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If the mean of the following data is 14.7, then the values of p and q are,

Classes
0-6
6-12
12-18
18-24
24-30
30-36
36-42
Total
Frequency
10
p
4
7
q
4
1
40

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11,3

12,8

10,2

11,5

answer is A.

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Detailed Solution

Given table is,

Classes	0-6	6-12	12-18	18-24	24-30	30-36	36-42	Total
Frequency	10	p	4	7	q	4	1	40

We know,
The mean is given by,

Mean = ∑ f i x i ∑ f i

Consider the table,

Classes	Class Mark ( $x_{i}$ )	Frequency ( $f_{i}$ )	$f_{i} x_{i}$
0-6	3	10	30
6-12	9	p	9p
12-18	15	4	60
18-24	21	7	147
24-30	27	q	27q
30-36	33	4	132
36-42	39	1	39
Total		∑ $f_{i} = 26 + p + q = 40$	∑ $f_{i} x_{i} = 408 + 9 p + 27 q$

The relation between p and q using total frequency,

26 + p + q = 40 ⇒ p + q = 40 − 26 ⇒ p + q = 14

......(i)

Mean = ∑ f i x i ∑ f i ⇒ M e a n = 408 + 9 p + 27 q 40

It is given that the mean is 14.7.
Therefore,

408 + 9 p + 27 q 40 = 14.7

⇒ 408 + 9 p + 27 q = 588 ⇒ 9 p + 27 q = 180 ⇒ p + 3 q = 20

......(ii)
Subtracting equation (i) from equation (ii), we get,

p + 3 q = 20 p + q = 14 (−) (−) (−) 2 q = 6 ⇒ q = 62 ⇒ q = 3

Putting the value q = 3 on equation (i), we get,

p + 3 = 14 ⇒ p = 14 − 3 ⇒ p = 11

Therefore, the required p = 11 and q = 3.
Hence, option 1 is correct.

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