Central tendency, dispersion and probability distributions
Averages locate the centre; dispersion measures spread, and CV = σ/mean × 100 compares consistency. Binomial, Poisson, normal and exponential distributions each model a different chance process.
Measures of central tendency
| Measure | How to find it | Best use | Watch out |
|---|---|---|---|
| Arithmetic mean | Σx ÷ n | Most data; uses every value | Pulled by extreme values; Σ(x − x̄) = 0 |
| Median | Middle value of ordered data; average of the two middle values when n is even | Skewed data and open-end classes, such as incomes | Positional; not affected by extremes |
| Mode | Most frequent value | Nominal data, e.g. the best-selling shirt size | May not exist, or there may be more than one |
| Geometric mean | nth root of the product of n values | Average growth rates, ratios, index numbers | Not defined if any value is zero or negative |
| Harmonic mean | n ÷ Σ(1/x) | Average speed or rate over equal distances | Gives the most weight to small values |
Rules NET loves
- AM ≥ GM ≥ HM for positive values, equal only when all values are equal. For two values, GM² = AM × HM.
- Symmetric data: mean = median = mode. Positive skew: mean > median > mode. Negative skew: mean < median < mode.
- SD is unaffected by a change of origin (adding a constant) but not by a change of scale: multiplying every value by k multiplies σ by |k|.
- For a normal distribution, QD ≈ ⅔σ and MD ≈ ⅘σ, so QD : MD : SD ≈ 10 : 12 : 15.
- Absolute measures (range, QD, MD, SD) carry units. Coefficients and CV are pure numbers, so they compare series in different units.
Worked example 1: mean, median and mode
Daily orders at a Bengaluru cloud kitchen over 10 days: 6, 8, 8, 9, 10, 12, 13, 14, 19, 21. Find the mean, median and mode, and the direction of skew.
- Sum = 6 + 8 + 8 + 9 + 10 + 12 + 13 + 14 + 19 + 21 = 120. Mean = 120 ÷ 10 = 12.
- n = 10 is even, so the median is the average of the 5th and 6th values: (10 + 12) ÷ 2 = 11.
- 8 occurs twice and every other value once, so the mode = 8.
- Mean (12) > median (11) > mode (8): the data are positively skewed, with a long right tail from 19 and 21.
- The empirical rule gives only an estimate: 3 × 11 − 2 × 12 = 9, close to the actual mode of 8.
Mean 12, median 11, mode 8; positively skewed.
Measures of dispersion
| Measure | Formula | Note |
|---|---|---|
| Range | H − L; coefficient = (H − L) ÷ (H + L) | Quickest; uses only two values |
| Quartile deviation | (Q3 − Q1) ÷ 2; coefficient = (Q3 − Q1) ÷ (Q3 + Q1) | Semi-interquartile range; ignores the outer 50% |
| Mean deviation | Σ|x − A| ÷ n, about the mean or median | Smallest when taken about the median |
| Standard deviation | σ = √[Σ(x − x̄)² ÷ n] | Karl Pearson's term; root-mean-square deviation; never negative. This is the population SD; the sample SD (s), used in t tests, divides by n − 1 |
| Variance | σ² | For independent variables, the variance of a sum is the sum of the variances |
| Coefficient of variation | σ ÷ x̄ × 100 | Relative and unit-free; use it to compare consistency |
Worked example 2: SD and CV
Monthly sales (₹ lakh) of two HUL sales officers over five months. P: 14, 18, 20, 22, 26. Q: 26, 30, 32, 34, 38. Who is more consistent? Use the population SD (divide by n).
- Mean of P = 100 ÷ 5 = 20. Deviations: −6, −2, 0, 2, 6. Squares: 36, 4, 0, 4, 36; sum = 80.
- Variance of P = 80 ÷ 5 = 16, so σ = 4. CV of P = 4 ÷ 20 × 100 = 20%.
- Mean of Q = 160 ÷ 5 = 32. Deviations are again −6, −2, 0, 2, 6, so σ = 4.
- CV of Q = 4 ÷ 32 × 100 = 12.5%.
Both have σ = ₹4 lakh, but Q's CV (12.5%) is lower than P's (20%), so Q is more consistent. When means differ, compare CVs, not SDs.
