In earlier classes you summarised data with a single central value — mean, median or mode. But a central value alone is a half-told story: two groups can share the same average and still be utterly different, one tightly clustered and the other wildly scattered. Measures of dispersion answer the question the average ignores — how spread out are the values?
The dispersion measures of this chapter fall into a natural ladder, each one a refinement of the last:
The crudest measure is the range, the gap between the largest and smallest observation:
$$\text{Range} = \text{Maximum value} - \text{Minimum value}$$
Range is quick but fragile — it depends only on the two extreme values and ignores everything in between, so a single outlier can blow it up. For a continuous frequency distribution it is read as the upper limit of the highest class minus the lower limit of the lowest class. A more honest measure looks at how far every value sits from the centre. That is the idea behind mean deviation: average the distances of all observations from a central value $a$ (the mean or the median). Distances are taken as absolute values so that values below and above the centre do not cancel.
Mean deviation for ungrouped (raw) data about a central value $a$:
$$\text{M.D.}(a)=\dfrac{1}{n}\sum_{i=1}^{n}|x_i-a|$$
Taking $a=\bar{x}$ gives mean deviation about the mean; taking $a=M$ (the median) gives mean deviation about the median. The working is always the same three steps: find the centre, list the absolute deviations, then average them.
For a discrete frequency distribution with distinct values $x_i$ repeated $f_i$ times, where $N=\sum f_i$, each deviation is weighted by how often its value occurs:
$$\text{M.D.}(\bar{x})=\dfrac{1}{N}\sum f_i|x_i-\bar{x}|,\qquad \text{M.D.}(M)=\dfrac{1}{N}\sum f_i|x_i-M|$$
Here $\bar{x}=\dfrac{\sum f_i x_i}{N}$. To locate the median of a discrete distribution, build the cumulative frequency column and pick the value of $x$ whose cumulative frequency first reaches or exceeds $\dfrac{N}{2}$.
For a continuous (grouped) frequency distribution the same two formulae apply, but now $x_i$ is the class mark (midpoint) of each class, $x_i=\dfrac{\text{lower limit}+\text{upper limit}}{2}$. The median is found by the standard formula on the median class:
$$M=\ell+\dfrac{\frac{N}{2}-C}{f}\times h$$
where $\ell$ is the lower limit of the median class, $C$ the cumulative frequency just before it, $f$ its frequency and $h$ its width.
A fact worth remembering for fast checks: among all choices of the central value $a$, the median minimises the sum of absolute deviations, so M.D. about the median is never larger than M.D. about the mean for the same data.
Deeper Insight — why absolute values, and why the median is special: The whole point of a deviation is to record distance, and distance has no sign — a value $4$ below the centre is just as "far away" as a value $4$ above it. If we summed the raw signed deviations $\sum(x_i-\bar{x})$ we would always get exactly zero, because the mean is the precise balance point where positive and negative deviations cancel. That zero tells us nothing about spread, so we strip the signs with $|\cdot|$ before averaging. The absolute value also explains a subtler property: the function $\sum|x_i-a|$ is smallest when $a$ is the median, not the mean. Intuitively, the median has as many observations on its left as on its right, so nudging $a$ in either direction lengthens as many distances as it shortens. This is exactly why mean deviation about the median is the natural "minimum total distance" measure, and it is the reason the NCERT singles it out.
Find the range of the data: $32, 41, 28, 54, 35, 26, 23, 33, 38, 40$.
Solution- Identify the largest value: maximum $= 54$.
- Identify the smallest value: minimum $= 23$.
- Range $= 54 - 23 = 31$.
Answer: Range $= 31$.
Find the mean deviation about the mean for the data: $6, 7, 10, 12, 13, 4, 8, 12$.
Solution- There are $n=8$ values. Mean $\bar{x}=\dfrac{6+7+10+12+13+4+8+12}{8}=\dfrac{72}{8}=9$.
- Absolute deviations $|x_i-9|$: $3, 2, 1, 3, 4, 5, 1, 3$.
- Sum of absolute deviations $=3+2+1+3+4+5+1+3=22$.
- $\text{M.D.}(\bar{x})=\dfrac{1}{8}\times 22=2.75$.
Answer: M.D. about the mean $= 2.75$.
Find the mean deviation about the median for: $3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21$.
Solution- Arrange in order: $3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21$. Here $n=11$ (odd).
- Median $=$ the $\left(\dfrac{11+1}{2}\right)$th $=6$th value $=9$.
- Absolute deviations $|x_i-9|$: $6, 6, 5, 4, 2, 0, 1, 3, 9, 10, 12$.
- Sum $=6+6+5+4+2+0+1+3+9+10+12=58$.
- $\text{M.D.}(M)=\dfrac{1}{11}\times 58=\dfrac{58}{11}\approx 5.27$.
Answer: M.D. about the median $=\dfrac{58}{11}\approx 5.27$.
Find the mean deviation about the mean for the discrete distribution: $x_i = 5, 10, 15, 20, 25$ with frequencies $f_i = 7, 4, 6, 3, 5$.
Solution- $N=\sum f_i=7+4+6+3+5=25$.
- $\sum f_i x_i = 5(7)+10(4)+15(6)+20(3)+25(5)=35+40+90+60+125=350$.
- Mean $\bar{x}=\dfrac{350}{25}=14$.
- Absolute deviations $|x_i-14|$: $9, 4, 1, 6, 11$.
- $\sum f_i|x_i-14| = 9(7)+4(4)+1(6)+6(3)+11(5)=63+16+6+18+55=158$.
- $\text{M.D.}(\bar{x})=\dfrac{158}{25}=6.32$.
Answer: M.D. about the mean $= 6.32$.
