Question

Let the mean and variance of the observations \( 2, 3, 3, 4, 5, 7, a, b \) be 4 and 2, respectively. Then, the mean deviation about the mode of the observation is:

• A. 2
• B. 3
• C. 4
• D. 5

Hint:

To calculate the mean deviation about the mode, find the mode, then compute the average of the absolute deviations from the mode.

Answer 1

The Correct Option is B

We are given the following information: - The observations are \( 2, 3, 3, 4, 5, 7, a, b \), - The mean of the observations is 4, - The variance of the observations is 2. ### Step 1: Use the information about the mean The mean \( \mu \) is given by: \[ \mu = \frac{2 + 3 + 3 + 4 + 5 + 7 + a + b}{8} = 4 \] Thus, we have: \[ 2 + 3 + 3 + 4 + 5 + 7 + a + b = 32 \] \[ 24 + a + b = 32 \quad \Rightarrow \quad a + b = 8 \] ### Step 2: Use the information about the variance The variance \( \sigma^2 \) is given by: \[ \sigma^2 = \frac{(x_1 - \mu)^2 + (x_2 - \mu)^2 + \cdots + (x_8 - \mu)^2}{8} = 2 \] \[ \frac{(2 - 4)^2 + (3 - 4)^2 + (3 - 4)^2 + (4 - 4)^2 + (5 - 4)^2 + (7 - 4)^2 + (a - 4)^2 + (b - 4)^2}{8} = 2 \] Simplifying: \[ \frac{4 + 1 + 1 + 0 + 1 + 9 + (a - 4)^2 + (b - 4)^2}{8} = 2 \] \[ \frac{16 + (a - 4)^2 + (b - 4)^2}{8} = 2 \] \[ 16 + (a - 4)^2 + (b - 4)^2 = 16 \] Thus: \[ (a - 4)^2 + (b - 4)^2 = 0 \] This implies: \[ a - 4 = 0 \quad \text{and} \quad b - 4 = 0 \] Therefore, \( a = 4 \) and \( b = 4 \). ### Step 3: Determine the mode The mode of the observations is the most frequent value. The observations are now: \[ 2, 3, 3, 4, 4, 5, 7 \] The mode is \( 3 \) (since it appears twice). ### Step 4: Calculate the mean deviation about the mode The mean deviation about the mode is the average of the absolute deviations from the mode: \[ MD = \frac{|2 - 3| + |3 - 3| + |3 - 3| + |4 - 3| + |4 - 3| + |5 - 3| + |7 - 3|}{7} \] \[ MD = \frac{1 + 0 + 0 + 1 + 1 + 2 + 4}{7} = \frac{9}{7} \approx 3 \] Thus, the mean deviation about the mode is approximately 3. Therefore, the correct answer is (2) 3.