Question

Let the Mean and Variance of five observations $ x_i $, $ i = 1, 2, 3, 4, 5 $ be 5 and 10 respectively. If three observations are $ x_1 = 1, x_2 = 3, x_3 = a $ and $ x_4 = 7, x_5 = b $ with $ a>b $, then the Variance of the observations $ n + x_n $ for $ n = 1, 2, 3, 4, 5 $ is

• A. 17
• B. 16.4
• C. 17.4
• D. 16

Hint:

Use the formulas for mean and variance to set up equations based on the given information. Solve these equations to find the values of the unknown observations. Then, apply the transformation to the original observations to get the new set of observations and calculate their variance.

Answer 1

The Correct Option is D

Given, Mean \( \bar{x} = \frac{\sum x_i}{n} = \frac{1 + 3 + a + 7 + b}{5} = 5 \) \[ 11 + a + b = 25 \] \[ a + b = 14 \] Given, Variance \( \sigma^2 = \frac{\sum x_i^2}{n} - \left( \bar{x} \right)^2 = 10 \) \[ \frac{1^2 + 3^2 + a^2 + 7^2 + b^2}{5} - (5)^2 = 10 \] \[ \frac{1 + 9 + a^2 + 49 + b^2}{5} - 25 = 10 \] \[ 59 + a^2 + b^2 = 175 \] \[ a^2 + b^2 = 116 \] We are given: \[ a + b = 14 \quad \text{and} \quad a^2 + b^2 = 116 \] Using the identity: \[ (a + b)^2 = a^2 + b^2 + 2ab \] Substitute the known values: \[ 14^2 = 116 + 2ab \] \[ 196 = 116 + 2ab \] \[ 2ab = 80 \Rightarrow ab = 40 \] Now solve the system: \[ a + b = 14, \quad ab = 40 \] Form the quadratic: \[ t^2 - (a + b)t + ab = 0 \] \[ t^2 - 14t + 40 = 0 \] \[ (t - 10)(t - 4) = 0 \] So, \[ t = 10 \quad \text{or} \quad t = 4 \] Given \( a>b \), we have: \[ a = 10, \quad b = 4 \] The observations \( x_i \) are: \[ 1, 3, 10, 7, 4 \] The new observations \( n + x_n \) for \( n = 1 \) to \( 5 \) are: \[ 1 + x_1 = 2, \quad 2 + x_2 = 5, \quad 3 + x_3 = 13, \quad 4 + x_4 = 11, \quad 5 + x_5 = 9 \] New set: \[ 2, 5, 13, 11, 9 \] Mean: \[ \frac{2 + 5 + 13 + 11 + 9}{5} = \frac{40}{5} = 8 \] Variance: \[ \frac{2^2 + 5^2 + 13^2 + 11^2 + 9^2}{5} - (8)^2 \] \[ = \frac{4 + 25 + 169 + 121 + 81}{5} - 64 \] \[ = \frac{400}{5} - 64 = 80 - 64 = 16 \]