Common Distributions and the Normal Distribution

Stage 33 has handed you a random variable X, its distribution table, and two summary numbers — the expectation E(X) and the variance D(X). Now we meet the named distributions that show up again and again, because the same trial keeps happening in the world. A single yes-or-no trial is the Bernoulli block; repeat it n times and count the successes and you have the binomial B(n, p), one term of a binomial expansion from Stage 32. Draw without replacement and the binomial bends into the hypergeometric. Then we cross a real border: when X can be any real number, its distribution is no longer a list of bars but a smooth density curve, and a probability becomes the area beneath it — ending at the famous normal bell and the 68–95–99.7 rule.

33.8.1 The two-point (Bernoulli) distribution

Start with the smallest possible random experiment: one trial with exactly two outcomes — heads or tails, defect or good, a free throw made or missed. Score it 1 for "success" and 0 for "failure." Then X ∈ {0, 1}, and the whole distribution table is two columns:

where p = P(success). This is the atom every later model is built from. Its two summaries come straight out of the formulas from 33.7:

E(X) = 0·(1−p) + 1·p = p,    D(X) = (0−p)²(1−p) + (1−p)²·p = p(1 − p).

Notice that the variance p(1−p) is largest at p = 0.5 (most uncertainty — a fair coin) and shrinks to 0 as p nears 0 or 1 (a near-sure outcome barely wobbles).

33.8.2 The binomial distribution B(n, p)

Now repeat the same Bernoulli trial n times, independently, with the same success-probability p every time. Let X = the number of successes. Then X follows the binomial distribution, written X ~ B(n, p).

How likely is exactly k successes? One particular pattern of k successes and n − k failures — say SSF…F — has probability p^k(1 − p)^n−k by independence (multiply the trials). But there are C(n, k) different patterns with the same count, so we add them all:

P(X = k) = C(n, k) p^k(1 − p)^n−k,   for k = 0, 1, …, n.

That is exactly the k-th term of the expansion of [(1−p) + p]ⁿ — so the probabilities are non-negative and sum to [(1−p) + p]ⁿ = 1ⁿ = 1 for free, which is precisely why this list is a valid distribution. The binomial theorem from Stage 32 walks straight in.

33.8.3 The expectation and variance of B(n, p)

Here is the payoff of building the binomial out of Bernoulli blocks. Write X = X₁ + X₂ + … + Xₙ, where each Xᵢ is one Bernoulli trial scored 0 or 1. Expectation adds (linearity, from 33.7), and for independent pieces variance adds too:

E(X) = n·p = np,    D(X) = n·p(1−p) = np(1 − p).

So you never need to push the long sum Σ k·C(n,k)pᵏ(1−p)ⁿ⁻ᵏ through by hand. For the 8 free throws at p = 0.5: E(X) = 8·0.5 = 4 makes on average, with D(X) = 8·0.5·0.5 = 2, so the standard deviation is σ = √2 ≈ 1.41 shots of spread. The widget above prints both np and np(1−p) live so you can confirm the centre and the width.

33.8.4 The hypergeometric distribution — drawing without replacement

The binomial demanded independent, fixed-p trials. That holds when you draw with replacement (put each card back). But most real sampling is without replacement: pull n items from a batch of N that contains M "good" ones, and count the good. Now each draw changes what is left, so the trials are not independent and p is not fixed — this is the hypergeometric distribution. We count instead with combinations (the classical model from 33.2):

P(X = k) = C(M, k)·C(N−M, n−k)C(N, n)  (choose the k good, the rest from the bad, over all ways to choose n).

As the batch N grows huge relative to the draw n, removing one item barely changes the proportion, and the hypergeometric closes in on the binomial — which is why opinion polls of millions are modelled as binomial. Conclusion: with replacement → binomial; without replacement → hypergeometric, narrower.

33.8.5 Continuous variables — probability becomes area

Every distribution so far has been discrete: X lands on separate values, each with its own bar. But a person's height, a battery's lifetime, the error in a measurement — these take any real value in a range. You cannot list them. Instead we describe X by a density curve f(x) ≥ 0, and the probability that X lands in an interval is the area under the curve over that interval:

P(a ≤ X ≤ b) = area under f from a to b,   and the total area under the whole curve is 1.

This forces a startling fact. The probability that X equals one exact value is the area of a strip of zero width: P(X = a) = 0. A continuous variable never hits a single number "on the nose" — it lands near it. (So for continuous X it makes no difference whether you write ≤ or <.)

