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Question
The following is the cumulative frequency distribution ( of less than type ) of 1000 persons each of age 20 years and above . Determine the mean age .
| Age below (in years): | 30 | 40 | 50 | 60 | 70 | 80 |
| Number of persons : | 100 | 220 | 350 | 750 | 950 | 1000 |
Solution
Let the assumed mean A = 45 and h = 10.
| Marks | Mid-Value(xi) | Frequency (fi) | `u_1 = (x_1 - 45)/10` | `f_iu_i` |
| 20–30 | 25 | 100 | -2 | -200 |
| 30–40 | 35 | 120 | -1 | -120 |
| 40–50 | 45 | 130 | 0 | -0 |
| 50–60 | 55 | 400 | 1 | 400 |
| 60–70 | 65 | 200 | 2 | 400 |
| 70–80 | 75 | 50 | 3 | 150 |
| N - 1000 | `sum"f"_"i""u"_"i" = 630` |
We know that mean, `barX = "A"+"h" (1/"N" sum"f"_"i""u"_"i")`
Now, we have
N = `sum`fi = 1000, h = 10, A = 45, `sum`fi ui = - 370\]
Mean = `bar"x"` = a + h `((sum"f"_"i""u"_"i")/(sum"f"_"i"))`
= 45 + 10`(630/1000)`
= 45 + 6.3
= 51.3 years
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