Find the mean deviation about the mean for the continuous distribution: classes $0\text{-}10, 10\text{-}20, 20\text{-}30, 30\text{-}40, 40\text{-}50$ with frequencies $5, 8, 15, 16, 6$.
Solution- Class marks $x_i$: $5, 15, 25, 35, 45$. Total $N=5+8+15+16+6=50$.
- $\sum f_i x_i = 5(5)+15(8)+25(15)+35(16)+45(6)=25+120+375+560+270=1350$.
- Mean $\bar{x}=\dfrac{1350}{50}=27$.
- Absolute deviations $|x_i-27|$: $22, 12, 2, 8, 18$.
- $\sum f_i|x_i-27| = 22(5)+12(8)+2(15)+8(16)+18(6)=110+96+30+128+108=472$.
- $\text{M.D.}(\bar{x})=\dfrac{472}{50}=9.44$.
Answer: M.D. about the mean $= 9.44$.
For the data $4, 7, 8, 9, 10, 12, 13, 17$, find the mean deviation about the median.
Solution- The data is already ordered and $n=8$ (even).
- Median $=$ mean of the $4$th and $5$th values $=\dfrac{9+10}{2}=9.5$.
- Absolute deviations $|x_i-9.5|$: $5.5, 2.5, 1.5, 0.5, 0.5, 2.5, 3.5, 7.5$.
- Sum $=5.5+2.5+1.5+0.5+0.5+2.5+3.5+7.5=24$.
- $\text{M.D.}(M)=\dfrac{1}{8}\times 24=3$.
Answer: M.D. about the median $= 3$.
Find the mean deviation about the median for the discrete distribution: $x_i = 3, 6, 9, 12, 15$ with frequencies $f_i = 4, 5, 8, 5, 3$.
Solution- $N=\sum f_i=4+5+8+5+3=25$, so $\dfrac{N}{2}=12.5$.
- Cumulative frequencies: $4, 9, 17, 22, 25$. The first to reach $12.5$ is $17$, against $x=9$, so the median $M=9$.
- Absolute deviations $|x_i-9|$: $6, 3, 0, 3, 6$.
- $\sum f_i|x_i-9| = 6(4)+3(5)+0(8)+3(5)+6(3)=24+15+0+15+18=72$.
- $\text{M.D.}(M)=\dfrac{72}{25}=2.88$.
Answer: M.D. about the median $= 2.88$.
Find the mean deviation about the median for the continuous distribution: classes $0\text{-}10, 10\text{-}20, 20\text{-}30, 30\text{-}40, 40\text{-}50, 50\text{-}60$ with frequencies $6, 8, 14, 16, 4, 2$.
Solution- $N=6+8+14+16+4+2=50$, so $\dfrac{N}{2}=25$. Cumulative frequencies: $6, 14, 28, 44, 48, 50$; the median class is $20\text{-}30$ (where cf first passes $25$).
- Here $\ell=20$, $C=14$, $f=14$, $h=10$, so $M=20+\dfrac{25-14}{14}\times 10=20+\dfrac{110}{14}\approx 27.86$.
- Class marks $x_i$: $5, 15, 25, 35, 45, 55$; deviations $|x_i-27.86|$: $22.86, 12.86, 2.86, 7.14, 17.14, 27.14$.
- $\sum f_i|x_i-M| = 22.86(6)+12.86(8)+2.86(14)+7.14(16)+17.14(4)+27.14(2)$.
- $=137.16+102.88+40.04+114.24+68.56+54.28=517.16$.
- $\text{M.D.}(M)=\dfrac{517.16}{50}\approx 10.34$.
Answer: Median $\approx 27.86$ and M.D. about the median $\approx 10.34$.
The mean deviation about the mean of $4$ observations $a, a, a, a$ (all equal) is asked. What is it, and what does this tell you about dispersion?
Solution- Mean $\bar{x}=\dfrac{a+a+a+a}{4}=a$.
- Every absolute deviation $|x_i-a|=0$.
- $\text{M.D.}(\bar{x})=\dfrac{1}{4}(0)=0$.
Answer: M.D. $=0$. A dispersion measure is zero exactly when all values are identical — there is no spread at all.
For the data $38, 70, 48, 40, 42, 55, 63, 46, 54, 44$, find both the mean deviation about the mean and the range.
Solution- $n=10$. Sum $=38+70+48+40+42+55+63+46+54+44=500$, so $\bar{x}=\dfrac{500}{10}=50$.
- Absolute deviations $|x_i-50|$: $12, 20, 2, 10, 8, 5, 13, 4, 4, 6$.
- Sum $=12+20+2+10+8+5+13+4+4+6=84$, so $\text{M.D.}(\bar{x})=\dfrac{84}{10}=8.4$.
- Range $=$ max $-$ min $=70-38=32$.
Answer: M.D. about the mean $= 8.4$ and range $= 32$. Note the range, using only the two extremes, is far larger than the typical deviation.
Find the mean deviation about the mean for the discrete distribution: $x_i = 10, 30, 50, 70, 90$ with frequencies $f_i = 4, 24, 28, 16, 8$.
Solution- $N=4+24+28+16+8=80$.
- $\sum f_i x_i = 10(4)+30(24)+50(28)+70(16)+90(8)=40+720+1400+1120+720=4000$, so $\bar{x}=\dfrac{4000}{80}=50$.
- Absolute deviations $|x_i-50|$: $40, 20, 0, 20, 40$.
- $\sum f_i|x_i-50| = 40(4)+20(24)+0(28)+20(16)+40(8)=160+480+0+320+320=1280$.
- $\text{M.D.}(\bar{x})=\dfrac{1280}{80}=16$.
Answer: M.D. about the mean $= 16$.