Probability Distributions

A distribution is a formula that gives you the probability of each outcome. Two matter here: the normal (smooth bell curve, for measurements) and the binomial (jagged bars, for counting successes).

Which one do I need?

Are you counting how many times something happened? → Binomial (a bar chart).

Are you measuring something that varies continuously (height, weight, time, error)? → Normal (a bell curve).

The third topic — the normal approximation to the binomial — exists purely because binomial tables get impossibly wide when n is large. You use the bell curve to avoid them.

On this page

Random variables & the big picture
Normal distribution
The z-table and how to use it
Binomial distribution
Normal approximation to the binomial
Self-check

1. Random Variables and the Big Picture

A random variable turns an outcome into a number, so you can do arithmetic on it. Roll a die: the outcome "3" becomes the number X = 3. Then the distribution of X is just a list of probabilities.

Left: exact probabilities sit on bars. Right: probabilities sit on areas. This one word — bars vs areas — explains every difference in this unit.

Bars vs areas — the root of everything

Binomial: P(X = 3) is the height of one bar.

Normal: P(X < 3) is an area between two lines. A single point has area zero — you can never ask for P(X = 3) on a continuous distribution. That is why the normal curve never gives an exact single value, only a range.

2. The Normal Distribution

Standard form X ~ N(μ, σ²) meaning mean μ, variance σ² (or standard deviation σ)

The normal curve is completely described by two numbers: the mean μ (where the peak sits) and the standard deviation σ (how wide the bell is).

The 68–95–99.7 rule. In the exam you often only need this and no z-table at all.

Range	Contains	Where you see it
μ ± 1σ	68%	most data
μ ± 2σ	95%	the usual "95% confidence"
μ ± 3σ	99.7%	outliers live outside here

Three facts you will be asked for 1. The curve is symmetric about x = μ (bell shape, mirror halves)
2. The total area under the curve is exactly 1
3. The curve approaches the x-axis but never touches it (so P(X = c) = 0 for any single c)

Trick — the four-part memory hook

Mean = middle (symmetry) Sigma = spread (width) Area = 1 (total probability) 68-95-99.7 (the rule)

Std-dev trick: bigger σ means a wider, flatter bell. Smaller σ means a narrower, taller one. The total area stays 1 either way.

3. Standardising and the z-table

Different questions have different μ and σ, which makes a shared table impossible. The z-score fixes that by converting any normal variable into a single standard normal distribution with μ = 0, σ = 1.

z-score — the single most important formula in this unit z = (x − μ) / σ

After standardising, every normal question becomes the same question: "how far from the middle, in σ units?"

Example — standardising

Heights of students: X ~ N(170, 25). What z-value corresponds to 180 cm? (Note: 25 is the variance σ², so σ = 5.)

σ = √25 = 5
z = (180 − 170) / 5 = 10/5 = 2.0

z = 2. So 180 cm is 2 standard deviations above the mean → about the 97.5th percentile.

The most common slip in Unit 4

Check whether the number you were given is σ or σ². The notation N(μ, σ²) means the second number is the variance. If the variance is 25, then σ = 5, not 25. Taking a square root when you should not (or the reverse) is worth more lost marks than anything else in this unit.

Reading the standard normal table

A z-table gives the area from the mean (0) up to z — the cumulative area in the positive direction. Everything else is derived from symmetry.

Φ(z) = the area from 0 to z. Every probability you will ever need is built from Φ(z) plus or minus 0.5.

Trick — remember only three z-values

z = 0 → 0.5000 (half the data) z = 1.645 → 0.9500 z = 1.96 → 0.9750 z = 2.576 → 0.9950

If your exam is open-book you only need the rule "add or subtract 0.5 from the table". If it is closed-book, memorise 1.645 and 1.96 — they are the critical values for 90% and 95% two-tailed tests, and they carry Unit 5 too.

Example — find a probability

X ~ N(100, 15). Find P(90 < X < 110).

Step 1 — standardise both ends.

z₁ = (90 − 100)/15 = −0.667
z₂ = (110 − 100)/15 = +0.667

Step 2 — look up. Φ(0.667) ≈ 0.7475, and by symmetry Φ(−0.667) = 1 − 0.7475 = 0.2525.

Step 3 — subtract.

P(90 < X < 110) = 0.7475 − 0.2525 = 0.4950

Alternative (no table needed): this is μ ± 0.667σ, which is well inside μ ± σ, so it is a bit under 68%. 0.495 matches — a good sanity check.

0.495

4. The Binomial Distribution

Requirements — all four MUST hold 1. fixed number of trials n
2. only two outcomes per trial: success or failure
3. independent trials
4. constant probability p on every trial

Requirement 4 is the one that catches people. "Without replacement" quietly changes p every time.

The binomial probability formula P(X = x) = ⁿC x · pˣ · (1 − p)ⁿ⁻ˣ , 0 ≤ x ≤ n

Notice it is just Unit 2 multiplied by Unit 3:

Three factors, three jobs. This sentence is worth more than the formula.

Example — 3 heads in 5 coin tosses

n = 5, p = ½, asking P(X = 3).

P(X=3) = ⁵C 3 (½)³ (½)²
= 10 × (1/8) × (1/4)
= 10/32 = 0.3125

The (½)² came from n − x = 5 − 3 = 2 failures. Getting ⁵C 3 vs ⁵C 2 makes no difference because they are equal — but never write ⁵C 5.

0.3125

Example — "at least" always means subtract from 1

A machine produces 10% defective items. In a sample of 5, find P(at least 2 defective).

