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

Step 1: Understanding the Concept:
We need to find the missing two observations using the given mean and variance, then determine the median and calculate the mean deviation.
Step 2: Key Formula or Approach:
Mean \(\bar{x} = \frac{\sum x_i}{n}\).
Variance \(\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2\).
Mean Deviation about Median = \(\frac{\sum |x_i - M|}{n}\).
Step 3: Detailed Explanation:
Let the missing observations be \(x\) and \(y\).
Sum of 8 observations = \(2+3+5+10+11+13+15+21 = 80\).
Total sum = \(10 \times 9 = 90 \Rightarrow 80 + x + y = 90 \Rightarrow x + y = 10\).
Sum of squares of 8 observations = \(4+9+25+100+121+169+225+441 = 1094\).
Variance: \(34.2 = \frac{1094 + x^2 + y^2}{10} - 9^2\).
\(34.2 = \frac{1094 + x^2 + y^2}{10} - 81 \Rightarrow 115.2 = \frac{1094 + x^2 + y^2}{10}\).
\(1152 = 1094 + x^2 + y^2 \Rightarrow x^2 + y^2 = 58\).
Using \((x+y)^2 = x^2 + y^2 + 2xy\):
\(100 = 58 + 2xy \Rightarrow 2xy = 42 \Rightarrow xy = 21\).
The numbers are 3 and 7.
The 10 observations in ascending order: \(2, 3, 3, 5, 7, 10, 11, 13, 15, 21\).
Median \(M = \frac{7+10}{2} = 8.5\).
Mean Deviation = \(\frac{|2-8.5| + |3-8.5| + |3-8.5| + |5-8.5| + |7-8.5| + |10-8.5| + |11-8.5| + |13-8.5| + |15-8.5| + |21-8.5|}{10}\).
MD = \(\frac{6.5 + 5.5 + 5.5 + 3.5 + 1.5 + 1.5 + 2.5 + 4.5 + 6.5 + 12.5}{10} = \frac{50}{10} = 5\).
Step 4: Final Answer:
The mean deviation about the median is 5.