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प्रश्न
If the mode of the following distribution is 55, then find the value of x. Hence, find the mean.
| Class interval | 0 – 15 | 15 – 30 | 30 – 45 | 45 – 60 | 60 – 75 | 75 – 90 |
| Frequency | 10 | 7 | x | 15 | 10 | 12 |
बेरीज
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उत्तर
| Class interval | Frequency |
| 0 – 15 | 10 |
| 15 – 30 | 7 |
| 30 – 45 | x → f0 |
| 45 – 60 | 15 → f1 |
| 60 – 75 | 10 → f2 |
| 75 – 90 | 12 |
Mode of the frequency distribution = 55
∴ Modal class is 45 – 60.
Mode = `l + ((f_1 - f_0)/(2f_1 - f_0 - f_2)) xx h`
Where l = Lower limit of modal class = 45
h = Class interval = 15 – 0 = 15
f1 = Frequency of the modal class = 15
f0 = Frequency of the class before modal class = x
f2 = Frequency of the class after modal class = 10
Now, Mode = `l + ((f_1 - f_0)/(2f_1 - f_0 - f_2)) xx h`
`55 = 45 + (15 - x)/(2(15) - x - 10) xx 15`
`55 − 45 = (15 - x)/(30 - x - 10) xx 15`
`10 = (15 - x)/(20 - x) xx 15`
10(20 – x) = (15 – x) × 15
200 – 10x = 225 – 15x
15x – 10x = 225 – 200
5x = 25
x = `25/5`
x = 5
| Class interval | Frequency (fi) |
Class mark (xi) |
fixi |
| 0 – 15 | 10 | 7.5 | 75 |
| 15 – 30 | 7 | 22.5 | 157.5 |
| 30 – 45 | 5 | 37.5 | 187.5 |
| 45 – 60 | 15 | 52.5 | 787.5 |
| 60 – 75 | 10 | 67.5 | 675 |
| 75 – 90 | 12 | 82.5 | 990 |
| ∑fi = 59 | ∑fixi = 2872.5 |
Mean `(barx)` = `(sumf_ix_i)/(sumf_i)`
= `2872.5/59`
= 48.68
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