Calculate the mean deviation for the following grouped data: Marks (Class) 1-20 (f=8), 21-40 (f=14), 41-60 (f=8), 61-80 (f=48), 81-100 (f=12).

The midpoint is calculated as $\frac{\text{lower limit} + \text{upper limit}}{2}$.

• For 1-20: $x_1 = \frac{1+20}{2} = 10.5$
• For 21-40: $x_2 = \frac{21+40}{2} = 30.5$
• For 41-60: $x_3 = \frac{41+60}{2} = 50.5$
• For 61-80: $x_4 = \frac{61+80}{2} = 70.5$
• For 81-100: $x_5 = \frac{81+100}{2} = 90.5$

| Marks (Class) | Frequency ($f$) | Midpoint ($x$) | $fx$ | $|x - \bar{x}|$ | $f|x - \bar{x}|$ | | :------------ | :-------------- | :------------- | :--- | :-------------- | :--------------- | | 1-20 | 8 | 10.5 | 84 | | | | 21-40 | 14 | 30.5 | 427 | | | | 41-60 | 8 | 50.5 | 404 | | | | 61-80 | 48 | 70.5 | 3384 | | | | 81-100 | 12 | 90.5 | 1086 | | | | Total | $\sum f = 90$ | | $\sum fx = 5385$ | | |

The formula for the mean of grouped data is $\bar{x} = \frac{\sum fx}{\sum f}$.

We will use $\bar{x} \approx 59.83$ for calculations, keeping more precision in mind.

• $|10.5 - 59.8333| = |-49.3333| = 49.3333$
• $|30.5 - 59.8333| = |-29.3333| = 29.3333$
• $|50.5 - 59.8333| = |-9.3333| = 9.3333$
• $|70.5 - 59.8333| = |10.6667| = 10.6667$
• $|90.5 - 59.8333| = |30.6667| = 30.6667$

| Marks (Class) | Frequency ($f$) | Midpoint ($x$) | $fx$ | $|x - \bar{x}|$ | $f|x - \bar{x}|$ | | :------------ | :-------------- | :------------- | :--- | :-------------- | :--------------- | | 1-20 | 8 | 10.5 | 84 | 49.3333 | $8 \times 49.3333 = 394.6664$ | | 21-40 | 14 | 30.5 | 427 | 29.3333 | $14 \times 29.3333 = 410.6662$ | | 41-60 | 8 | 50.5 | 404 | 9.3333 | $8 \times 9.3333 = 74.6664$ | | 61-80 | 48 | 70.5 | 3384 | 10.6667 | $48 \times 10.6667 = 512.0016$ | | 81-100 | 12 | 90.5 | 1086 | 30.6667 | $12 \times 30.6667 = 368.0004$ | | Total | $\sum f = 90$ | | $\sum fx = 5385$ | | $\sum f|x - \bar{x}| = 1760.001$ |

The formula for mean deviation of grouped data is $MD = \frac{\sum f|x - \bar{x}|}{\sum f}$.

Rounding to two decimal places:

The mean deviation is 19.56.