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In the following data the median of the runs scored by 60 top batsmen of the world in one-day international cricket matches is 5000. Find the missing frequencies x and y.
| Runs scored | 2500 – 3500 | 3500 – 4500 | 4500 – 5500 | 5500 – 6500 | 6500 – 7500 | 7500 - 8500 |
| Number of batsman | 5 | x | y | 12 | 6 | 2 |
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We prepare the cumulative frequency table, as shown below:
| Runs scored | Number of batsman `bb((f_i))` | Cumulative Frequency (cf) |
| 2500 – 3500 | 5 | 5 |
| 3500 – 4500 | x | 5 + x |
| 4500 – 5500 | y | 5 + x + y |
| 5500 – 6500 | 12 | 17 + x + y |
| 6500 – 7500 | 6 | 23 + x + y |
| 7500 – 8500 | 2 | 25 + x + y |
| Total | N = ΣЁЭСУЁЭСЦ = 60 |
Let x and y be the missing frequencies of class intervals 3500 – 4500 respectively. Then,
25 + x + y = 60 ⇒ x + y = 35 ……(1)
Median is 5000, which lies in 4500 – 5500.
So, the median class is 4500 – 5500.
∴ l = 4500, h = 1000, N = 60, f = y and cf = 5 + x
Now,
Median, `"M" = "i" + (("N"/2−"cf")/"f") × "h"`
`⇒ 5000 = 4500 + ((60/2 −(5+x))/ "y")× 1000`
`⇒ 5000 - 4500 = ((30−5−x)/"y") × 1000`
`⇒ 500 = ((25−x)/"y") × 1000`
⇒ y = 50 – 2x
⇒ 35 – x = 50 – 2x [From (1)]
⇒ 2x – x = 50 – 35
⇒ x = 15
∴ y = 35 – x
⇒ y = 35 – 15
⇒ y = 20
Hence, x = 15 and y = 20.
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