The Normal Distribution

— Mathematics · AS & A Level (9709) · Probability & Statistics 1 · Chapter 8 —

★ Exam Tip: The book's own advice: always sketch a normal curve and shade the area you want. Most mistakes (wrong sign of z, wrong tail) disappear once you can see the shaded region.

0. Why are errors “normal”?

Every measurement of a continuous quantity (time, mass, distance…) has error — tools aren't perfect, humans add unreliability.
Small errors are more likely than big ones; over- and under-estimates are about equally likely.
Repeat the measurement ⇒ errors cancel ⇒ average error ≈ 0.
Found by Gauss (late 18th c.) from errors in astronomical observations; Quételet later saw height, weight, girth, strength are ≈ normal too.

Key idea: values near the average are most likely; the further away, the less likely; symmetric about the average.

Prerequisite: X ~ B(n,p) ⇒ E(X) = np, Var(X) = npq

1. Continuous Random Variables

Definition: a continuous random variable can take infinitely many values in a range (sugar in an orange, marathon time, height, temperature). It is defined over an interval, not at single values.

P(X = any single value) = 0
probabilities exist only for intervals: P(a ≤ X ≤ b)

e.g. mass of an apple X g: it could be 111.2233…, 137.8642… — infinite values, so one exact value has probability 0, but P(130 ≤ X < 140) can be found.

	Discrete RV	Continuous RV
What we find	P(X = r)	P(a ≤ X ≤ b)
Distribution shows	specific values + their probabilities	range of values + probabilities of intervals

From histogram → probability density function (PDF)

Continuous data → histogram (column areas ∝ frequency).
Change frequency density to relative frequency density = relative frequencyclass width.
Now column areas = relative frequencies (estimates of probability). Total area = 1 (instead of Σf). Vertical axis: “probability density”.
Draw a smooth curve over the columns (many equal-width classes ⇒ better) ⇒ curve y = f(x) = the PDF. Area under it = 1.

a bell ⇒ “a bit more normal”

two peaks ⇒ “a bit odd”

Tip: say “function” only for a random variable (PDF). For data, say curve / graph.

Shapes of curves

negatively skewed

symmetric

positively skewed

Mode / median: mode = peak of the graph. Median = value splitting the area into two equal halves (calculate from histogram, or estimate from cumulative frequency graph).

Explore 8.1 — three frequency tables (classes 3–6 … 21–24)

w: 13 in every class | x: 3, 9, 18, 24, 18, 9, 3 | y: 8, 19, 10, 4, 10, 19, 8

w: flat (uniform)

x: bell-shaped

y: U-shaped, 2 peaks

Common: all three are symmetric about 13.5 (centre of 3–24) ⇒ mean = median = 13.5 for each.
Differences: only x has mean = median = mode (a normal-type curve). w has no single mode; y has two modes (the end classes).
Spread: y is most spread out, x least.

2. The Normal Curve

The distribution of x above is a symmetric, bell-shaped curve = a normal curve.

If a distribution is a normal curve:
① Mean = median = mode
② Peak is at the mean μ — this is the line of symmetry
③ Density falls as you move away from μ on both sides (further ⇒ less likely)
④ Bigger σ ⇒ curve wider & lower (area stays 1)

same μ, different σ: B = A stretched vertically & horizontally

same σ, different μ: slide sideways (translation)

Note: Larger σ ⇒ lower, wider curve. Different μ ⇒ same shape, new location. The area under every normal curve is always 1, whatever μ and σ.

3. The Normal Distribution X ~ N(μ, σ²)

A normal random variable’s curve is its PDF. If X is normally distributed with mean μ and variance σ²:

f(x) = 1σ√2π exp{ − (x − μ)²2σ² } for all real x

X ~ N(μ, σ²)

“X has a normal distribution with mean μ and variance σ²”. Parameters: μ and σ².

Tip: exp{ } means e = 2.71828… raised to the power in the bracket; e^p > 0 always.

P(a ≤ X ≤ b) = area under curve from x = a to x = b = ∫_a^b f(x) dx. This integral can't be done exactly ⇒ we use tables.

