Mathematics Advanced • Year 12 • Module 5 • Lesson 13

Normal Distribution Applications

Build fluency in left/right/interval probability calculations and the inverse-normal flow x = μ + zσ.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Complete the three probability rules for X ~ N(μ, σ²):

P(X < x) = P(Z < ____________)

P(X > x) = ____________ − P(Z < z)

P(a < X < b) = P(Z < z_b) − P(Z < ____________)

Q1.2 Complete the inverse-normal flow (given probability, find x):

1. Translate the question to P(Z ____ z) = ____________ (a known value).

2. Look up z using ____________ tables or invNorm on a calculator.

3. Un-standardise: x = ____________ + z × ____________

Q1.3 Fill in the key z-values:

P(Z < z) = 0.90 ⇒ z ≈ ____________ P(Z < z) = 0.95 ⇒ z ≈ ____________ P(Z < z) = 0.975 ⇒ z ≈ ____________

Stuck? Revisit lesson § Formula Reference and § Copy Into Your Books — key z-values.

2. Worked example — chocolate bar weights X ~ N(250, 5²)

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

Problem. Bars are X ~ N(250, 25) grams. Find P(X < 242), P(X > 258), P(240 < X < 260) and the weight x such that P(X > x) = 0.10.

Step 1 — Standardise each boundary.

z₂₄₂ = (242 − 250)/5 = −1.6 z₂₅₈ = (258 − 250)/5 = +1.6
z₂₄₀ = (240 − 250)/5 = −2 z₂₆₀ = (260 − 250)/5 = +2

Reason: every problem becomes a standard-normal problem after z = (x − μ)/σ.

Step 2 — Read each probability from standard normal tables / calculator.

P(X < 242) = P(Z < −1.6) = 1 − P(Z < 1.6) = 1 − 0.9452 = 0.0548
P(X > 258) = P(Z > 1.6) = 1 − 0.9452 = 0.0548
P(240 < X < 260) = P(Z < 2) − P(Z < −2) ≈ 0.9772 − 0.0228 = 0.9544

Reason: by symmetry, P(Z < −a) = 1 − P(Z < a).

Step 3 — Inverse-normal: find x such that P(X > x) = 0.10.

P(X > x) = 0.10 ⇒ P(Z > z) = 0.10 ⇒ P(Z < z) = 0.90
z ≈ 1.282 (from tables)
x = μ + zσ = 250 + 1.282(5) = 250 + 6.41 = 256.41 g

Reason: invert by reading z from the 0.90 cumulative entry, then un-standardise.

Conclusion. About 5.5% of bars are below 242 g; about 5.5% above 258 g; about 95.4% lie between 240 and 260 g; and only the heaviest 10% exceed 256.4 g.

3. Faded example — fill in the missing steps

Lengths of bolts are L ~ N(100, 0.8²) mm. Fill in each blank. 4 marks

Step 1 — Find P(L < 99):

z = (99 − 100)/0.8 = ____________

P(L < 99) = P(Z < ____________) = 1 − P(Z < 1.25) ≈ 1 − 0.8944 = ____________

Step 2 — Find P(L > 101.5):

z = (101.5 − 100)/0.8 = ____________

P(L > 101.5) = 1 − P(Z < ____________) ≈ 1 − 0.9696 = ____________

Step 3 — Find x such that P(L < x) = 0.95 (95th percentile of bolt length):

z ≈ ____________ (from tables)

x = 100 + ____________ × 0.8 = ____________ mm

Conclusion. P(L < 99) ≈ ____________, P(L > 101.5) ≈ ____________, 95th percentile ≈ ____________ mm.

Stuck? Revisit lesson § Common phrasings table — "90th / 95th percentile".

4. Graduated practice — left / right / interval / inverse

Show the line standardising to z, then state the probability or x value.

Foundation — direct probability lookup (4 questions)

Q	Find	z and answer
4.1 1	X ~ N(100, 144). Find z for x = 112.
4.2 1	For the same X, find P(X < 112). [Use P(Z < 1) ≈ 0.8413.]
4.3 1	For the same X, find P(X > 88). [By symmetry.]
4.4 1	For X ~ N(50, 100), find P(X > 70). [Use P(Z < 2) ≈ 0.9772.]

Standard — typical HSC difficulty (6 questions)

Show working in the space below each question.

4.5 Fish lengths are L ~ N(35, 4) cm. Find P(L > 38). 2 marks

4.6 Components are M ~ N(50, 0.4²) mm. Rejected if outside 49.2-50.8 mm. Find the percentage rejected. 2 marks

4.7 Exam marks E ~ N(68, 100). Top 15% receive a Distinction. Find the minimum Distinction mark using z ≈ 1.036 for P(Z < z) = 0.85. 2 marks

4.8 IQ ~ N(100, 15²). Find the 98th percentile using z ≈ 2.054. 2 marks

4.9 X ~ N(100, 144). Find P(94 < X < 106). [Use P(Z < 0.5) ≈ 0.6915.] 2 marks

4.10 A scholarship goes to the top 8% of applicants. Scores are S ~ N(65, 100). Find the cut-off using z ≈ 1.405 for P(Z < z) = 0.92. 2 marks

