Which statement about the normal distribution is true?

Mean, median and mode are equal

In a joint distribution of discrete X and Y, P(X = 1, Y = 1) = 0.2, P(X = 1) = 0.5 and P(Y = 1) = 0.4. What can be concluded about this cell?

It fits independence; other cells must be checked

X and Y are certainly dependent

The table cannot be a valid joint distribution

It fits independence; other cells must be checked

Random Variables and Probability Distributions

A random variable turns the outcomes of an experiment into numbers, and a probability distribution says how likely each number is. This chapter covers discrete and continuous random variables, probability mass and density functions, expectation and variance, higher moments, the four named distributions of the syllabus — binomial, Poisson, normal and exponential — and the joint distribution of two discrete random variables.

Random variables, expectation and variance

A discrete random variable takes countable values with a probability mass function p(x) = P(X = x), where every p(x) ≥ 0 and Σp(x) = 1. A continuous one takes values in an interval with a probability density function f(x) ≥ 0 whose total area is 1; then P(a ≤ X ≤ b) is the area under f between a and b, and P(X = a) = 0 for any single point. The distribution function F(x) = P(X ≤ x) rises from 0 to 1.

The expectation E(X) = Σx·p(x) (or ∫x f(x)dx) is the long-run average. The variance is Var(X) = E(X − μ)² = E(X²) − [E(X)]². The rules used most: E(aX + b) = aE(X) + b; Var(aX + b) = a²Var(X); E(X + Y) = E(X) + E(Y) always; and Var(X + Y) = Var(X) + Var(Y) only when X and Y are independent (more generally, + 2Cov(X, Y)). The r-th raw moment is E(Xʳ) and the r-th central moment E(X − μ)ʳ, exactly as in Chapter 4.

Binomial and Poisson distributions

Binomial B(n, p): n independent trials, each a success with the same probability p (q = 1 − p); X counts successes. P(X = r) = ⁿCᵣ pʳ qⁿ⁻ʳ, r = 0, 1, …, n. Mean = np, variance = npq, so the variance is always less than the mean. It is symmetric when p = 1/2, positively skewed when p < 1/2.

B(5, 0.4): the whole distribution
r	P(X = r)
0	0.0778
1	0.2592
2	0.3456
3	0.2304
4	0.0768
5	0.0102

Poisson P(λ): the count of rare events in a fixed interval — calls per minute, misprints per page. P(X = r) = e^(−λ) λʳ ÷ r!, r = 0, 1, 2, …. Its defining feature: mean = variance = λ. It is the limit of the binomial when n is large, p small and np = λ fixed. With λ = 2, P(X = 0) = e⁻² ≈ 0.1353 and P(X = 1) ≈ 0.2707; successive terms follow P(r + 1) = P(r)·λ/(r + 1).

⚠️ Exam trap

A question that gives 'mean 4, variance 5' for a binomial has no answer — a binomial's variance npq is always smaller than its mean np. If mean equals variance, think Poisson.

Normal and exponential distributions

The normal distribution N(μ, σ²) is the bell curve: symmetric about μ, with mean = median = mode = μ, β₁ = 0 and β₂ = 3, and tails that never touch the axis. Any normal variable is turned into the standard normal by Z = (X − μ)/σ, which has mean 0 and SD 1, so one table serves every normal curve. The areas to memorise: within μ ± σ about 68.27%, within μ ± 2σ about 95.45%, within μ ± 3σ about 99.73%; and ±1.96σ cuts off exactly 95%.

Worked: marks are N(50, 10²). P(X > 70) = P(Z > 2) = 0.5 − 0.4772 = 0.0228, about 2.3% of candidates.

The exponential distribution with rate λ has f(x) = λe^(−λx) for x ≥ 0. It models waiting times between Poisson events. Mean = 1/λ, variance = 1/λ², so the SD equals the mean. P(X > x) = e^(−λx), and it is memoryless: P(X > s + t | X > s) = P(X > t) — a component that has survived s hours is as good as new.

Joint distribution of two discrete random variables

For two discrete variables the joint probability function p(x, y) = P(X = x, Y = y) is a table whose entries sum to 1. Its row totals give the marginal distribution of X, its column totals the marginal of Y, and the conditional distribution is P(X = x | Y = y) = p(x, y)/P(Y = y). X and Y are independent if p(x, y) = P(X = x)·P(Y = y) for every cell. Cov(X, Y) = E(XY) − E(X)E(Y).

Worked joint table
X \ Y	Y = 0	Y = 1	P(X = x)
X = 0	0.1	0.2	0.3
X = 1	0.3	0.4	0.7
P(Y = y)	0.4	0.6	1

E(X) = 0.7, E(Y) = 0.6, E(XY) = 0.4, so Cov(X, Y) = 0.4 − 0.42 = -0.02. Since P(X = 0, Y = 0) = 0.1 ≠ 0.3 × 0.4 = 0.12, X and Y are not independent — one failing cell is enough.

