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The mean of the following frequency data is 42, Find the missing frequencies x and y if the sum of frequencies is 100
|
Class interval |
0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
| Frequency | 7 | 10 | x | 13 | y | 10 | 14 | 9 |
Find x and y.
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The given data is shown as follows:
| Class interval | Frequency `(f_i)` | Class mark `(x_i)` | `f_ix_i` |
| 0-10 | 7 | 5 | 35 |
| 10-20 | 10 | 15 | 150 |
| 20-30 | x | 25 | 25x |
| 30-40 | 13 | 35 | 455 |
| 40-50 | y | 45 | 45y |
| 50-60 | 10 | 55 | 550 |
| 60-70 | 14 | 65 | 910 |
| 70-80 | 9 | 75 | 675 |
| Total | `sum f_i = 63 + x+y` | `sum f_i x_i = 2775 + 25x+45y` |
Sum of the frequencies = 100
⇒ ΣЁЭСЦ ЁЭСУЁЭСЦ = 100
⇒ 63 + x + y = 100
⇒ x + y = 100 – 63
⇒ x + y = 37
⇒ y = 37 – x ……..(1)
Now, the mean of the given data is given by
x = ` (sum _(i) f_i x_i )/(sum_(i) f_i )`
⇒ 42 =`(2775+25x+45y)/100`
⇒ 4200 = 2775 + 25x + 45y
⇒ 4200 – 2775 = 25x + 45y
⇒ 1425 = 25x + 45(37 – x) [from (1)]
⇒ 1425 = 25x + 1665 – 45x
⇒ 20x = 1665 – 1425
⇒ 20x = 240
⇒ x = 12
If x = 12, then y = 37 – 12 = 25
Thus, the value of x is 12 and y is 25.
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