Convert the inclusive classes to continuous class boundaries, find the median class and median via the median formula, then compute $\text{M.D.} = \dfrac{\sum f_i|x_i-\text{median}|}{N}$ using class midpoints. (Alternative: use the step-deviation method to find mean and variance.)

Mean deviation about the median:

Since the classes are printed as $16$–$20$, $21$–$25$, etc. (a gap of 1 between classes), we first convert to continuous class boundaries by subtracting/adding $0.5$:

$N=100$, so $N/2=50$. The cumulative frequency first exceeds 50 in the class 35.5–40.5 (cf jumps from 37 to 63), so this is the median class: $l=35.5$, $cf=37$ (cumulative freq. before the median class), $f=26$, $h=5$.

$\text{Median} = l + \frac{\frac N2 - cf}{f}\times h = 35.5 + \frac{50-37}{26}\times5 = 35.5+\frac{13}{26}\times5 = 35.5+2.5 = 38$

Now compute $|x_i - 38|$ and $f_i|x_i-38|$ for each class:

| $x_i$ | $|x_i-38|$ | $f_i$ | $f_i|x_i-38|$ |

|---|---|---|---|

| 18 | 20 | 5 | 100 |

| 23 | 15 | 6 | 90 |

| 28 | 10 | 12 | 120 |

| 33 | 5 | 14 | 70 |

| 38 | 0 | 26 | 0 |

| 43 | 5 | 12 | 60 |

| 48 | 10 | 16 | 160 |

| 53 | 15 | 9 | 135 |

$\sum f_i|x_i-38| = 100+90+120+70+0+60+160+135 = 735$

$\text{M.D. about median} = \frac{\sum f_i|x_i-\text{median}|}{N} = \frac{735}{100} = 7.35$

Or — Mean and variance of the frequency distribution (classes 0–10,…,40–50; frequencies 5,8,15,16,6):

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