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

Given observations: \( 2, 3, 3, 4, 5, 7, a, b \). Mean (\( \mu \)) = 4. Variance (\( \sigma^2 \)) = 2. Step 1: Utilize the mean information The mean formula is \( \mu = \frac{\sum x_i}{n} \). \[ \mu = \frac{2 + 3 + 3 + 4 + 5 + 7 + a + b}{8} = 4 \] \[ 2 + 3 + 3 + 4 + 5 + 7 + a + b = 32 \] \[ 24 + a + b = 32 \] \[ a + b = 8 \] Step 2: Utilize the variance information The variance formula is \( \sigma^2 = \frac{\sum (x_i - \mu)^2}{n} \). \[ \sigma^2 = \frac{(2 - 4)^2 + (3 - 4)^2 + (3 - 4)^2 + (4 - 4)^2 + (5 - 4)^2 + (7 - 4)^2 + (a - 4)^2 + (b - 4)^2}{8} = 2 \] \[ \frac{(-2)^2 + (-1)^2 + (-1)^2 + 0^2 + 1^2 + 3^2 + (a - 4)^2 + (b - 4)^2}{8} = 2 \] \[ \frac{4 + 1 + 1 + 0 + 1 + 9 + (a - 4)^2 + (b - 4)^2}{8} = 2 \] \[ \frac{16 + (a - 4)^2 + (b - 4)^2}{8} = 2 \] \[ 16 + (a - 4)^2 + (b - 4)^2 = 16 \] \[ (a - 4)^2 + (b - 4)^2 = 0 \] This implies \( a - 4 = 0 \) and \( b - 4 = 0 \). Therefore, \( a = 4 \) and \( b = 4 \). Step 3: Determine the mode The complete set of observations is \( 2, 3, 3, 4, 4, 5, 7 \). The mode is the most frequent value, which is \( 3 \). Step 4: Calculate the mean deviation about the mode Mean Deviation (MD) is the average of absolute deviations from the mode. The observations are \( 2, 3, 3, 4, 4, 5, 7 \). The mode is 3. MD = \( \frac{|2 - 3| + |3 - 3| + |3 - 3| + |4 - 3| + |4 - 3| + |5 - 3| + |7 - 3|}{7} \) MD = \( \frac{1 + 0 + 0 + 1 + 1 + 2 + 4}{7} = \frac{9}{7} \) MD \( \approx 3 \) The mean deviation about the mode is approximately 3. The correct answer is (2) 3.