Probability distributions at a glance
| Distribution | Type | Mean | Variance | Use when |
|---|---|---|---|---|
| Binomial (James Bernoulli) | Discrete | np | npq | A fixed number n of independent trials, each a success or failure, with constant p. Variance is less than the mean. |
| Poisson (S. D. Poisson, 1837) | Discrete | λ | λ | Rare events counted in a fixed interval: calls a minute, defects a sheet. Limit of the binomial when n is large, p is small and np = λ. |
| Normal (de Moivre, Gauss) | Continuous | μ | σ² | Measurements such as heights, weights and errors. Bell-shaped and symmetric; mean = median = mode; total area 1. |
| Exponential | Continuous | 1/λ | 1/λ² | Time between Poisson events, e.g. time between customer arrivals. Memoryless; its SD equals its mean. |
Probability formulas
Worked example 3: binomial
10% of LED bulbs from a line are defective. A buyer tests 4 bulbs at random. Find P(exactly 1 defective), P(none defective), and the mean and variance of the number defective.
- n = 4, p = 0.1, q = 0.9.
- P(X = 1) = 4C1 × 0.1 × 0.9³ = 4 × 0.1 × 0.729 = 0.2916.
- P(X = 0) = 0.9⁴ = 0.6561, so P(at least 1) = 1 − 0.6561 = 0.3439.
- Mean = np = 4 × 0.1 = 0.4. Variance = npq = 0.4 × 0.9 = 0.36.
P(1) = 0.2916; P(0) = 0.6561; mean 0.4, variance 0.36. As always in a binomial, the variance is less than the mean.
Worked example 4: Poisson
An SBI branch receives 2 complaints a day on average. Assuming a Poisson distribution, find P(no complaint), P(exactly 3) and P(at most 1) on a given day. Use e^−2 = 0.135335.
- λ = 2. P(0) = e^−2 × 2⁰ ÷ 0! = 0.1353.
- P(3) = e^−2 × 2³ ÷ 3! = 0.135335 × 8 ÷ 6 = 0.1804.
- P(1) = e^−2 × 2 = 0.2707. P(at most 1) = P(0) + P(1) = 0.135335 × 3 = 0.4060.
P(0) = 0.1353, P(3) = 0.1804 and P(X ≤ 1) = 0.4060, all rounded to 4 decimal places. Mean = variance = 2.
Worked example 5: normal distribution and z-scores
Monthly salaries in a firm are normal with mean ₹40,000 and SD ₹5,000. What share earn (a) more than ₹50,000, (b) between ₹35,000 and ₹45,000, (c) less than ₹47,500?
- (a) z = (50,000 − 40,000) ÷ 5,000 = 2. Area from 0 to 2 is 0.4772, so P(X > 50,000) = 0.5 − 0.4772 = 0.0228.
- (b) z = −1 and +1. Area within ±1σ = 2 × 0.34134 = 0.6827.
- (c) z = (47,500 − 40,000) ÷ 5,000 = 1.5. Area from 0 to 1.5 is 0.4332, so P(X < 47,500) = 0.5 + 0.4332 = 0.9332.
(a) 2.28%, (b) 68.27%, (c) 93.32%.
NET sets one-step calculations: the mode from the mean and median, the mean or CV from the SD, n and p from a binomial's mean and variance, or a Poisson probability. It also asks you to match each distribution with its mean and variance.
Typical MCQ: “A binomial distribution has mean 6 and SD 2. Find n and p.” Variance = 4, so q = 4 ÷ 6 = 2/3, p = 1/3 and n = 6 ÷ (1/3) = 18. Trap: using the SD (2) instead of the variance (4) as npq.
Typical MCQ: “A car covers two equal distances at 40 km/h and 60 km/h. Which average gives its average speed?” The harmonic mean: 2 ÷ (1/40 + 1/60) = 48 km/h. Trap: the arithmetic mean (50 km/h) overstates it.
Typical MCQ: “Series A has mean 50 and SD 10; series B has mean 80 and SD 12. Which is more consistent?” CV of A = 20%, CV of B = 15%, so B is more consistent. Trap: choosing A because its SD is smaller.
Poisson is the 'equal' one: mean = variance = λ. Binomial is the 'shy' one: its variance npq is always smaller than its mean np.