Question

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.

Answer 1

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