Mathematics Advanced • Year 12 • Module 5 • Lesson 12

The Normal Distribution

Build fluency in z-scores and the 68-95-99.7 empirical rule for any normal distribution.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Complete the z-score formula for X ~ N(μ, σ²):

z = ( ____________ − ____________ ) / ____________

Q1.2 Fill in the empirical (68-95-99.7) rule percentages:

P(μ − σ < X < μ + σ) ≈ ____________ %

P(μ − 2σ < X < μ + 2σ) ≈ ____________ %

P(μ − 3σ < X < μ + 3σ) ≈ ____________ %

Q1.3 The standard normal Z ~ N(μ, σ²) has μ = ____________ and σ = ____________.

Stuck? Revisit lesson § Formula Reference and § Empirical Rule.

2. Worked example — using the empirical rule for X ~ N(170, 8²)

Follow each step; the reasoning is on the right of each line.

Problem. Year 12 students' heights are H ~ N(170, 8²) cm. Find P(162 < H < 178), P(H > 186) and the height below which 97.5% of students fall.

Step 1 — Express each boundary as μ ± kσ.

162 = 170 − 8 = μ − σ (k = −1)
178 = 170 + 8 = μ + σ (k = +1)
186 = 170 + 2(8) = μ + 2σ (k = +2)

Reason: writing boundaries as multiples of σ lets the empirical rule speak directly.

Step 2 — Apply the empirical rule to each interval.

P(μ − σ < H < μ + σ) ≈ 68% → P(162 < H < 178) ≈ 68%
P(H > μ + 2σ) ≈ 2.5% → P(H > 186) ≈ 2.5%

Reason: 68% lies within μ ± σ; the upper tail beyond μ + 2σ holds half of the outer 5% = 2.5%.

Step 3 — Convert "97.5% below" into μ + kσ.

97.5% below ⇒ 2.5% above ⇒ cut-off at μ + 2σ
x = 170 + 2(8) = 186 cm

Reason: by symmetry, 50% + 47.5% = 97.5% accumulates up to μ + 2σ.

Conclusion. P(162 < H < 178) ≈ 68%; P(H > 186) ≈ 2.5%; the 97.5%-percentile height is 186 cm.

3. Faded example — fill in the missing steps

IQ scores are I ~ N(100, 15²). Fill in each blank. 4 marks

Step 1 — Express the boundaries as μ ± kσ:

70 = 100 − ____________ × 15 = μ − ____________ σ

130 = 100 + ____________ × 15 = μ + ____________ σ

Step 2 — Use the empirical rule on the interval 70 to 130:

P(70 < I < 130) = P(μ − 2σ < I < μ + 2σ) ≈ ____________ %

Step 3 — Find P(I > 145) using the upper-tail value:

145 = 100 + ____________ × 15 = μ + ____________ σ

P(I > μ + 3σ) ≈ ____________ %

Step 4 — Compute the z-score for I = 124:

z = (124 − 100) / 15 = ____________

Conclusion. P(70 < I < 130) ≈ ____________ %, P(I > 145) ≈ ____________ %, and z = ____________ for I = 124.

Stuck? Revisit lesson § Empirical Rule — the IQ example.

4. Graduated practice — z-scores and the empirical rule

For each item, show the substitution into z = (x − μ)/σ or identify the kσ boundary clearly before stating the answer.

Foundation — single z-score (4 questions)

Q	X ~ N(μ, σ²)	z-score
4.1 1	μ = 60, σ = 3; x = 66	z =
4.2 1	μ = 50, σ = 10; x = 30	z =
4.3 1	μ = 100, σ = 15; x = 100	z =
4.4 1	μ = 80, σ = 6; x = 71	z =

Standard — typical HSC difficulty (6 questions)

Show working in the space below each part.

