Question

If x_i | f_i, i = 1, 2,...n is a frequency distribution with variance 2, mode 24 and arithmetic mean 25, then the mean square deviation from the mode is:
1. 3
2. 1
3. 4
4. 2

Answer 1

Correct Answer - Option 1 : 3

Given

Variance = 2

Mode = 24

Arithmetic mean = 25

Formula

Var(X) = (1/n)∑(x_i - x̅)²

Calculation

Var(X) =  (1/n)∑(xi - x̅)²

⇒ 2 = (1/n)∑(x_i - 25)²

⇒ 2n = ∑x_i² + 625n - 50∑x_i

⇒ 2n = ∑x_i² + 625n - 50(∑x_i/n) × n

⇒ 2n = ∑x_i² + 625n - 50n × 25       {∑x_i/n = mean = 25}

⇒ 2n = ∑x_i² - 625n

⇒ ∑x_i² = 627n

Mean square deviation = MSD = (1/n)∑(x_i - Z)²  here Z = Mode

⇒ MSD = (1/n)∑(x_i - 24)²

⇒ (1/n)∑(x_i² + 576n - 48x_i)

⇒ (1/n)∑x_i + 576n - 48 (∑x_i/n) × n)

⇒ (1/n)(627n + 576 - 48 × 25)

⇒ (1/n)( 627n - 624n)

⇒ (1/n)(3n)

∴ The value of mean squarer deviation from mode is 3