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Class 12 Statistics Notes · GSEB

Properties of Binomial Distribution

Properties of Binomial Distribution — master the mean, variance, standard deviation, and skewness patterns of the binomial distribution. GSEB Class 12 Commerce Statistics notes with visual explorer.

Last updated: 22 Sep 2026

Notes

Properties of Binomial Distribution

The binomial distribution B(n, p) has several important properties that help us understand its shape, centre, and spread. These are frequently tested in GSEB exams.

Mean, Variance, and Standard Deviation

Mean

μ=np\mu = np

Variance

σ2=npq\sigma^2 = npq

Standard Deviation

σ=npq\sigma = \sqrt{npq}

Mean > Variance always

Since q = 1 − p and 0 < q < 1 (for non-trivial cases), we have np > npq. The mean is always greater than the variance in a binomial distribution. This is a key property GSEB tests.

Real-life check

If 200 students take an exam and each has 70% chance of passing (n=200, p=0.7):
• Expected passes = np = 200 × 0.7 = 140
• Variance = npq = 200 × 0.7 × 0.3 = 42
• SD = √42 ≈ 6.48 — so typically 140 ± 6.48 students pass.

Skewness and Shape

The shape of the binomial distribution depends on p. Drag the slider to see how the distribution changes.

Skewness Explorer — B(10, p)
0.050.50 (symmetric)0.95
0.00
0
0.01
1
0.04
2
0.12
3
0.21
4
0.25
5
0.21
6
0.12
7
0.04
8
0.01
9
0.00
10

Mean μ = np

5.0

Variance σ² = npq

2.50

Mean > Variance?

Yes ✓

Skewness

Symmetric

Skewness pattern

When p = 0.5, the distribution is perfectly symmetric around the mean.

p = 0.5

Perfectly symmetric. Left half mirrors right half.

p < 0.5

Positively skewed. Tail to the right. Most values are small.

p > 0.5

Negatively skewed. Tail to the left. Most values are large.

Properties at a Glance

Key Properties of B(n, p)
AspectPropertyFormula / Value
Meanμ = npLong-run average number of successes
Varianceσ² = npqMeasure of spread (always < mean)
Standard Deviationσ = √npqSame units as X
Mean > Variancenp > npq (since q < 1)Always true for 0 < p < 1
Symmetric whenp = 0.5Distribution mirrors around mean
Positive skewp < 0.5Tail extends to the right
Negative skewp > 0.5Tail extends to the left
Sum of probabilitiesΣP(x) = 1For x = 0 to n

Solved Examples

Solved Example

Problem

A die is thrown 5 times. Find the mean and variance of the number of sixes.

Solution

Mean = 0.833, Variance = 0.694, SD = 0.833

Solved Example

Problem

If X ~ B(8, 0.4), find P(X = 3) and the mean and variance of X.

Solution

P(X=3) = 0.2787, Mean = 3.2, Variance = 1.92

Solved Example

Problem

The mean of a binomial distribution is 12 and the variance is 3. Find n and p.

Solution

n = 16, p = 0.75

Key Takeaways

Key Takeaways

  • Mean = np, Variance = npq, Standard Deviation = √npq — memorize these formulas.
  • Mean is always greater than variance (np > npq since q < 1).
  • When p = 0.5, the distribution is perfectly symmetric.
  • When p < 0.5, it is positively skewed (right tail). When p > 0.5, negatively skewed (left tail).
  • From mean and variance, you can recover n and p: q = variance/mean, then p = 1 − q, then n = mean/p.

Practice

  1. X ~ B(10, 0.3). Find the mean, variance, and SD.
  2. If mean = 15 and variance = 5 of a binomial distribution, find n and p.
  3. For what value of p is B(6, p) symmetric? What is P(X = 3) in that case?

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