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

To solve this problem, we need to find the mean deviation about the mean for the 6 observations given. Let's break down the solution in a step-by-step manner:

1. Understanding Mean and Variance: The formula for the mean (average) of observations is: 

\[\text{Mean} = \frac{\sum x_i}{n}\]

1.  Given, the mean of observations \(-3, 4, 7, -6, a, b\) is 2: 

\[\frac{-3 + 4 + 7 - 6 + a + b}{6} = 2\]

1.  Simplifying, we have: 

\[2 = \frac{2 + a + b}{6}\]

1.  Solving for \( a + b \): 

\[2 = \frac{2 + a + b}{6} \Rightarrow 12 = 2 + a + b \Rightarrow a + b = 10\]

1. Calculating Variance: The formula for variance is: 

\[\text{Variance} = \frac{\sum (x_i - \text{mean})^2}{n}\]

1.  Given that the variance is 23: 

\[\frac{(-3-2)^2 + (4-2)^2 + (7-2)^2 + (-6-2)^2 + (a-2)^2 + (b-2)^2}{6} = 23\]

1.  Simplifying individual terms: 

\[(x_i - 2)^2 = [(-3 - 2)^2 = 25, \, (4 - 2)^2 = 4, \, (7 - 2)^2 = 25,\, (-6 - 2)^2 = 64]\]

1.  Plug in values: 

\[\frac{25 + 4 + 25 + 64 + (a-2)^2 + (b-2)^2}{6} = 23\]

1.  

\[\Rightarrow 118 + (a-2)^2 + (b-2)^2 = 138\]

1.  

\[\Rightarrow (a-2)^2 + (b-2)^2 = 20\]

1. Solving Simultaneous Equations: Now, solve for \(a\) and \(b\) using: 

\[a + b = 10 \quad \text{(equation 1)}\]

1.  

\[(a-2)^2 + (b-2)^2 = 20 \quad \text{(equation 2)}\]

1.  Substitute \(b = 10 - a\) in equation 2: 

\[(a-2)^2 + ((10-a)-2)^2 = 20\]

1.  

\[(a-2)^2 + (8-a)^2 = 20\]

1.  Solving this: 

\[(a-2)^2 + (8-a)^2 = 20 \Rightarrow a^2 - 4a + 4 + 64 - 16a + a^2 = 20\]

1.  

\[2a^2 - 20a + 68 = 20\]

1.  

\[2a^2 - 20a + 48 = 0 \Rightarrow a^2 - 10a + 24 = 0\]

1.  Solving quadratic \(a^2 - 10a + 24 = 0\): 

\[a = \frac{10 \pm \sqrt{100 - 96}}{2} = \frac{10 \pm 2}{2}\]

1.  So, \(a = 6 \, \text{or} \, 4\). 
   If \(a = 6\), then \(b = 4\). If \(a = 4\), then \(b = 6\). Either way, \(a\) and \(b\) take the values 4 and 6.
2. Calculating Mean Deviation: Mean deviation about a mean is given by: 

\[\text{Mean Deviation} = \frac{\sum |x_i - \text{mean}|}{n}\]

1.  Compute mean deviation: 

\[\sum |x_i - 2| = |-3-2| + |4-2| + |7-2| + |-6-2| + |a-2| + |b-2|\]

1.  

\[|-3-2| + |4-2| + |7-2| + |-6-2| + |4-2| + |6-2|\]

1.  

\[|5| + |2| + |5| + |8| + |2| + |4| = 26\]

1.  

\[\text{Mean Deviation} = \frac{26}{6} = \frac{13}{3}\]

Answer 2

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 \(x_6 = 6\):
\[ \frac{| -3 - 2| + |4 - 2| + |7 - 2| + |-6 - 2| + |4 - 2| + |6 - 2|}{6} = \frac{5 + 2 + 5 + 8 + 2 + 4}{6} = \frac{26}{6} = \frac{13}{3} \]