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प्रश्न
Find the mean, median and mode of the following data:
| Classes: | 0 – 50 | 50 – 100 | 100 – 150 | 150 – 200 | 200 – 250 | 250 – 300 | 300 – 350 |
| Frequency: | 2 | 3 | 5 | 6 | 5 | 3 | 1 |
बेरीज
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उत्तर
| Class interval |
Mid value()x | Frequency(fi) | fixi | Cumulative frequency |
| 0 – 50 | 25 | 2 | 50 | 2 |
| 50 – 100 | 75 | 3 | 225 | 5 |
| 100 – 150 | 125 | 5 | 625 | 10 |
| 150 – 200 | 175 | 6 | 1050 | 16 |
| 200 – 250 | 225 | 5 | 1125 | 21 |
| 250 – 300 | 275 | 3 | 825 | 24 |
| 300 – 350 | 325 | 1 | 325 | 25 |
| N = 25 | `sumfx=4225` |
Here, the maximum frequency is 6 so the modal class 150 − 200.
Therefore,
l = 150
h = 50
f = 6
f1 = 5
f2 = 5
F = 10
Mean `=(sumfx)/N=4225/25=169`
Thus, the mean of the data is 169.
We have N = 25 then N/2 = 12.5
Median `=l+(N/2-F)/fxxh`
`=150+(12.5-10)/6xx50`
`=150+2.5/6xx50`
`=150+125/6`
= 150 + 20.83
= 170.83
Thus, the median of the data is 170.83.
Mode `=l+(f-f1)/(2f-f1-f2)xxh`
`=150+(6-5)/(2xx6-5-5)xx50`
`=150+1/(12-10)xx50`
`=150+1/2xx50`
`=150+50/2`
= 150 + 25
= 175
Thus, the mode of the data is 175.
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