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प्रश्न
Find the mean, median and mode of the given data:
| Class | 65 – 85 | 85 – 105 | 105 – 125 | 125 – 145 | 145 – 165 | 165 – 185 | 185 –205 |
| Frequency | 8 | 7 | 22 | 17 | 13 | 5 | 3 |
उत्तर
| Class `(f_i)` |
Frequency `(x_i)` |
Class mark | `d_i = x_i - A` `h = 20` |
`u_i = d_i/h` | `f_iu_i` |
| 65 – 85 | 8 | 75 | – 60 | – 3 | – 24 |
| 85 – 105 | 7 | 95 | – 40 | – 2 | – 14 |
| 105 –125 | 22 | 115 | – 20 | – 1 | – 22 |
| 125 – 145 | 17 | 135 = A | 0 | 0 | 0 |
| 145 –165 | 13 | 155 | 20 | 1 | 13 |
| 165 – 185 | 5 | 175 | 40 | 2 | 10 |
| 185 –205 | 3 | 195 | 60 | 3 | 9 |
| `sumf_i = 75` | `sumf_i u_i = - 28` |
We know, Mean `bar"X" = "A" + ((sumf_i u_i)/(sumf_i)) xx h`
= `135 + ((-28)/75) xx 20`
= `135 + ((-28)/15) xx 4`
= 135 – 7.47
= 127.53
Thus, the mean of the data is 127.53.
Now, the maximum frequency is 22 belonging to class interval 105 – 125.
∴ Modal class = 105 – 125
So, L = 105, f1 = 22, f0 = 7, f2 = 17, h = 20.
Mode = `"L" + ((f_1 - f_0)/(2f_1 - f_0 - f_2)) xx h"`
= `105 + ((22 - 7)/(2(22) - 7 - 17)) xx 20`
= `105 + (15/(44 - 24)) xx 20`
= `105 + (15/20) xx 20`
= 105 + 15
= 120
Thus, the mode is 120.
| Class | Frequency `(f_i)` | Cumulative frequency |
| 65 – 85 | 8 | 8 |
| 85 – 105 | 7 | 8 + 7 = 15 |
| 105 –125 | 22 | 15 + 22 = 37 |
| 125 – 145 | 17 | 37 + 17 = 54 |
| 145 –165 | 13 | 54 + 13 = 67 |
| 165 – 185 | 5 | 67 + 5 = 72 |
| 185 –205 | 3 | 72 + 3 = 75 |
Here, N = 75
∴ `(N/2)^("th") "term" = 75/3 = (37.5)^("th") "term"`
Since, 37.5th term lies in the class interval 125 – 145.
∴ Median class =125 – 145
`"So", L = 125, f = 17, N/2 = 37.5, c.f. = 37, h = 20.`
Median = `L + (("N"/2 - c.f.)/f) xx h`
= `125 + ((37.5 - 37)/17) xx 20`
= `125 + (0.5/17) xx 20`
= `125 + (5/17) xx 2`
= 125 + 0.588
= 125.588
As a result, the median is 125.588.
As a result, the data's mean, median, and mode are 127.53, 125.588, and 120, respectively.
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