33.8.6 The normal distribution N(μ, σ²)

One density curve dominates all of statistics. When a quantity is the sum of many small independent effects — measurement errors, exam scores, heights across a population — those wobbles average out into a symmetric, single-peaked bell curve: the normal distribution, written X ~ N(μ, σ²). Two numbers fix it completely:

• μ (the mean) is where the peak sits — slide μ and the whole bell shifts left or right without changing shape.
• σ (the standard deviation) is the width — a small σ gives a tall, skinny, certain curve; a large σ a short, wide, uncertain one. The area stays 1, so taller means narrower.

The curve is symmetric about μ, so P(X ≤ μ) = 0.5: the mean splits the area in half. And because total area is 1, the curve must get taller as it narrows — confirming again that a tall density height is not a large probability, just a concentrated one.

33.8.7 The 3σ rule — 68 · 95 · 99.7

Because every normal curve is the same shape just rescaled, the area inside a band of k standard deviations is always the same number, whatever μ and σ are. These three are worth memorising — each is the area of the matching band:

Roughly 68% of the data sits within one σ of the mean, 95% within two, and 99.7% within three. Only about 3 values in 1000 ever land outside 3σ — which is why, in quality control, a point beyond 3σ is flagged as "unusual," a likely signal that something has changed rather than ordinary luck.

★ What to carry forward

Three things to keep: a binomial is n Bernoulli atoms, so E = np and D = np(1−p); without replacement you switch to the (narrower) hypergeometric; and for a continuous variable a probability is the area under a density — never the height — with the normal's 68–95–99.7 rule the doorway into Statistics in Stage 34.

✎ Exercises

1. A loaded die shows a six with probability p = 1/4. Let X = 1 if it shows a six and X = 0 otherwise. Find E(X) and D(X).
   Show answer

   This is a Bernoulli trial with p = 1/4. So E(X) = p = 1/4 and D(X) = p(1−p) = (1/4)(3/4) = 3/16 = 0.1875.

2. A fair coin is flipped 6 times. Find P(X = 4), the probability of exactly 4 heads, and E(X).
   Show answer

   X ~ B(6, 0.5). P(X = 4) = C(6, 4)(0.5)⁴(0.5)² = 15·(0.5)⁶ = 15/64 ≈ 0.234. And E(X) = np = 6·0.5 = 3 heads on average.

3. A factory line produces parts that are defective with probability p = 0.1, independently. In a box of n = 20, find E(X) and D(X) for the number of defectives.
   Show answer

   X ~ B(20, 0.1). E(X) = np = 20·0.1 = 2 defectives expected. D(X) = np(1−p) = 20·0.1·0.9 = 1.8, so σ = √1.8 ≈ 1.34.

4. A drawer has 10 socks: 6 black and 4 white. You pull 3 without looking. What is the probability that exactly 2 are black? Is this binomial?
   Show answer

   No replacement, so it is hypergeometric with N = 10, M = 6 black, n = 3: P(X = 2) = C(6,2)·C(4,1)C(10,3) = 15·4/120 = 0.5. It is not binomial — the draws are dependent.

5. Heights in a population are N(170, 6²) cm. About what fraction are between 164 cm and 182 cm?
   Show answer

   μ = 170, σ = 6. The range 164 to 182 is μ − 1σ to μ + 2σ. Half of 68% plus half of 95%: 0.341 + 0.477 ≈ 0.818, so about 82%. (Check: the area from 164 to 182 under N(170, 6²) ≈ 0.819.)

6. For a continuous variable with density f, a student writes "P(X = 5) = f(5)." What is wrong, and what is P(X = 5) really?
   Show answer

   A density height f(5) is not a probability — it can even exceed 1. For a continuous variable the probability of a single exact value is the area of a zero-width strip: P(X = 5) = 0. Probability only appears as the area over an interval.

🎯 Quick check

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§ For teachers and parents

This lesson closes Stage 33's probability ladder by naming the standard models. Pupils first meet the two-point (Bernoulli) and binomial B(n, p) distributions, derive E = np and D = np(1−p), contrast the hypergeometric (sampling without replacement), and then make the conceptual leap from discrete bars to a continuous density where probability is area, ending at the normal curve and the empirical 68–95–99.7 rule. It builds directly on HSS-MD.A.3 (develop a probability distribution for a random variable and compute expected values) and HSS-ID.A.4 (use the mean and standard deviation to fit a data set to a normal distribution and estimate population percentages, recognising that areas under the normal curve are the probabilities). The recurring caution worth repeating at home: a tall density curve does not mean a large probability — only the shaded area is the probability, and a single exact value of a continuous variable has probability 0. This is the doorway into Statistics in Stage 34.