e = l + (\frac{f_{1} - f_{0}}{{2 f}_{1} - f_{0 -} f_{2}}) \times h l = l o w e r l i m i t o f m o d a l c l a s s h = s i z e o f c l a s s i n t e r v a l f_{1} = f r e q u e n c y o f t h e m o d a l c l a s s f_{0} = f r e q u e n c y o f t h e c l a s s p r e c e e d i n g t h e m o d a l c l a s s f_{2} = f r e q u e n c y o f t h e c l a s s s u c c e e d i n g t h e m o d a l c l a s s M a x i m u m F r e q u e n c y = 11 M o d a l c l a s s = 21 - 25 l = 21, f_{1} = 11 f_{0} = 3 f_{2} = 0, h = 2 M o d e = 21 + (\frac{11 - 3}{22 - 3 - 0}) \times 2 = 21 + (\frac{8}{19}) \times 2 = 21 + \frac{16}{19} = 21 + 0.84 = 21.84 A n s$

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Similar questions

:

X	$5 - 9$	$9 - 13$	$13 - 17$	$17 - 21$
F	$3$	$8$	$9$	$4$

18

19 - 21

Class	$11 - 13$	$13 - 15$	$15 - 17$	$17 - 19$	$19 - 21$	$21 - 23$	$23 - 25$
Frequency	$3$	$6$	$9$	$13$	$f$	$5$	$4$

5, 9, 13, 17, 21, 25, 29, . . .

18

f

Colour Pencils	$0 - 5$	$5 - 10$	$10 - 15$	$15 - 20$	$20 - 25$	$25 - 30$
Number of students	$14$	$16$	$17$	$13$	$8$	$12$

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Over	$1 - 5$	$5 - 9$	$9 - 13$	$13 - 17$	$17 - 21$	$21 - 25$	$25 - 29$
Runs	$0$	$2$	$4$	$10$	$3$	$11$	$0$