Measures of Dispersion

How spread out is the data?

Dr. Arun Mitra

Department of Community Medicine, AIIMS Bibinagar

2026-10-27

Question

Two wards. Each has 10 women.

Both wards have a mean Hb of 11 g/dL.

Are the two wards the same? Which ward has the larger spread?

Guess the SD of Ward B: (a) 0.3 (b) 1 (c) 2.5 g/dL

Show hands. We check it in a few minutes.

Same mean, different wards

The mean is the same. The spread is not.

Why we need dispersion

  • A mean tells you the centre.
  • It does not tell you how far people sit from the centre.
  • Doctors care about the spread: BP, Hb, birth weight.

Dispersion = how much the values vary from each other or from the centre.

Today’s toolbox

Measure Use it with
Range Quick look
Interquartile range (IQR) Median, skewed data
Mean deviation, MAD, MSD Steps toward the SD
Variance and SD Mean, symmetric data
Normal curve and z-score Where a value sits (SD)
Coefficient of variation (CV) Compare different units

1. Range and IQR

Range

Range = largest − smallest

A different sample: Hb of 10 women (g/dL), mean 10.8.

8.2   9.1   9.8   10.2   10.6   11.0   11.4   11.9   12.5   13.3

Range = 13.3 − 8.2 = 5.1 g/dL

Range uses only two values. Change 13.3 to 18.0 (a lab error?) and the range becomes 9.8. It nearly doubles.

Quartiles and IQR

Split the sorted data into four equal parts.

  • Q1 = value with 25% below it
  • Q2 = median
  • Q3 = value with 75% below it

IQR = Q3 − Q1 (the spread of the middle 50%)

IQR worked example

8.2   9.1   9.8   10.2   10.6   |   11.0   11.4   11.9   12.5   13.3

  • Median = (10.6 + 11.0) / 2 = 10.8
  • Q1 = middle of lower half = 9.8
  • Q3 = middle of upper half = 11.9

IQR = 11.9 − 9.8 = 2.1 g/dL

. . .

With the outlier (18.0), Q3 stays 11.9. IQR stays 2.1.

We use the “median of each half” method. Software (R, SPSS) may give slightly different quartiles. The idea is the same.

Dot plot and boxplot

  • Box = Q1 to Q3. Line inside = median.
  • Whiskers reach the furthest values within 1.5 × IQR of the box. Dots lie beyond.

Trap: the box looks wide. Does it hold more patients? No. Each quarter of the data has the same number of patients. A wide box means more spread, not more people.

2. From deviations to the SD

Deviation from the mean

Days of fever in 5 children: 4, 4, 6, 8, 8. Mean = 6 days.

Deviation = value − mean.

Days (x) Deviation (x − 6)
4 −2
4 −2
6 0
8 2
8 2
Sum 0

Deviations always add up to zero. Keep this table. We use it four times.

1. Mean deviation

Mean deviation = (sum of deviations) ÷ n

= 0 ÷ 5 = 0

It is always zero. It tells us nothing. The plus and minus signs cancel.

Some textbooks use “mean deviation” for the next measure (with signs dropped). We keep the names apart here.

2. Mean absolute deviation (MAD)

Drop the minus signs. Then take the average.

Deviation Absolute deviation
−2 2
−2 2
0 0
2 2
2 2
8

MAD = 8 ÷ 5 = 1.6 days. On average, a child is 1.6 days from the mean.

Easy to understand. But absolute values are awkward in later statistics, so MAD is used less.

3. Mean squared deviation (MSD)

Square each deviation. Then take the average.

Deviation Squared deviation
−2 4
−2 4
0 0
2 4
2 4
16

MSD = 16 ÷ 5 = 3.2 days².

Squaring removes the minus signs. Bigger gaps count more. The unit is squared (days²), which is hard to picture.

4. Variance and SD

\[ s^2 = \frac{\sum (x - \bar{x})^2}{n-1} \qquad s = \sqrt{s^2} \]

Variance is the MSD with n − 1 instead of n.

  • Variance = 16 ÷ (5 − 1) = 4 days²
  • SD = √4 = 2 days (back to the original unit)

Why n − 1? Values sit closer to their own sample mean than to the true mean. So dividing by n underestimates the variance. Dividing by n − 1 corrects this.

The ladder

Same 5 children (4, 4, 6, 8, 8):

Measure Value Unit
Mean deviation 0 days
Mean absolute deviation 1.6 days
Mean squared deviation 3.2 days²
Variance 4 days²
Standard deviation 2 days

SD is the one we use most. It has the original unit. It works well with the normal curve.

Back to the two wards

Ward A Ward B
Mean Hb 11.0 11.0
SD 0.29 2.31
Lowest to highest 10.4 to 11.4 7.2 to 14.1
Range 1.0 6.9

Ward B has the wider spread. Did you guess 2.5? Its lowest woman has Hb 7.2.