⚠️ Exam trap

Zero covariance does not prove independence; independence implies zero covariance, not the reverse.

Exam traps and a quick revision

The four distributions at a glance
Distribution	Mean	Variance	Key fact
Binomial	np	npq	variance < mean
Poisson	λ	λ	mean = variance
Normal	μ	σ²	mean = median = mode; β₂ = 3
Exponential	1/λ	1/λ²	memoryless; SD = mean

Key takeaways

Var(X) = E(X²) − [E(X)]²; Var(aX + b) = a²Var(X).
Binomial: P(X = r) = ⁿCᵣpʳqⁿ⁻ʳ, mean np, variance npq.
Poisson: P(X = r) = e^(−λ)λʳ/r!, mean = variance = λ.
Normal: standardise with Z = (X − μ)/σ; 68–95–99.7 for ±1, 2, 3σ.
Exponential: mean 1/λ, variance 1/λ², memoryless; independence of X, Y needs p(x, y) = p(x)p(y) in every cell.

Practice questions (10)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

For a binomial distribution with n = 10 and p = 0.4, the mean and variance are:
1. 4 and 2.4
2. 2.4 and 4
3. 4 and 4
4. 6 and 2.4
Show answer
Answer: A — 4 and 2.4
Mean = np = 4 and variance = npq = 10 × 0.4 × 0.6 = 2.4. Equal mean and variance would be a Poisson.
A binomial distribution has mean 6 and variance 2.4. Its parameters n and p are:
1. n = 12, p = 0.5
2. n = 15, p = 0.4
3. n = 10, p = 0.6
4. n = 10, p = 0.4
Show answer
Answer: C — n = 10, p = 0.6
q = variance/mean = 2.4/6 = 0.4, so p = 0.6 and n = mean/p = 6/0.6 = 10. Taking p = 0.4 confuses p with q.
For a Poisson distribution with mean 3, the standard deviation is:
1. √3
2. 9
3. 1/3
4. 3
Show answer
Answer: A — √3
In a Poisson distribution mean = variance = λ = 3, so the SD is √3 ≈ 1.73. 3 is the variance.
If X follows a Poisson distribution with λ = 2, then P(X = 0) is:
1. 1 − e⁻²
2. e²
3. 2e⁻²
4. e⁻²
Show answer
Answer: D — e⁻²
P(X = 0) = e^(−λ)λ⁰/0! = e⁻² ≈ 0.135. 2e⁻² is P(X = 1), and 1 − e⁻² is P(X ≥ 1).
Which statement about the normal distribution is true?
1. Mean, median and mode are equal
2. Its β₂ is 0
3. Its variance equals its mean
4. It is positively skewed
Show answer
Answer: A — Mean, median and mode are equal
The normal curve is symmetric and unimodal, so its mean, median and mode coincide at μ. Its β₂ is 3 (excess kurtosis 0), and mean = variance is a Poisson property.
Marks are normally distributed with mean 50 and SD 10. Given that the area from Z = 0 to Z = 2 is 0.4772, the proportion scoring above 70 is:
1. 0.4772
2. 0.0228
3. 0.9772
4. 0.0456
Show answer
Answer: B — 0.0228
Z = (70 − 50)/10 = 2, and P(Z > 2) = 0.5 − 0.4772 = 0.0228. 0.0456 counts both tails beyond ±2.
For an exponential distribution with rate λ = 0.5, the mean and variance are:
1. 2 and 2
2. 2 and 4
3. 0.5 and 0.25
4. 0.5 and 2
Show answer
Answer: B — 2 and 4
Mean = 1/λ = 2 and variance = 1/λ² = 4. Using λ itself as the mean is the usual slip.
A random variable X has E(X) = 3 and Var(X) = 4. The value of E(X²) is:
1. 13
2. 25
3. 7
4. 5
Show answer
Answer: A — 13
Var(X) = E(X²) − [E(X)]², so E(X²) = Var(X) + [E(X)]² = 4 + 9 = 13. Adding E(X) instead of its square gives 7, and subtracting gives 5.
If Var(X) = 5, then Var(2X + 3) equals:
1. 23
2. 20
3. 10
4. 13
Show answer
Answer: B — 20
Var(aX + b) = a²Var(X) = 4 × 5 = 20. The constant 3 shifts X without spreading it.
In a joint distribution of discrete X and Y, P(X = 1, Y = 1) = 0.2, P(X = 1) = 0.5 and P(Y = 1) = 0.4. What can be concluded about this cell?
1. X and Y are certainly dependent
2. The table cannot be a valid joint distribution
3. P(X = 1 | Y = 1) = 0.2
4. It fits independence; other cells must be checked
Show answer
Answer: D — It fits independence; other cells must be checked
0.5 × 0.4 = 0.2 matches this cell, which is consistent with independence; independence needs p(x, y) = p(x)p(y) in every cell. The conditional P(X = 1 | Y = 1) = 0.2/0.4 = 0.5.