The mode of the following series is 36. Find the missing frequency in it
Class interval 0-10 10-20 20-30 30-40 40-50 50-60 60-70
Frequency 8 10 ...... 16 12 6 7

10

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15

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16

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12

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Solution

The correct option is A $10$
We know that, Mode $= l + \frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}} \times h$
Where $l$ is lower limit of the modal class, $h$ is size of the class interval, $f_{1} =$ frequency of the modal class, $f_{0} =$ frequency of the class preceding the modal class, $f_{2} =$ frequency of the class succeeding the modal class
Since the class $30 - 40$ has the highest frequency $16$ , it is the modal class.
Now, Mode $= 30 + \frac{16 - f}{32 - f - 12} \times 10 = 36$
$⟹ 30 + \frac{16 - f}{20 - f} \times 10 = 36$
$⟹ \frac{16 - f}{20 - f} \times 10 = 6$
$⟹ 160 - 10 f = 120 - 6 f$
$⟹ 40 = 4 f$
$⟹ f = 10$

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Similar questions

χ

CLASS	0-10	10-20	20-30	30-40	40-50	50-60	60-70
FREQUENCY	8	10	$χ$	16	12	6	7

42

x

y

100

Class interval	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	7	10	x	13	y	10	14	9

Class interval	0-10	10-20	20-30	30-40	40-50	50-60	60-70
Frequency	2	3	x	6	5	3	2

Class Interval	0-10	10-20	20-30	30-40	40-50	50-60
Frequency	6	9	14	2	19	10

Class Interval	0-10	10-20	20-30	30-40	40-50	50-60
Frequency	12	20	26	18	10	6

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