Question

Let \( a, b \in \mathbb{R} \). Let the mean and the variance of 6 observations \(-3, 4, 7, -6, a, b\) be 2 and 23, respectively. The mean deviation about the mean of these 6 observations is:

• A. \( \frac{13}{3} \)
• B. \( \frac{16}{3} \)
• C. \( \frac{11}{3} \)
• D. \( \frac{14}{3} \)

Answer 1

The Correct Option is A

Set Up the Equations for Mean and Variance: 
Let the six observations be \(x_1 = -3\), \(x_2 = 4\), \(x_3 = 7\), \(x_4 = -6\), \(x_5 = a\), and \(x_6 = b\). Given that the mean of these observations is 2, 

we have: \[ \frac{-3 + 4 + 7 - 6 + a + b}{6} = 2 \] 
Simplifying, we get: \[ 2 + a + b = 12 \implies a + b = 10 \] 
 

Calculate the Variance: 
The variance of the observations is given as 23. We know that:
\[ \text{Variance} = \frac{\sum_{i=1}^6 x_i^2}{6} - \left(\frac{\sum_{i=1}^6 x_i}{6}\right)^2 \] 
Substitute the mean (2) and solve for the sum of squares:
\[ \frac{(-3)^2 + 4^2 + 7^2 + (-6)^2 + a^2 + b^2}{6} - 2^2 = 23 \] 
Calculating each term, we find:
\[ \frac{9 + 16 + 49 + 36 + a^2 + b^2}{6} - 4 = 23 \] 
Simplifying: \[ 110 + a^2 + b^2 = 162 \implies a^2 + b^2 = 52 \] 
Solve for \(a\) and \(b\): 
We now have two equations: \[ a + b = 10 \quad \text{and} \quad a^2 + b^2 = 52 \]
Using the identity \((a + b)^2 = a^2 + b^2 + 2ab\):
\[ 10^2 = 52 + 2ab \implies 100 = 52 + 2ab \implies ab = 24 \] 
Solving these equations, we find \(a = 4\) and \(b = 6\) (or vice versa). 

Calculate the Mean Deviation about the Mean: 
The mean deviation about the mean (2) is given by:
\[ \frac{|x_1 - 2| + |x_2 - 2| + |x_3 - 2| + |x_4 - 2| + |x_5 - 2| + |x_6 - 2|}{6} \]
Substitute the values \(x_1 = -3\), \(x_2 = 4\), \(x_3 = 7\), \(x_4 = -6\), \(x_5 = 4\), and \(x_6 = 6\):
\[ \frac{| -3 - 2| + |4 - 2| + |7 - 2| + |-6 - 2| + |4 - 2| + |6 - 2|}{6} = \frac{5 + 2 + 5 + 8 + 2 + 4}{6} = \frac{26}{6} = \frac{13}{3} \]