To calculate the mean deviation about the mean for grouped data, we follow these steps:

1. Calculate the mid-point ($x_i$) for each income class.

The mid-point is the average of the upper and lower class boundaries.

$x_i = \frac{\text{Lower class limit} + \text{Upper class limit}}{2}$

For each income class:

• $80 - 100$: Mid-point = $\frac{80 + 100}{2} = 90$
• $100 - 120$: Mid-point = $\frac{100 + 120}{2} = 110$
• $120 - 140$: Mid-point = $\frac{120 + 140}{2} = 130$
• $140 - 160$: Mid-point = $\frac{140 + 160}{2} = 150$
• $160 - 180$: Mid-point = $\frac{160 + 180}{2} = 170$

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

The formula for the mean is:

$\bar{x} = \frac{\sum (f_i x_i)}{\sum f_i}$

Where:

• $f_i$ is the frequency (number of persons) in each class.
• $x_i$ is the mid-point of each class.

We multiply the mid-point by the frequency for each class:

(90 \times 5) & = 450 \\ (110 \times 6) & = 660 \\ (130 \times 2) & = 260 \\ (150 \times 3) & = 450 \\ (170 \times 4) & = 680 \\ \end{aligned}$$ Now sum these values: $$\sum (f_i x_i) = 450 + 660 + 260 + 450 + 680 = 2500$$ And sum the frequencies: $$\sum f_i = 5 + 6 + 2 + 3 + 4 = 20$$ Thus, the mean is: $$\bar{x} = \frac{2500}{20} = 125$$ ### 3. Calculate the deviations from the mean ($$|x_i - \bar{x}|$$). Now, we find the absolute deviations from the mean for each class: $$\begin{aligned} |90 - 125| & = 35 \\ |110 - 125| & = 15 \\ |130 - 125| & = 5 \\ |150 - 125| & = 25 \\ |170 - 125| & = 45 \\ \end{aligned}$$ ### 4. Calculate the mean deviation. The formula for the mean deviation is: $$\text{Mean deviation} = \frac{\sum (f_i |x_i - \bar{x}|)}{\sum f_i}$$ Now, multiply the deviations by the frequencies: $$\begin{aligned} (35 \times 5) & = 175 \\ (15 \times 6) & = 90 \\ (5 \times 2) & = 10 \\ (25 \times 3) & = 75 \\ (45 \times 4) & = 180 \\ \end{aligned}$$ Now sum these values: $$\sum (f_i |x_i - \bar{x}|) = 175 + 90 + 10 + 75 + 180 = 530$$ Finally, calculate the mean deviation: $$\text{Mean deviation} = \frac{530}{20} = 26.5$$ ### Conclusion: The mean deviation about the mean is **26.5**. Would you like any further details or have any questions? Here are 5 related questions: 1. How does mean deviation differ from standard deviation? 2. Can you calculate the variance and standard deviation for this data set? 3. What is the importance of mean deviation in data analysis? 4. How would the mean deviation change if the income class intervals were smaller? 5. What is the effect of outliers on mean deviation? **Tip:** In skewed distributions, the mean deviation is often smaller than the standard deviation, as it doesn't square the deviations.