Statistics

Two data sets can have the same mean yet look completely different if one set of values is tightly bunched together and the other is widely scattered. Measures of dispersion tell us how spread out the observations are around a central value. The simplest of these is the range.

The range of a data set is the difference between the largest (maximum) and the smallest (minimum) observation:

Range = X_max − X_min

A larger range indicates greater variability, while a smaller range indicates that the data is more closely clustered. However, the range uses only the two extreme values and ignores how the remaining observations are distributed, so it is considered a crude (rough) measure of dispersion.

Worked Example. The marks obtained by six students in a test are: 25, 18, 32, 40, 22, 15. Find the range.

Arranging in order: 15, 18, 22, 25, 32, 40. Here X_max = 40 and X_min = 15.

Range = 40 − 15 = 25.

So the marks are spread over a range of 25.

The mean deviation measures the average amount by which observations differ from a fixed central value (the mean or the median), using absolute values so that positive and negative deviations do not cancel out.

For n ungrouped observations x₁, x₂, …, x_n:

Mean deviation about the mean: M.D.(x̄) = (1/n) Σ|x_i − x̄|, where x̄ is the mean of the data.

Mean deviation about the median: M.D.(M) = (1/n) Σ|x_i − M|, where M is the median (for n odd, the middle value after arranging in order; for n even, the average of the two middle values).

Worked Example. Find the mean deviation about the mean and about the median for the data: 4, 6, 7, 10, 12, 12, 13, 16, 19.

Mean deviation about the mean. Here n = 9 and the sum of observations is 4+6+7+10+12+12+13+16+19 = 99, so x̄ = 99/9 = 11.

The absolute deviations |x_i − 11| are 7, 5, 4, 1, 1, 1, 2, 5, 8, whose sum is 34.

M.D.(x̄) = 34/9 ≈ 3.78.

Mean deviation about the median. The data is already arranged in ascending order, so with n = 9 (odd), the median is the middle (5th) observation, M = 12.

The absolute deviations |x_i − 12| are 8, 6, 5, 2, 0, 0, 1, 4, 7, whose sum is 33.

M.D.(M) = 33/9 ≈ 3.67.

Notice that the mean deviation about the median (3.67) is smaller than the mean deviation about the mean (3.78) — the median always gives the least possible value of mean deviation among all central values.

When data is arranged as a frequency distribution, each value x_i occurs with a frequency f_i, and N = Σf_i is the total number of observations. The mean deviation formulas change to weighted averages that use the frequencies.

For a discrete frequency distribution: Mean x̄ = (Σf_ix_i)/N; Mean deviation about mean = (1/N)Σf_i|x_i − x̄|; Mean deviation about median = (1/N)Σf_i|x_i − M|.

For a continuous (grouped) frequency distribution, x_i is taken as the mid-point of each class, and the median is located using M = l + [(N/2 − cf)/f] × h, where l is the lower limit of the median class, cf is the cumulative frequency of the class preceding the median class, f is the frequency of the median class, and h is the class width.

Worked Example. Find the mean deviation about the mean and about the median for the continuous distribution with classes 0–10, 10–20, 20–30, 30–40, 40–50 and frequencies 5, 8, 15, 16, 6 (N = 50).

Mean deviation about mean. Taking mid-points x_i as 5, 15, 25, 35, 45, Σf_ix_i = 25+120+375+560+270 = 1350, so x̄ = 1350/50 = 27.

The absolute deviations |x_i − 27| are 22, 12, 2, 8, 18; the weighted values f_i|x_i − 27| are 110, 96, 30, 128, 108, summing to 472.

M.D.(x̄) = 472/50 = 9.44.

Mean deviation about median. The cumulative frequencies are 5, 13, 28, 44, 50. Since N/2 = 25 first reaches the class 20–30 (cumulative frequency 28 there), this is the median class, with l = 20, cf = 13, f = 15, h = 10.

M = 20 + [(25 − 13)/15] × 10 = 20 + 8 = 28.

The absolute deviations |x_i − 28| are 23, 13, 3, 7, 17; the weighted values are 115, 104, 45, 112, 102, summing to 478.

M.D.(M) = 478/50 = 9.56.

Since mean deviation uses absolute values, it is not always algebraically convenient. Variance instead squares each deviation from the mean (removing the sign automatically) and averages the squares; the standard deviation is the positive square root of the variance, bringing the units back in line with the original data.

For n ungrouped observations with mean x̄:

Variance: σ² = (1/n) Σ(x_i − x̄)²

Standard deviation: σ = √[(1/n) Σ(x_i − x̄)²]

A useful shortcut (avoids computing every deviation separately) is: σ² = (1/n)Σx_i² − (x̄)².

Worked Example. Find the variance and standard deviation of the data: 6, 8, 10, 12, 14.

Here n = 5 and the mean is x̄ = (6+8+10+12+14)/5 = 50/5 = 10.

The deviations (x_i − x̄) are −4, −2, 0, 2, 4, and their squares are 16, 4, 0, 4, 16, which sum to 40.

Variance σ² = 40/5 = 8.

Standard deviation σ = √8 = 2√2 ≈ 2.83.

For grouped data, each squared deviation is weighted by its frequency f_i, with N = Σf_i:

Variance: σ² = (1/N) Σf_i(x_i − x̄)², and the standard deviation σ is its square root.

When the mid-points are large, it is easier to use the step-deviation method. Choose an assumed mean A (usually a middle class mid-point) and class width h, and let d_i = (x_i − A)/h. Then:

x̄ = A + h × (Σf_id_i/N)

σ² = h² × [ (Σf_id_i²/N) − (Σf_id_i/N)² ]

Worked Example. Using the step-deviation method, find the variance and standard deviation for the distribution with classes 0–10, 10–20, 20–30, 30–40, 40–50 and frequencies 5, 8, 15, 16, 6 (N = 50).

Take assumed mean A = 25 and h = 10, so mid-points 5, 15, 25, 35, 45 give d_i = −2, −1, 0, 1, 2.

f_id_i are −10, −8, 0, 16, 12, summing to Σf_id_i = 10.

f_id_i² are 20, 8, 0, 16, 24, summing to Σf_id_i² = 68.

Mean x̄ = 25 + 10 × (10/50) = 25 + 2 = 27.

Variance σ² = 10² × [68/50 − (10/50)²] = 100 × [1.36 − 0.04] = 132.

Standard deviation σ = √132 ≈ 11.49.

Standard deviation is an absolute measure of dispersion — it carries the same units as the data, so it cannot directly tell us which of two differently-scaled or differently-centred distributions is more variable. The coefficient of variation (C.V.) is a relative measure, expressed as a percentage, that allows a fair comparison:

C.V. = (σ/x̄) × 100, where x̄ ≠ 0.

A distribution with a smaller C.V. is said to be more consistent (less variable), while a distribution with a larger C.V. is more variable (less consistent).

Worked Example. The share prices of two firms show the following statistics: Firm A has mean price ₹60 and standard deviation ₹4; Firm B has mean price ₹50 and standard deviation ₹6. Which firm's share price is more consistent?

C.V. of Firm A = (4/60) × 100 ≈ 6.67%.

C.V. of Firm B = (6/50) × 100 = 12%.

Since the C.V. of Firm A (6.67%) is smaller than that of Firm B (12%), the share price of Firm A is more consistent (less variable) than that of Firm B.