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

Answer: (A)

$\frac{13}{3}$

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 $x_6 = 6$:
$\frac{| -3 - 2| + |4 - 2| + |7 - 2| + |-6 - 2| + |4 - 2| + |6 - 2|}{6} = \frac{5 + 2 + 5 + 8 + 2 + 4}{6} = \frac{26}{6} = \frac{13}{3}$