4.5 For X ~ N(50, 100), use the empirical rule to find P(40 < X < 60). 2 marks

4.6 A machine fills bottles with μ = 500 mL, σ = 5 mL. Between what two volumes do approximately 95% of bottles fall? 2 marks

4.7 For X ~ N(50, 100), use the empirical rule to estimate P(X > 70). 2 marks

4.8 Find the raw score x for z = −1.5 if μ = 80 and σ = 6. 2 marks

4.9 For IQ ~ N(100, 15²), use the empirical rule to estimate P(I < 115). 2 marks

4.10 A student scores 75% in Maths (μ = 65, σ = 10) and 70% in English (μ = 60, σ = 8). Calculate the z-score for each and state which subject they performed better in relative to the cohort. 2 marks

Extension — combine concepts (2 questions)

4.11 Test scores X ~ N(60, 100). State P(X > 50) and P(X < 50) without computing any decimal, justifying using symmetry of the normal curve about the mean. 3 marks

4.12 Show that for X ~ N(μ, σ²), the value 1 standard deviation below the mean (x = μ − σ) always corresponds to z = −1, regardless of μ and σ. Then state in one line why z-scores let you compare values from different normal distributions. 3 marks

Stuck on 4.12? Substitute x = μ − σ directly into z = (x − μ)/σ.

5. Self-check the easy 3

Tick the first three once you've checked your method works.

For 4.1 I obtained z = (66 − 60)/3 = +2 and noted z > 0 means above the mean.

For 4.2 I obtained z = (30 − 50)/10 = −2 and noted z < 0 means below the mean.

For 4.3 I obtained z = 0 because x = μ — a score exactly at the mean always has z = 0.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — z-score formula

z = (x − μ) / σ.

Q1.2 — Empirical rule

68%, 95%, 99.7%.

Q1.3 — Standard normal

μ = 0; σ = 1. (So Z ~ N(0, 1).)

Q3 — Faded example, IQ ~ N(100, 15²)

Step 1: 70 = 100 − 2 × 15 = μ − 2σ; 130 = 100 + 2 × 15 = μ + 2σ.
Step 2: P(70 < I < 130) ≈ 95%.
Step 3: 145 = 100 + 3 × 15 = μ + 3σ; P(I > μ + 3σ) ≈ 0.15%.
Step 4: z = (124 − 100)/15 = 24/15 = 1.6.

Q4.1 — z for x = 66, μ = 60, σ = 3

z = (66 − 60)/3 = 2. (Two SDs above the mean.)

Q4.2 — z for x = 30, μ = 50, σ = 10

z = (30 − 50)/10 = −2. (Two SDs below the mean.)

Q4.3 — z for x = 100, μ = 100, σ = 15

z = (100 − 100)/15 = 0. (Exactly at the mean — that's what z = 0 means.)

Q4.4 — z for x = 71, μ = 80, σ = 6

z = (71 − 80)/6 = −9/6 = −1.5.

Q4.5 — P(40 < X < 60) for X ~ N(50, 100)

σ = √100 = 10. 40 = 50 − 10 = μ − σ; 60 = 50 + 10 = μ + σ. So P(40 < X < 60) ≈ 68%.

Q4.6 — Middle 95% of bottle volumes

95% ⇒ μ ± 2σ = 500 ± 10. So between 490 mL and 510 mL.

Q4.7 — P(X > 70) for X ~ N(50, 100)

70 = 50 + 20 = μ + 2σ. The upper tail beyond μ + 2σ is half of (100% − 95%) = 2.5%.

Q4.8 — Raw score for z = −1.5, μ = 80, σ = 6

x = μ + zσ = 80 + (−1.5)(6) = 80 − 9 = 71.

Q4.9 — P(I < 115) for IQ ~ N(100, 15²)

115 = 100 + 15 = μ + σ. P(I < μ + σ) = 50% + 34% = 84% (half the curve plus half of the 68% band).

Q4.10 — Maths vs English z-scores

z_Maths = (75 − 65)/10 = 1.0. z_English = (70 − 60)/8 = 10/8 = 1.25. English z is higher, so the student performed better in English relative to the cohort.

Q4.11 — Symmetry argument for P(X > 50) and P(X < 50)

For X ~ N(60, 100), σ = 10 and 50 = 60 − 10 = μ − σ. The normal curve is symmetric about μ. So P(X > 50) consists of the 34% inside μ − σ to μ plus the 50% above μ — giving P(X > 50) ≈ 84%. The complementary tail is P(X < 50) ≈ 16%. (No table or calculator needed — symmetry plus the 68% rule.)

Q4.12 — Why z = −1 always means "1 SD below"

Substitute x = μ − σ into z = (x − μ)/σ: z = ((μ − σ) − μ)/σ = −σ/σ = −1, regardless of μ and σ. This is why z-scores enable cross-distribution comparison: once you've standardised, "z = −1" carries the same meaning ("one SD below the mean") whether the original variable was IQ, height, or exam marks.