Your turn: think first

Days of fever: 4, 4, 6, 8, 8 (SD = 2)

Step 1 (1 min, no calculator). Add 10 to every value. Does the SD change? Now double every value. Does it change?

Adding 10: SD stays 2 (the whole group just moves).
Doubling: SD doubles to 4.

Step 2 (next slides). Calculate an SD yourself, one step at a time.

Hands-on: SD in 6 steps

Diarrhoea episodes in 5 children in one year: 1, 6, 7, 7, 9

Work in pairs. Write each answer on paper. No calculator needed.

  1. Add the 5 values. You should see: 30.
  2. Mean = sum ÷ 5. You should see: 6.
  3. Deviation = each value − mean. You should see: −5, 0, 1, 1, 3.
  4. Square each deviation. You should see: 25, 0, 1, 1, 9.
  5. Variance = sum of squares ÷ (n − 1). You should see: 36 ÷ 4 = 9.
  6. SD = square root of the variance. You should see: 3.

Check: the deviations in step 3 must add up to 0. If not, go back to step 2.

Hands-on: the answer

Episodes (x) Mean x − mean (x − mean)²
1 6 −5 25
6 6 0 0
7 6 1 1
7 6 1 1
9 6 3 9
30 0 36

Variance = 36 ÷ 4 = 9 episodes². SD = √9 = 3 episodes.

Challenge (if time): find the MAD and the MSD for this data.
MAD = (5 + 0 + 1 + 1 + 3) ÷ 5 = 2. MSD = 36 ÷ 5 = 7.2.

3. SD and the normal curve

About 68 – 95 – 99.7

Mean ± 1 SD ≈ 68%. Mean ± 2 SD ≈ 95% (exactly 1.96 SD). Mean ± 3 SD ≈ 99.7%.

Check it on real data

Birth weight of 189 babies (MASS::birthwt, a US hospital, 1986).

Mean = 2945 g, SD = 729 g. Mean ± 2 SD = 1486 to 4403 g. About 96% of babies lie inside.

Z-score

How many SDs is a value from the mean?

\[ z = \frac{x - \bar{x}}{s} \]

A baby of 1500 g: z = (1500 − 2945) / 729 = −1.98.

About 2 SD below the mean. Only a few babies are this small.

The rule works for symmetric, bell-shaped data only. Next, a skewed example.

4. Coefficient of variation

Problem: different units

Who varies more: Hb or body weight?

Mean SD
Hb (g/dL) 12 1.5
Weight (kg) 60 9

You cannot compare 1.5 g/dL with 9 kg. The units differ.

CV

\[ CV = \frac{SD}{\text{Mean}} \times 100 \]

  • Hb: 1.5 / 12 × 100 = 12.5%
  • Weight: 9 / 60 × 100 = 15%

Weight varies more than Hb, relative to its mean. CV has no unit.

Try: days of fever, mean 6, SD 2. CV = 33%. Does fever length vary more or less than Hb (12.5%)?

5. Which one to use?

A skewed example

Weight of the same 189 mothers (kg), from MASS::birthwt.

A few heavy mothers pull the right tail.

Mean and SD, or median and IQR?

Mean (kg) Median (kg) SD (kg) IQR (kg)
58.9 54.9 13.9 13.6

Quartiles from R’s default method (Q1 49.9, Q3 63.5).

Mean ± 2 SD = 31.1 to 86.6 kg. Real weights go up to 113 kg and down to 36 kg. The lower limit is too low and the upper is too tight.

Mean and SD are pulled by the long tail. Median and IQR describe the typical mother.

Choose the measure

Data shape Centre Spread
Symmetric Mean SD
Skewed or outliers Median IQR
Compare different units n/a (a ratio) CV

Report the pair together: mean (SD) or median (IQR). CV needs a true zero and a mean well above zero. Do not use it for temperature in °C.

Check yourself

Q1

Which measure uses only two values?

Range.

Q2

Mean BP = 120 mmHg, SD = 10. About 95% of people lie between:

100 and 140 mmHg (mean ± 2 SD), if BP is roughly bell-shaped.

Q3

Which spread measure goes with the median?

IQR.

Q4

You want to compare variation in height (cm) and weight (kg). Use:

Coefficient of variation.

Q5

Can an SD be negative? Can it be zero?

Never negative. Zero only if every value is the same.

Take home

  1. A mean without a spread can mislead.
  2. Mean goes with SD. Median goes with IQR.
  3. About 95% of bell-shaped data lie within mean ± 2 SD.
  4. Use CV to compare different units.

Back to the question: Ward B has the larger spread (SD 2.31, not 0.3).

Questions?