P(X≥2) = 1 − P(0) − P(1)
     = 1 − [⁵C 0 (0.1)⁰(0.9)⁵ + ⁵C 1 (0.1)¹(0.9)⁴]
     = 1 − [1(0.59049) + 5(0.1)(0.6561)]
     = 1 − [0.59049 + 0.32805]
     = 1 − 0.91854 = 0.08146

0.08146. Note (0.1)⁰ = 1 — an "impossible" outcome still counts as one way.

Mean, variance and standard deviation of a binomial

These three formulas give free marks — learn them by heart Mean: μ = np
Variance: σ² = np(1 − p)
Std dev: σ = √[ np(1−p) ]

Trick — the shape tells you everything

p close to ½ (near 0.5) → the curve is balanced and tall (σ is largest).
p close to 0 or 1 → the curve is lopsided and flat, hugging a corner.

Proof of the second point: p(1−p) is largest when p = 0.5, and it falls to 0 as p approaches 0 or 1. So the standard deviation is biggest in the middle.

Example — mean and variance

Binomial with n = 40, p = 0.25.

μ = 40(0.25) = 10
σ² = 40(0.25)(0.75) = 7.5
σ = √7.5 ≈ 2.739

μ = 10, σ ≈ 2.74. Most counts will land within 10 ± 5, i.e. 5 to 15.

5. Normal Approximation to the Binomial

When n is large (say n > 30) the binomial bars become so many and so thin that they look like a smooth bell curve. The binomial is then approximately normal, which means you can use the far easier normal formula and z-table.

Approximate the binomial as normal with μ = np, σ = √[np(1−p)]

When is the approximation good enough? (check BOTH) np ≥ 5 and n(1−p) ≥ 5 (equivalently n(1−p) ≥ 5 and np ≥ 5)

THE CONTINUITY CORRECTION — do not skip this P(a ≤ X ≤ b) ≈ P( a − 0.5 < Y < b + 0.5 ) where Y ~ N(μ, σ²)

The left picture counts the two bars. The right picture, without the correction, counts the empty space between them.

How to apply the correction correctly

Move 0.5 outward at every boundary:

P(X ≥ 5) → P(Y > 4.5) (lower bound goes down)
P(X ≤ 5) → P(Y < 5.5) (upper bound goes up)
P(4 < X < 8) → P(3.5 < Y < 7.5)
P(X = 5) → P(4.5 < Y < 5.5) (both ends move)

Mnemonic: "shave half off the outside" — widen the region to cover the full bars.

Example — full normal approximation

Question: a coin with P(head) = 0.6 is tossed 50 times. Find P(at least 30 heads).

Step 1 — check the approximation is allowed.

np = 50(0.6) = 30 ≥ 5 ✓ n(1−p) = 50(0.4) = 20 ≥ 5 ✓

Step 2 — compute the normal parameters.

μ = 30
σ = √(30 × 0.4) = √12 ≈ 3.464

Step 3 — apply the continuity correction.

P(X ≥ 30) ≈ P(Y > 29.5)

Step 4 — standardise.

z = (29.5 − 30) / 3.464 = −0.5 / 3.464 = −0.1443

Step 5 — read the answer.

P(Y > 29.5) = 1 − P(Y < 29.5) = 1 − 0.4427 = 0.5573

≈ 0.557 (about 56%). Sanity check: the mean is 30, so "at least 30" should be a bit above 50%. ✓

Trick — the 5-step routine for any approximation question

Check np ≥ 5 and n(1−p) ≥ 5. Write this down, it is a mark.
Write μ = np and σ = √np(1−p). Keep them visible.
Correct the boundary by 0.5 — write the corrected version in full.
Standardise to z.
Read the table, remembering to add or subtract 0.5.

Sanity check every time: the answer must be between 0 and 1, and if the question is symmetric about the mean the answer should be near 0.5.

Binomial vs Normal — how to choose

n small (say < 30)	use binomial formula
n large (≥ 30)	use normal approximation
np < 5 or n(1−p) < 5	approximation invalid — go back to binomial
continuous measurement	normal from the start

Formula summary

Normal z = (x−μ)/σ · P(X<x) = Φ(z)+0.5

Binomial P(X=x) = ⁿC x pˣ(1−p)ⁿ⁻ˣ
μ = np · σ² = np(1−p)

Approximation continuity: a → a−0.5, b → b+0.5

Self-check

Open to see the answers

1. X ~ N(50, 16). Find P(46 < X < 52).

16 is the variance → σ = 4
z₁ = (46−50)/4 = −1, z₂ = (52−50)/4 = +1
By the 68–95–99.7 rule, μ ± 1σ contains 68%.
P = ≈ 0.68 (exactly 0.6827 from the table)

2. Binomial: n = 10, p = 0.2. Find P(X = 3).

P(X=3) = ¹⁰C 3 (0.2)³ (0.7)⁷
= 120 × 0.008 × 0.0823543
= 0.96 × 0.0823543 = 0.0850
Check the conditions: 10(0.2) = 2 < 5, so the normal approximation would NOT be valid here. Good that you used the exact formula.

3. Binomial n = 100, p = 0.4. Find P(X ≤ 45).

Valid? np = 40 ≥ 5 ✓ and n(1−p) = 60 ≥ 5 ✓
μ = 40, σ = √(100 × 0.4 × 0.6) = √24 = 4.899
Continuity: P(X ≤ 45) ≈ P(Y < 45.5)
z = (45.5 − 40)/4.899 = 5.5/4.899 = 1.1227
P = 0.5 + Φ(1.1227) ≈ 0.5 + 0.3694 = 0.8694

4. For a normal curve, what is P(X = μ)?

0. A single point has zero width, so zero area. You can only ever ask about ranges on a continuous distribution. This is worth memorising — it is a common one-mark question.