Mean Deviation About Mean for Frequency Distribution Data

To find the mean deviation about the mean for the given frequency distribution, we follow a step-by-step calculation.

Step 1: Calculate the total frequency ($N$)

The total frequency $N$ is given by: $N = \sum_{i=1}^{6} f_i = 8 + 6 + 2 + 2 + 2 + 6 = 26$

Step 2: Calculate the mean ($\bar{x}$)

First, compute $\sum_{i=1}^{6} f_i x_i$:

$$\begin{aligned} \sum_{i=1}^{6} f_i x_i &= (5 \times 8) + (7 \times 6) + (9 \times 2) + (10 \times 2) + (12 \times 2) + (15 \times 6) \\ &= 40 + 42 + 18 + 20 + 24 + 90 \\ &= 234 \end{aligned}$$

The mean $\bar{x}$ is: $\bar{x} = \frac{\sum f_i x_i}{N} = \frac{234}{26} = 9$

Step 3: Calculate the absolute deviations $|x_i - \bar{x}|$ and their weighted sum $\sum f_i |x_i - \bar{x}|$

Now, find the absolute deviation of each $x_i$ from the mean $\bar{x} = 9$:

• For $x_1 = 5$: $|5 - 9| = 4 \implies f_1 |x_1 - 9| = 8 \times 4 = 32$
• For $x_2 = 7$: $|7 - 9| = 2 \implies f_2 |x_2 - 9| = 6 \times 2 = 12$
• For $x_3 = 9$: $|9 - 9| = 0 \implies f_3 |x_3 - 9| = 2 \times 0 = 0$
• For $x_4 = 10$: $|10 - 9| = 1 \implies f_4 |x_4 - 9| = 2 \times 1 = 2$
• For $x_5 = 12$: $|12 - 9| = 3 \implies f_5 |x_5 - 9| = 2 \times 3 = 6$
• For $x_6 = 15$: $|15 - 9| = 6 \implies f_6 |x_6 - 9| = 6 \times 6 = 36$

Summing these up: $\sum_{i=1}^{6} f_i |x_i - \bar{x}| = 32 + 12 + 0 + 2 + 6 + 36 = 88$

Step 4: Calculate the Mean Deviation about Mean

The mean deviation about the mean is given by: $\text{Mean Deviation} = \frac{\sum_{i=1}^{6} f_i |x_i - \bar{x}|}{N} = \frac{88}{26} = \frac{44}{13}$

Thus, the correct option is C.