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: \[ \frac{2^2 + 5^2 + 13^2 + 11^2 + 9^2}{5} - (8)^2 \] \[ = \frac{4 + 25 + 169 + 121 + 81}{5} - 64 \] \[ = \frac{400}{5} - 64 = 80 - 64 = 16 \]

Answer 2

To solve this problem, we need to analyze the given data about the observations and compute the variance after transformation. Let's break it down step-by-step:

Step 1: Understand the given information

• The mean of the five observations \( x_1, x_2, x_3, x_4, x_5 \) is 5.
• The variance of the five observations is 10.
• We know three of the observations: \( x_1 = 1 \), \( x_2 = 3 \), \( x_4 = 7 \).
• Two observations are in terms of unknowns: \( x_3 = a \) and \( x_5 = b \) with \( a > b \).

Step 2: Calculate the unknowns using mean and variance

• Mean: 

\[\frac{1 + 3 + a + 7 + b}{5} = 5\]

•  

\[\Rightarrow 11 + a + b = 25 \quad \Rightarrow a + b = 14\]

• Variance of original observations: 

\[\frac{(1-5)^2 + (3-5)^2 + (a-5)^2 + (7-5)^2 + (b-5)^2}{5} = 10\]

• Simplifying, 

\[\Rightarrow \frac{16 + 4 + (a-5)^2 + 4 + (b-5)^2}{5} = 10\]

•  

\[\Rightarrow (a-5)^2 + (b-5)^2 = 26\]

Step 3: Transform and calculate new variance

• New observations are \( n + x_n \) for \( n = 1, 2, 3, 4, 5 \).
• Modified observations:
  • \( n + x_n \) for \( x_1 = 1 \): \( 1 + 1 = 2 \)
  • \( n + x_n \) for \( x_2 = 3 \): \( 2 + 3 = 5 \)
  • \( n + x_n \) for \( x_3 = a \): \( 3 + a \)
  • \( n + x_n \) for \( x_4 = 7 \): \( 4 + 7 = 11 \)
  • \( n + x_n \) for \( x_5 = b \): \( 5 + b \)
• Mean of new observations: 

\[\frac{2 + 5 + (3+a) + 11 + (5+b)}{5} = \frac{26 + a + b}{5}\]

• Since \( a + b = 14 \), mean simplifies to: 

\[\frac{26 + 14}{5} = 8\]

• Variance formula: 

\[\frac{(2-8)^2 + (5-8)^2 + ((3+a)-8)^2 + (11-8)^2 + ((5+b)-8)^2}{5}\]

•  

\[= \frac{36 + 9 + (a-5)^2 + 9 + (b-5)^2}{5}\]

• Knowing \( (a-5)^2 + (b-5)^2 = 26 \): 

\[\frac{36 + 9 + 26 + 9}{5} = \frac{80}{5} = 16\]

Thus, the variance of the transformed observations is 16, making the correct option 16.