Extension — combine concepts (2 questions)

4.11 Blood cholesterol in healthy adults is C ~ N(5.2, 0.8²) mmol/L. A doctor flags any reading above the 95th percentile as "elevated". Find the threshold and state in one sentence what proportion of healthy adults will be flagged in error if it is used as a screening rule. Use z ≈ 1.645. 3 marks

4.12 A manufacturer wants fewer than 1% of products below the minimum label weight. If μ = 500 g and σ = 8 g, what is the highest minimum weight they can claim? Use z ≈ −2.326 for P(Z < z) = 0.01. 3 marks

Stuck on 4.12? Set P(X < x_min) = 0.01 and solve x_min = μ + zσ with z = −2.326.

5. Self-check the easy 3

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

For 4.1 I computed z = (112 − 100)/12 = 1 (positive — above the mean).

For 4.2 I read P(Z < 1) directly as 0.8413 (≈ 84.1%).

For 4.3 I used symmetry: P(X > 88) = P(Z > −1) = P(Z < 1) = 0.8413.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Three probability rules

P(X < x) = P(Z < z); P(X > x) = 1 − P(Z < z); P(a < X < b) = P(Z < z_b) − P(Z < z_a).

Q1.2 — Inverse-normal flow

(1) P(Z < z) = (the cumulative probability). (2) Inverse normal / standard normal tables. (3) x = μ + z × σ.

Q1.3 — Key z-values

P(Z < z) = 0.90 ⇒ z ≈ 1.282. P(Z < z) = 0.95 ⇒ z ≈ 1.645. P(Z < z) = 0.975 ⇒ z ≈ 1.96.

Q3 — Faded example, L ~ N(100, 0.64)

Step 1: z = (99 − 100)/0.8 = −1.25; P(L < 99) = P(Z < −1.25) = 1 − 0.8944 ≈ 0.1056.
Step 2: z = (101.5 − 100)/0.8 = 1.875; P(L > 101.5) = 1 − P(Z < 1.875) ≈ 1 − 0.9696 ≈ 0.0304.
Step 3: z ≈ 1.645; x = 100 + 1.645 × 0.8 = 101.32 mm.

Conclusion: P(L < 99) ≈ 10.56%, P(L > 101.5) ≈ 3.04%, 95th percentile ≈ 101.3 mm.

Q4.1 — z for x = 112, X ~ N(100, 144)

σ = √144 = 12. z = (112 − 100)/12 = 1.

Q4.2 — P(X < 112)

P(Z < 1) ≈ 0.8413.

Q4.3 — P(X > 88) for X ~ N(100, 144)

z = (88 − 100)/12 = −1. P(X > 88) = P(Z > −1) = P(Z < 1) ≈ 0.8413.

Q4.4 — P(X > 70) for X ~ N(50, 100)

z = (70 − 50)/10 = 2. P(X > 70) = 1 − P(Z < 2) ≈ 1 − 0.9772 = 0.0228 (≈ 2.3%).

Q4.5 — Fish, L ~ N(35, 4), P(L > 38)

σ = √4 = 2. z = (38 − 35)/2 = 1.5. P(L > 38) = 1 − P(Z < 1.5) ≈ 1 − 0.9332 = 0.0668 (about 6.7%).

Q4.6 — Components rejection rate

z_lower = (49.2 − 50)/0.4 = −2; z_upper = (50.8 − 50)/0.4 = 2. P(−2 < Z < 2) ≈ 0.9545. Rejected = 1 − 0.9545 = 0.0455 (≈ 4.6%).

Q4.7 — Distinction cut-off (top 15%)

P(E > x) = 0.15 ⇒ P(Z < z) = 0.85 ⇒ z ≈ 1.036. x = 68 + 1.036(10) = 68 + 10.36 = 78.36 ≈ 78.

Q4.8 — IQ 98th percentile

z ≈ 2.054. x = 100 + 2.054(15) = 100 + 30.81 = 130.8 ≈ 131.

Q4.9 — P(94 < X < 106) for X ~ N(100, 144)

z₉₄ = −0.5; z₁₀₆ = +0.5. P = P(Z < 0.5) − P(Z < −0.5) = 0.6915 − 0.3085 = 0.383 (about 38.3%).

Q4.10 — Scholarship cut-off (top 8%)

P(Z < z) = 0.92 ⇒ z ≈ 1.405. x = 65 + 1.405(10) = 65 + 14.05 = 79.05 ≈ 79.

Q4.11 — Cholesterol 95th-percentile threshold

z ≈ 1.645. x = 5.2 + 1.645(0.8) = 5.2 + 1.316 = 6.516 mmol/L. 5% of healthy adults will exceed this threshold and be flagged — a 5% false-positive rate by construction.

Q4.12 — Highest minimum label weight

P(X < x_min) = 0.01 ⇒ z = −2.326. x_min = μ + zσ = 500 + (−2.326)(8) = 500 − 18.61 = 481.4 g. The company can claim "at least 481 g" with fewer than 1% of bottles falling short.