Calculate arithmetic mean of the following series.
Take 15 as assumed average and use Short-cut Method.

Size 20 19 18 17 16 15 14 13 12 11

Frequency 1 2 4 8 11 10 7 4 2 1

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Solution

Size
(X) Frequency
(f) Deviation
d = X − A
(A = 15) Multiple of Deviation and Frequency
(fd)

20
19
18
17
16
15
14
13
12
11 1
2
4
8
11
10
7
4
2
1 5
4
3
2
1
0
−1
−2
−3
−4 5
8
12
16
11
0
−7
−8
−6
−4

Σf = 50 Σfd = 52 − 25 = 27

$Mean size (X) = A + \frac{Σ f d}{Σ f} or, \bar{X} = 15 + \frac{27}{50} or, \bar{X} = 15 + 0.54 \Rightarrow \bar{X} = 15.54$

So the average size is 15.54.

Suggest Corrections

Class Interval	11−13	13−15	15−17	17−19	19−21	21−23	23−25
Frequency	3	6	9	13	?	5	4

Daily wages (in ₹)	2−4	4−6	6−8	8−10	10−12	12−14	14−16	16−18
No. of Workers	11	14	20	32	25	7	5	2

Size	30	29	28	27	26	25	24	23	22	21
Frequency (f)	2	4	5	3	2	7	1	4	5	7

C.I	5-6	8-10	11-13	14-16	17-19
Frequency	10	8	7	3	2

Size (X)	Frequency (f)	Deviation d = X − A (A = 15)	Multiple of Deviation and Frequency (fd)
20 19 18 17 16 15 14 13 12 11	1 2 4 8 11 10 7 4 2 1	5 4 3 2 1 0 −1 −2 −3 −4	5 8 12 16 11 0 −7 −8 −6 −4
	Σf = 50		Σfd = 52 − 25 = 27

Weekly Wages	5	6	7	8	12	?	15	17	18
Frequency	3	10	5	10	12	15	11	13	1