Boundaries don't matter! P(3≤X≤7) = P(3<X<7) = P(3<X≤7) … (area is the same)

Remember: continuous ⇒ “<” and “≤” give the same probability. (Not so for discrete — see continuity correction!)

Fixed properties (true for every normal distribution)

Property	Probability
Half the values below μ, half above	P(X<μ) = P(X>μ) = 0.5
≈ 68.26% within 1 s.d. of mean	P(μ−σ ≤ X ≤ μ+σ) = 0.6826
≈ 95.44% within 2 s.d.	P(μ−2σ ≤ X ≤ μ+2σ) = 0.9544
≈ 99.72% within 3 s.d.	P(μ−3σ ≤ X ≤ μ+3σ) = 0.9972

μ±σ

μ±2σ

μ±3σ

Tip: the probability of lying within a given number of s.d. of the mean is fixed — it doesn't depend on μ or σ.

Use symmetry to get other probabilities (values 0, ±1, ±2 = number of s.d. from mean):

P(X ≤ μ+σ)

P(X ≤ μ+2σ)

P(−1 ≤ X ≤ 1) = 0.6826P(X ≤ 1) = 12 × 0.6826 + 0.5P(X ≤ 1) = 0.8413 = P(X ≥ −1)

P(−2 ≤ X ≤ 2) = 0.9544P(X ≤ 2) = 12 × 0.9544 + 0.5P(X ≤ 2) = 0.9772 = P(X ≥ −2)

Explore 8.3is it really normal? — check with the ±1σ rule

A: μ=40, var 64 ⇒ σ=8; 32–48 is μ±σ. B: μ=72, σ=12; 60–84 is μ±σ. C: μ=123, σ=11; 112–134 is μ±σ. If normal, ≈ 0.6826 of observations should fall inside.

Var.	Observed in μ±σ	Fraction	vs 0.6826
A	8060 / 13120	0.614	too low ✗
B	8475 / 12420	0.682	matches ✓
C	8013 / 10974	0.730	too high ✗

Conclusion: only B is consistent with a normal distribution — “A, B and C are all normal” is NOT supported.

4. The Standard Normal Variable Z

Infinite choices of (μ, σ²) — so we convert every normal variable into the standard normal variable Z.

Z ~ N(0, 1) φ(z) = 1√2π exp{ − z²2 }

Tip: φ (lower) and Φ (upper) are the Greek letter phi.

mean 0, variance 1, axis of symmetry z = 0

z = ±1, ±2, ±3 = 1, 2, 3 s.d. above/below the mean.
z < 0 ⇒ below the mean; z > 0 ⇒ above.
φ(z) ≈ 0 for z > 3 and z < −3; total area = 1.
A vertical line at z splits the area into P(Z ≤ z) and P(Z > z).

Φ(z) = P(Z ≤ z) (area to the left of z)

Φ(z) can’t be integrated, so use the Standard normal distribution function table (or calculator Φ and Φ⁻¹). Tables show only z ≥ 0 (z = 0.000 … 2.999).

Reading the table — 4-figure z

Rows: 1st & 2nd figures of z (left column).
Columns 0–9: 3rd figure.
“ADD” columns 1–9: 4th figure — add that number to the last digits of the value.

Find Φ(0.274):

row 0.2, column 7 ⇒ Φ(0.27) = 0.60644th figure 4 ⇒ “ADD 15” ⇒ + 0.0015Φ(0.274) = 0.6064 + 0.0015 = 0.6079

Inverse notation: Φ(0.274) = 0.6079 ⇔ Φ⁻¹(0.6079) = 0.274. Where possible use a 4-figure z.

Note: in the critical values table the probability Φ(z) is called p (75%, 90%, 95%… and complements 25%, 10%, 5%…).

Negative z — use symmetry

z = b > 0: Φ(b) = P(Z ≤ b)

z = −a: Φ(a) = P(Z ≥ −a)

Negative zP(Z ≤ −a) = 1 − Φ(a) P(Z ≥ b) = 1 − Φ(b)

One table value gives four probabilities. Φ(0.11) = 0.5438 ⇒ P(Z ≤ 0.11) = P(Z ≥ −0.11) = 0.5438; P(Z ≥ 0.11) = P(Z ≤ −0.11) = 1 − 0.5438 = 0.4562.

