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प्रश्न
The median of the following data is 525. Find the values of x and y, if the total frequency is 100.
| Class interval | Frequency |
| 0 – 100 | 2 |
| 100 – 200 | 5 |
| 200 – 300 | x |
| 300 – 400 | 12 |
| 400 – 500 | 17 |
| 500 – 600 | 20 |
| 600 – 700 | y |
| 700 – 800 | 9 |
| 800 – 900 | 7 |
| 900 – 1000 | 4 |
उत्तर
| Class interval | Frequency | Cumulative Frequency |
| 0 – 100 | 2 | 2 |
| 100 –200 | 5 | 7 |
| 200 – 300 | x | 7 + x |
| 300 –400 | 12 | 19 + x |
| 400 – 500 | 17 | 36 + x |
| 500 – 600 | 20 | 56 + x |
| 600 – 700 | y | 56 + x + y |
| 700 – 800 | 9 | 65 + x + y |
| 800 – 900 | 7 | 72 + x + y |
| 900 – 1000 | 4 | 76 + x + y |
Median = 525, so Median Class = 500 – 600
76 + x + y = 100
⇒ x + y = 24 ......(i)
Median = `l + (n/2 - cf)/f xx h`
Since, l = 500, h = 100, f = 20, cf = 36 + x and n = 100
Therefore, putting the value in the Median formula, we get;
525 = `500 + (50 - (-36 + x))/20 xx 100`
So x = 9
y = 24 – x ......[From equation (i)]
y = 24 – 9 = 15
Therefore, the value of x = 9 and y = 15.
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