Calculate arithmetic mean from the following data by Short-cut Method:

Class Interval 20−25 25−30 30−35 35−40 40−45 45−50 50−55

Frequency 10 12 8 20 11 4 5

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Solution

Class Interval Mid-Value
(m) Frequency
(f) Deviation
m − A
(d) fd

20 − 25
25 − 30
30 − 35
35 − 40
40 − 45
45 − 50
50 − 55 22.5
27.5
32.5
37.5 = A
42.5
47.5
52.5 10
12
8
20
11
4
5 −15
−10
−5
0
5
10
15 −150
−120
−40
0
55
40
75

∑f = 70 ∑fd = −140

Here, the assumed mean is decided from the mid-points. Let us take 37.5 as the assumed mean. So, we calculate the deviations of the mid-values from 37.5
$X = A + \frac{Σ f d}{Σ f} Here, A represents the assumed mean d represents the deviation of the mid - values from the assumed mean X = 37.5 + (\frac{- 140}{70}) or, X = 37.5 - 2 = 35.5$

Hence, the mean of the series is 35.5

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Similar questions

Calculation (i) Arithmetic Mean and (ii) Standard Deviation for the following frequency distribution.

Class Interval

Frequency

20 – 25

25 – 30

30 – 35

35 – 40

40 – 45

45 – 50

N = 50

Class Interval	10−20	20−30	30−40	40−50	50−60	60−70	70−80
Frequency	5	10	15	20	25	18	7

Class interval	20−25	25−30	30−35	35−40	40−45	45−50
Frequency	30	24	52	28	46	10

Age (in yrs)	20−25	25−30	30−35	35−40	40−45	45−50	50−55
No. of workers	17	11	8	5	4	3	2

Size	5−10	10−15	15−20	20−25	25−30	30−35	35−40	40−45	45−50
Frequency	6	10	18	30	15	12	10	6	4

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