Question

The mean and variance of 10 observations are 9 and 34.2, respectively. If 8 of these observations are \( 2, 3, 5, 10, 11, 13, 15, 21 \), then the mean deviation about the median of all the 10 observations is:

• A. \(4\)
• B. \(6\)
• C. \(5\)
• D. \(7\)

Hint:

For problems involving missing observations, always use the mean to find the sum and variance to find squared deviations.

Answer 1

The Correct Option is A

Concept: Mean is the average of observations, variance is the mean of squared deviations from the mean, and mean deviation about the median is the average of absolute deviations from the median.
Step 1: Use the given mean to find the total sum Mean \( = 9 \), number of observations \( = 10 \) \[ \text{Total sum} = 10 \times 9 = 90 \] Sum of the given 8 observations: \[ 2 + 3 + 5 + 10 + 11 + 13 + 15 + 21 = 80 \] Let the remaining two observations be \( x \) and \( y \). \[ x + y = 90 - 80 = 10 \quad \cdots (1) \]
Step 2: Use the variance formula Given variance: \[ \sigma^2 = 34.2 \] \[ \sum (x_i - \mu)^2 = 10 \times 34.2 = 342 \] Now compute squared deviations of the given 8 observations from the mean 9: \[ (2-9)^2 + (3-9)^2 + (5-9)^2 + (10-9)^2 + (11-9)^2 + (13-9)^2 + (15-9)^2 + (21-9)^2 \] \[ = 49 + 36 + 16 + 1 + 4 + 16 + 36 + 144 = 302 \] Thus, \[ (x-9)^2 + (y-9)^2 = 342 - 302 = 40 \quad \cdots (2) \]
Step 3: Solve for \( x \) and \( y \) From (1), \( y = 10 - x \) Substitute in (2): \[ (x-9)^2 + (1-x)^2 = 40 \] \[ x^2 - 18x + 81 + x^2 - 2x + 1 = 40 \] \[ 2x^2 - 20x + 42 = 0 \] \[ x^2 - 10x + 21 = 0 \] \[ (x-3)(x-7) = 0 \] So, \[ x = 3,\; y = 7 \]
Step 4: Arrange all observations \[ 2, 3, 3, 5, 7, 10, 11, 13, 15, 21 \] Median (average of 5th and 6th terms): \[ \text{Median} = \frac{7 + 10}{2} = 8.5 \]
Step 5: Calculate mean deviation about the median \[ \sum |x_i - 8.5| = 6.5 + 5.5 + 5.5 + 3.5 + 1.5 + 1.5 + 2.5 + 4.5 + 6.5 + 12.5 = 40 \] \[ \text{Mean deviation about median} = \frac{40}{10} = 4 \]