Always: transfer the given probability to a sketch first. e.g. P(Z ≥ z) > 0.5 ⇒ P(Z ≤ z) < 0.5 ⇒ z < 0.

WE 8.1Z ~ N(0,1): P(Z < 1.23), P(Z ≥ 1.23)

Φ(1.23) = 0.8907 (area left of 1.23)P(Z < 1.23) = 0.8907P(Z ≥ 1.23) = 1 − 0.8907 = 0.1093

WE 8.2P(0.4 ≤ Z < 1.7) to 3 d.p.

(area left of 1.7) − (area left of 0.4); both found in main body of table (no ADD needed).

Φ(1.7) = 0.9554 , Φ(0.4) = 0.6554P = Φ(1.7) − Φ(0.4)= 0.9554 − 0.6554 = 0.300

WE 8.3P(−1 ≤ Z < 2.115) to 3 s.f.

P(Z < −1) = 1 − Φ(1) , soP = Φ(2.115) − [1 − Φ(1)]= Φ(2.115) + Φ(1) − 1= 0.9828 + 0.8413 − 1 = 0.824

WE 8.4find a: P(Z < a) = 0.9072 (tables in reverse)

Closest table value to 0.9072 is 0.9066 = Φ(1.32). Need +0.0006 ⇒ “ADD 6” ⇒ 4th figure is 4.

0.9066 = Φ(1.32)a = 1.32 + 0.004a = 1.324check: Φ(1.324) = 0.9066 + 0.0006 = 0.9072 ✓

Inverse notation: a = Φ⁻¹(0.9072) = Φ⁻¹(0.9066 + 0.0006) = Φ⁻¹(0.9066) + 0.004 = 1.324

WE 8.5find b: P(Z ≥ b) = 0.7713

P(Z ≥ b) > 0.5 ⇒ b is negative; write b = −a (mirror picture).

P(Z ≤ a) = 0.7713 ⇒ a = Φ⁻¹(0.7713)= Φ⁻¹(0.7704 + 0.0009)= Φ⁻¹(0.7704) + 0.003 = 0.740 + 0.003 = 0.743b = −a = −0.743

5. Standardising: X ~ N(μ, σ²) → Z

The same Z-tables work for any normal variable, because coding (shift & scale) keeps it normal:

subtract μ: curve slides to centre on 0 · divide by σ: s.d. becomes 1

Z = X − μσ ~ N(0, 1) z = x − μσ = number of s.d. x is from the mean

Careful: divide by σ = √variance. If N(20, 9) then σ = 3, so P(X < 23) = P(Z < 23 − 20√9).

Remember: the s.d.-probabilities in section 3 apply to ALL normal variables.

WE 8.6X ~ N(11, 25): P(X < 18)

z = 18 − 11√25 = 1.4 (1.4σ above mean)P(X < 18) = P(Z < 1.4) = Φ(1.4)= 0.919

WE 8.7X ~ N(20, 7): P(X ≤ 16.6)

z = 16.6 − 20√7 = −1.285 (1.285σ below mean)P(X ≤ 16.6) = P(Z ≤ −1.285) = 1 − Φ(1.285)= 0.0994

WE 8.8X ~ N(5, 5): P(2 ≤ X < 9)

x: 2, 5, 9 ↔ z: −1.342, 0, 1.789

WE 8.8working

z (x=2) = 2 − 5√5 = −1.342 , z (x=9) = 9 − 5√5 = 1.789right of 0: Φ(1.789) − 0.5 = 0.4633left of 0: 0.5 − [1 − Φ(1.342)] = 0.4102total = 0.4633 + 0.4102 = 0.8735P(2 ≤ X < 9) = 0.874

Alternative (as WE 8.3): Φ(1.789) − [1 − Φ(1.342)] — pick the method you prefer.

