Question

Let the mean and the variance of 5 observations x₁, x₂, x₃, x₄, x₅ be `24/5` and `194/25` respectively. If the mean and variance of the first 4 observation are `7/2` and a respectively, then (4a + x₅) is equal to ______.

पर्याय

• 13

• 15

• 17

• 18

Answer 1

Let the mean and the variance of 5 observations x₁, x₂, x₃, x₄, x₅ be `24/5` and `194/25` respectively. If the mean and variance of the first 4 observation are `7/2` and a respectively, then (4a + x₅) is equal to 15.

Explanation:

The observations are x₁, x₂, x₃, x₄, x₅  mean of x₁, x₂, x₃, x₄, x₅  = `24/5`

⇒ `(x_1 + x_2 + x_3 + x_4 + x_5)/5 = 24/5`

⇒ x₁ + x₂ + x₃ + x₄ + x₅ = 24  ...(i)

and mean of x₁, x₂, x₃, x₄ = `7/2`

⇒ `(x_1 + x_2 + x_3 + x_4)/4 = 7/2`

⇒ x₁ + x₂ + x₃ + x₄ = 14  ...(2)

From equation (1) and (2)

x₅ = 24 – 14 = 10

Variance of x₁, x₂, x₃, x₄, x₅ = `194/25`

⇒ `(x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2)/5 = 194/25 + (24/5)^2`

⇒ `x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 = 194/25 + 576/5` = 154

⇒ `x_1^2 + x_2^2 + x_3^2 + x_4^2` = 154 – (10)²

⇒ `x_1^2 + x_2^2 + x_3^2 + x_4^2` = 54

Variance of x₁, x₂, x₃, x₄ = a

⇒ `(x_1^2 + x_2^2 + x_3^2 + x_4^2)/4 - 49/4` = a

⇒ `54/4 - 49/4` = a

⇒ a = `5/4`

⇒ 4a + x₅ = 5 + 10 = 15