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प्रश्न
The mean and variance of seven observations are 8 and 16 respectively. If five of these are 2, 4, 10, 12 and 14, then find the remaining two observations.
उत्तर
Let the missing two observations be ‘a’ and ‘b’
Arithmetic mean = 8
| xi | xi2 |
| 2 | 4 |
| 4 | 16 |
| 10 | 100 |
| 12 | 144 |
| 14 | 196 |
| a | a2 |
| b | b2 |
| `sumx_"i"` = 42 + a + b | `sumx_"i"^2` = 460 + a2 + b2 |
`(2 + 4 + 10 + 12 + 14 + "a" + "b")/7` = 8
⇒ `(42 + "a" + "b")/7` = 8
a + b + 42 = 56
a + b = 56 − 42
a + b = 14 ...(1)
variance = 16
variance = `(sumx_"i"^2)/"n" - ((sumx_"i")/"n")^2`
16 = `(460 + "a"^2 + "b"^2)/7 - (8^2)`
⇒ 16 = `(460 + "a"^2 + "b"^2)/7 - 64`
16 + 64 = `(460 + "a"^2 + "b"^2)/7`
⇒ 80 = `(460 + "a"^2 + "b"^2)/7`
560 − 460 = a2 + b2
a2 + b2 = 100
⇒ (a + b)2 − 2ab = 100 ...[a2 + b2 = (a + b)2 − 2ab]
142 − 2ab = 100
⇒ 196 − 2ab = 100 ...[a + b = 14 (from (1)]
196 − 100 = 2ab
96 = 2ab
⇒ ab = `96/2` = 48
∴ b = `48/"a"` ...(2)
Substitute the value of b = `48/"a"` in (1)
`"a" + 48/"a"` = 14
⇒ a2 + 48 = 14a
a2 − 14a + 48 = 0
⇒ (a − 6) (a − 8) = 0
a = 6 or 8
|
When a = 6 b = `48/"a"` = `48/6` = 8 |
When a = 8 b = `48/"a"` = `48/8` = 6 |
∴ Missing observation is 8 and 6 or 6 and 8
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