Handy area results

0<a<b
Φ(b) − Φ(a)

−a<0<b
Φ(b) + Φ(a) − 1

−a<0<a
2Φ(a) − 1

WE 8.9Y ~ N(μ,σ²), P(Y < 10) = 0.75, P(Y ≥ 12) = 0.1 — find μ and σ

P(Y ≥ 12) < 0.5 ⇒ P(Y < 12) = 0.9 > 0.5 ⇒ 12 > μ; and P(Y<10) > 0.5 ⇒ 10 > μ. Both points sit right of the mean. 0.75 and 0.90 are critical values.

WE 8.9working

z_a = Φ⁻¹(0.75) = 0.674 , z_b = Φ⁻¹(0.90) = 1.28210 − μσ = 0.674 ⇒ 10 − μ = 0.674σ …[1]12 − μσ = 1.282 ⇒ 12 − μ = 1.282σ …[2][2] − [1]: 2 = 0.608σσ = 3.29 , μ = 10 − 0.674(3.29) = 7.78

Method — finding unknown μ and/or σ: ① sketch & locate the values vs μ ② convert each probability to z with tables in reverse ③ write (x − μ)/σ = z for each ④ solve (one unknown: direct; two unknowns: simultaneous equations).

6. Modelling with the Normal Distribution

Gauss: astronomical measurement errors; Quételet: height, weight, girth, strength ≈ normal. Real data are only modelled by the normal curve (e.g. theoretical P(pineapple mass < 0.2 kg) > 0, tiny but not impossible).

WE 8.10newborn mass ~ N(3.35, 0.0858 kg²); estimate how many of 1356 babies < 3.5 kg

z = 3.5 − 3.35√0.0858 = 0.512P(mass < 3.5) = Φ(0.512) = 0.6957 (relative frequency 69.57%)0.6957 × 1356 = 943.37≈ 943 babies

Tip: a model only gives estimates, but a number of babies must be an integer.

WE 8.11oil tins: X ~ N(506.18, 2.96²) ml

a) % of tins with less than 500 ml? b) P(exactly 1 of 3 tins has < 500 ml)?

z = −2.088

z = 500 − 506.182.96 = −2.088P(X < 500) = P(Z < −2.088) = 1 − Φ(2.088)= 1 − 0.9816 = 0.0184a) 1.84%

Y = number of tins < 500 ml ⇒ Y ~ B(3, 0.0184)P(Y = 1) = ³C₁ × 0.0184¹ × 0.9816²b) 0.0532

Tip: a probability found from a normal distribution can be used as p in a binomial distribution.

Pattern: “% / how many” ⇒ standardise, Φ, × total. “exactly k of n items” ⇒ get p from normal, then binomial B(n, p).

7. Normal Approximation to the Binomial

Binomial P(X ≥ 60) from 100 tosses needs 41 terms (60, 61, …). With a normal approximation it's just one Φ lookup. As n grows the binomial shape becomes bell-shaped (de Moivre found the curve — now the normal curve; also once called Gaussian. “Normal” does not mean other distributions are abnormal!).

n=12, p=0.5: nearly normal

n=25, p=0.35 (np=8.75, nq=16.25) OK ✓

n=25, p=0.15 (np=3.75) skewed ✗

Shape is lost when p is small and also when q is small (n=25, p=0.95 ⇒ nq = 1.25 ✗; p=0.75 ⇒ nq = 6.25 adequate but not great).

Key point 8.4X ~ B(n, p) ≈ N(μ, σ²) with μ = np , σ² = npq
condition: np > 5 and nq > 5 (larger ⇒ better)

Check: B(40, 0.9): nq = 4 < 5 ✗ no. B(250, 0.2): np = 50, nq = 200 ✓ yes.

Continuity correction (Key point 8.5 — must always be made!)

A discrete value such as X = 13 is represented by the continuous class 12.5 ≤ X < 13.5. So replace 13 by 12.5 or 13.5 depending on whether 13 is included.

