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प्रश्न
Find the mean, median and mode of the following data.
| Class | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 80 | 80 – 100 | 100 – 120 | 120 – 140 |
| Frequency | 6 | 8 | 10 | 12 | 6 | 5 | 3 |
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उत्तर
To find the mean, let us put the data in the table given below:
| Class | Frequency `(f_i)` | Class mark `(x_i)` | `f_i x_i` |
| 0 – 20 | 6 | 10 | 60 |
| 20 – 40 | 8 | 30 | 240 |
| 40 – 60 | 10 | 50 | 500 |
| 60 – 80 | 12 | 70 | 840 |
| 80 – 100 | 6 | 90 | 540 |
| 100 – 120 | 5 | 110 | 550 |
| 120 – 140 | 3 | 130 | 390 |
| Total | `Ʃ f_i` = 50 | `Ʃ f_i x_i` = 3120 |
Mean =`(Ʃ _i f_i x_i)/(Ʃ_i f_i)`
=`3120/50`
= 62.4
Thus, the mean of the given data is 62.4.
Now, to find the median, let us put the data in the table given below:
| Class | Frequency` (f_i)` | Cumulative Frequency (cf) |
| 0 – 20 | 6 | 6 |
| 20 – 40 | 8 | 14 |
| 40 – 60 | 10 | 24 |
| 60 – 80 | 12 | 36 |
| 80 – 100 | 6 | 42 |
| 100 – 120 | 5 | 47 |
| 120 – 140 | 3 | 50 |
| Total | `N = Σ f_i` = 50 |
Now, N = 50 ⇒`N/2 =25`
The cumulative frequency just greater than 25 is 36, and the corresponding class is 60 – 80.
Thus, the median class is 60 – 80.
∴ l = 60, h = 20, N = 50, f = 12 and cf = 24.
Now,
Median = l +`((N/2 - cf )/f) xx h`
=`60+ ((25-24)/12) xx 20`
= 60 + 1.67
= 61.67
Thus, the median is 61.67.
We know that,
The modal class is 60–80 (highest frequency = 12).
Using the formula,
Mode = `l + (f_1 - f_0)/(2f_1 - f_0 - f_2) xx h`
where: l = 60, f1 = 12, f0 = 10, f2 = 6, h = 20
= `60 + (12 - 10)/(2(12) - 10 - 6) xx 20`
= `60 + 2/8 xx 20`
= 60 + 5 = 65
Thus, the mode is 65.
