Question

If the mean deviation about the mean is \(m\) and the variance is \(\sigma^2\) for the following data, then \(m + \sigma^2 =\) % Data table \[ \begin{array}{|c|c|c|c|c|c|} \hline x & 1 & 3 & 5 & 7 & 9 \\ \hline f & 4 & 24 & 28 & 16 & 8 \\ \hline \end{array} \]

• A. \( 8 \)
• B. \( 7.2 \)
• C. \( \frac{28}{5} \)
• D. \( 6 \)

Hint:

- Remember to square the differences from the mean for variance, while taking absolute values for the mean deviation.
- Ensure all calculations are precise and check your work for errors in arithmetic.

Answer 1

The Correct Option is D

Step 1: Calculate the mean (\(\mu\)) of the data using the formula: \[ \mu = \frac{\sum (f_i x_i)}{\sum f_i}. \] - Substituting the values: \[ \mu = \frac{4\cdot1 + 24\cdot3 + 28\cdot5 + 16\cdot7 + 8\cdot9}{4 + 24 + 28 + 16 + 8} = \frac{4 + 72 + 140 + 112 + 72}{80} = \frac{400}{80} = 5. \] Step 2: Calculate the variance (\(\sigma^2\)) using the formula: \[ \sigma^2 = \frac{\sum f_i (x_i - \mu)^2}{\sum f_i}. \] - Substitute \(\mu = 5\) and compute \(\sigma^2\): \[ \sigma^2 = \frac{4(1 - 5)^2 + 24(3 - 5)^2 + 28(5 - 5)^2 + 16(7 - 5)^2 + 8(9 - 5)^2}{4 + 24 + 28 + 16 + 8}. \] - Compute each term: \[ \sigma^2 = \frac{4(16) + 24(4) + 28(0) + 16(4) + 8(16)}{80} = \frac{64 + 96 + 0 + 64 + 128}{80} = \frac{352}{80} = 4.4. \] Step 3: The mean deviation about the mean \(m\) is calculated using: \[ m = \frac{\sum f_i |x_i - \mu|}{\sum f_i}. \] - Substitute \(\mu = 5\) and compute \(m\): \[ m = \frac{4|1 - 5| + 24|3 - 5| + 28|5 - 5| + 16|7 - 5| + 8|9 - 5|}{4 + 24 + 28 + 16 + 8}. \] - Compute each term: \[ m = \frac{4(4) + 24(2) + 28(0) + 16(2) + 8(4)}{80} = \frac{16 + 48 + 0 + 32 + 32}{80} = \frac{128}{80} = 1.6. \] Step 4: Finally, compute \(m + \sigma^2\): \[ m + \sigma^2 = 1.6 + 4.4 = 6. \]