X = 43 is the bar from 42.5 to 43.5

want	use	why
P(X < 43)	x = 42.5	43 excluded
P(X ≤ 43)	x = 43.5	43 included
P(X > 43)	x = 43.5	43 excluded
P(X ≥ 43)	x = 42.5	43 included

Trick: if the boundary value is included (≤, ≥) go outward half a unit to take in the whole bar; if excluded (<, >) go inward. Ask: “is 43 in the answer?”

Reading inequalities: X < a: fewer than a · X > a: more than a · X ≤ a: at most a (not more than) · X ≥ a: at least a (not less than).

WE 8.12X ~ B(100, 0.4): a) P(X < 43) b) P(X > 43)

μ = np = 40 , σ² = npq = 24np = 40 > 5 , nq = 60 > 5 ✓ ⇒ X ≈ N(40, 24)

a) x = 42.5 ⇒ z = 42.5 − 40√24 = 0.5103P(X < 43) = P(Z < 0.510) = Φ(0.510) = 0.6950= 0.695

b) x = 43.5 ⇒ z = 43.5 − 40√24 = 0.7144P(X > 43) = P(Z > 0.714) = 1 − Φ(0.714) = 1 − 0.7623= 0.238

WE 8.138000 items per box, 0.2% yellow: a) P(< 20 yellow) b) P(exactly 2 of 3 boxes have < 20)

X = no. of yellow in a box ⇒ X ~ B(8000, 0.002)np = 16 , nq = 7984 ✓ ⇒ X ≈ N(16, 15.968)

P(X < 20): use x = 19.5z = 19.5 − 16√15.968 = 0.87587…P = Φ(0.876) = 0.8094a) ≈ 0.809

Y = no. of boxes with < 20 yellow ⇒ Y ~ B(3, 0.8094)P(Y = 2) = ³C₂ × 0.8094² × 0.1906b) 0.375

Don’t forget the continuity correction! And don’t use a rounded 0.8 in later steps — keep 0.8094.

WE 8.14fair coin tossed 888 times: P(at most 450 heads)

X ~ B(888, 0.5) ⇒ np = nq = 444 > 5 , npq = 222 ⇒ X ≈ N(444, 222)P(X ≤ 450): use x = 450.5z = 450.5 − 444√222 = 0.436P = Φ(0.436) = 0.6686≈ 0.669

Steps for any binomial → normal question: ① define X ~ B(n,p) ② check np > 5, nq > 5 ③ μ = np, σ² = npq ④ continuity correction ⑤ standardise ⑥ Φ / tables ⑦ round at the end.

★ Quick Recap — Full Chapter Cheat-Sheet

✓ Continuous RV & PDF

Continuous RV ⇒ P(X = value) = 0; use P(a ≤ X ≤ b) = area under PDF; total area = 1
Boundaries (< vs ≤) make no difference for continuous variables
Normal curve: symmetric bell, mean = median = mode = μ; larger σ ⇒ wider & lower

✓ Normal / standard normal

X ~ N(μ, σ²) (σ² is the variance); Z ~ N(0,1); Z = (X − μ)/σ
Within 1, 2, 3 s.d. of mean: 0.6826, 0.9544, 0.9972
Φ(z) = P(Z ≤ z); P(Z > z) = 1 − Φ(z); P(Z ≤ −a) = 1 − Φ(a)
P(a < Z < b) = Φ(b) − Φ(a); P(−a < Z < b) = Φ(b) + Φ(a) − 1; P(−a < Z < a) = 2Φ(a) − 1
Finding z from a probability: use tables in reverse (Φ⁻¹); P(Z ≥ b) > 0.5 ⇒ b negative
Unknown μ, σ: standardise each condition ⇒ (x − μ)/σ = z ⇒ solve simultaneously

✓ Modelling & binomial approximation

“How many” = probability × total (give a whole number); “exactly k of n” ⇒ binomial with p from the normal
B(n,p) ≈ N(np, npq) if np > 5 and nq > 5 — always make the continuity correction
< a → a − 0.5 | ≤ a → a + 0.5 | > a → a + 0.5 | ≥ a → a − 0.5
Sketch every time! Don’t use rounded intermediate probabilities.