Question

The mean and standard deviation of 50 observations are 15 and 2 respectively. It was found that one incorrect observation was taken such that the sum of correct and incorrect observations is 70. If the correct mean is 16, then the correct variance is equal to ______.

Options

• 10

• 36

• 43

• 60

Answer 1

The mean and standard deviation of 50 observations are 15 and 2 respectively. It was found that one incorrect observation was taken such that the sum of correct and incorrect observations is 70. If the correct mean is 16, then the correct variance is equal to 43.

Explanation:

Number of observations are = 50

Mean `(barx)` = 15

Standard deviation (σ) = 2

Let incorrect observation is x₁ and correct observation is (y₁)

Given x₁ + y₁ = 70

∵ `barx = (x_1 + x_2 + ... + x_50)/50` = 15 ...(Given)

⇒ x₁ + x₂ + ...... x₅₀ = 750  ...(i)

Now

Mean of correct observation is 16

`(y_1 + x_2 + ... + x_50)/50` = 16

y₁ + x₂ + x₃ + .... x₅₀ = 16 × 50  ...(ii)

Equation (ii) – Equation (i)

⇒ y₁ – x₁ = 16 × 50 – 15 × 50

y₁ – x₁ = 50 and x₁ – x₁ = 70

y₁ = 60

x₁ = 10

⇒ 4 = `(x_1^2 + x_2^2 + ... + x_50^2)/50 - 15^2` ...(iii)

⇒ σ² = `(y_1^2 + x_2^2 + x_50^2)/50 - 16^2`  ...(iv)

From (iii)

⇒ 4 = `(10)^2/50 + (x_2^2 + x_3^2 + .... + x_50^2)/50 - 225`

⇒ 4 = `2 - 225 + ((x_2^2 + x_3^2 + .... + x_50^2))/50`

⇒ 227 = `((x_2^2 + x_3^2 + .... + x_50^2))/50`

From (iv)

σ² = `(60)^2/50 + ((x_2^2 + x_3^2 + ... + x_50^2)/50) - (16)^2`

σ² = `(60 xx 60)/50 + 227 - 256`

σ² = 72 + 227 – 256

σ² = 43