Example: battery lifetimes
What is given?
Battery lifetimes follow a normal model. Mean = 100 hours. SD = 15 hours.
Question A: find a chance
“What is the chance a battery lasts less than 120 hours?”
120 hours is given in the question. Use normalcdf to find the chance below it.
Question B: find a lifetime
“What lifetime has 90% of batteries below it?”
Here 90% is given, but the lifetime is unknown. Use invNorm to find it, in hours.
Select Below 120: coral shows batteries lasting less than the given 120 hours.
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What are −1E99 and 1E99?
Calculator stand-ins for far left and far right.
Below 120: lower = −1E99; upper = 120.
Where do μ and σ go?
After the bounds in normalcdf.
Mean 100, SD 15: normalcdf(lower,upper,100,15).
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Find a probability
Read the question
Scores have mean 100, SD 15. Find the chance of a score below 120.
Enter the four inputs
normalcdf(−1E99,120,100,15)
Lower end · upper end · mean · SD. −1E99 means far left.
Read the result
0.909 ≈ 90.9%. Probability × 100 gives the percentage.
Below 120: coral is the area we want.
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Which part of the bell curve do I want?: Below → left. Above → right. Between → middle. Try each button below.
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| Question wording | Meaning | Shade |
|---|---|---|
| Below 120 | Less than 120 | Left of 120 |
| Above 120 | More than 120 | Right of 120 |
| Between 85 and 120 | From 85 to 120 | Middle |
| Lowest 8% | 8% below the cutoff | Left tail |
| Highest 2% | 2% above the cutoff | Right tail |
Highest 2%: where do I shade?
The highest values are on the right. Shade the small right-hand tail.
Lowest 8%: where do I shade?
The lowest values are on the left. Shade the small left-hand tail.
Find the height for the tallest 10%
Read the question
Mean height 170 cm, SD 6 cm. Where do the tallest 10% begin?
Use the area on the left
10% is above the cutoff. 100% − 10% = 90% below. Enter 0.90.
Find the cutoff
invNorm(0.90,170,6) = 177.689… cm.
Answer: 178 cm to 3 significant figures.
Set left area to 90%. The remaining 10% is the tallest group.
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Fastest or slowest?
For times: fastest = smaller values; slowest = larger values.
Slowest 15%: enter 0.85 as the left area.
Top 2%: which calculator area?
2% above the cutoff → 98% below it.
invNorm needs the area below, so enter 0.98.
Bottom 8%: which calculator area?
8% below the cutoff is already the left area.
Enter 0.08 in invNorm.
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Find the middle 80%
Read the question
Diameters: mean 40 mm, SD 1.5 mm. Find the middle 80% interval.
Split the outside
20% outside → 10% per tail. Left areas at the two ends: 0.10 and 0.90.
Calculate both ends
invNorm(0.10,40,1.5) ≈ 38.1
invNorm(0.90,40,1.5) ≈ 41.9
Answer: 38.1–41.9 mm.
Coral = middle 80%. Each tail = 10%.
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An interval is not centred?
Subtract its area from the area below its known upper endpoint.
Example: mean 100, SD 15; P(k < X < 120) = 0.68.
P(X < 120) = 0.908788…
Left area at k = 0.908788… − 0.68 = 0.228788…
k = invNorm(0.228788…,100,15) ≈ 88.9.
Do not assume every 68% interval is μ ± σ.
Find the half-width m?
Find one endpoint, then its distance from the mean.
Central 60%, mean 4, SD 0.25: upper left area = 0.80. Upper endpoint = 4.210405…; m = 4.210405… − 4 = 0.210.
Chance → expected number
Read the question
200 batteries. Lifetimes have mean 100 hours, SD 15 hours. How many last more than 120 hours, on average?
Find the chance for one battery
normalcdf(120,1E99,100,15) = 0.0912112…
About 9.12% last beyond 120 hours.
Multiply by 200
200 × 0.0912112… ≈ 18.2 batteries on average. Use the unrounded probability.
Coral = batteries lasting more than 120 hours.
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Can an expected number be a decimal?
Yes: it is a long-run average.
Example: 18.2 items is an expected count, even though an actual count is whole.
Add a transfer time?
Find the arrival cutoff, then add the transfer time.
Example: arrival mean 20 minutes late, SD 10; 99% cutoff = invNorm(0.99,20,10) = 43.263… minutes late. Add 10 minutes to change → 53.263… minutes after the scheduled arrival.
If asked for the earliest WHOLE minute meeting at least 99%, round UP to 54 minutes. A nearest-minute instruction is different.
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A group count → a cutoff
Read the question
80 animals. Masses have mean 30 kg, SD 4 kg. The heaviest 12 animals are above a cutoff. Find it.
Turn 12 animals into an area
12 ÷ 80 = 0.15 above the cutoff.
Left area = 1 − 0.15 = 0.85.
Find the mass
invNorm(0.85,30,4) ≈ 34.1 kg.
85% below the cutoff; the heaviest 15% above it.
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One example, four exam tasks
Read the situation
Bag masses: mean 250 g, SD 12 g. Use the same mean and SD in each part.
Chance above 270 g
normalcdf(270,1E99,250,12) ≈ 0.0478.
Chance between 240 and 260 g
normalcdf(240,260,250,12) ≈ 0.595.
Expected bags below 235 g, out of 150
150 × normalcdf(−1E99,235,250,12) ≈ 15.8 bags.
Cutoff for the heaviest 8%
Heaviest = right-hand tail: 8% is above the cutoff.
invNorm needs the percentage below it: 100% − 8% = 92% → 0.92.
invNorm(0.92,250,12) ≈ 267 g (3 s.f.). Shade above the cutoff, to the right.
Above 270 is shaded. Try Between to compare regions.
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What if it says lightest 8%?
Lightest = left-hand tail. 8% is already below the cutoff.
Enter 0.08, not 0.92:
invNorm(0.08,250,12) ≈ 233 g (3 s.f.).
Heaviest: subtract from 1. Lightest: use the percentage directly.
Two independent measurements?
If one meets the condition with probability q, both do with probability q².
This uses the independence rule from 4.6.
Normal probability AND another event?
Use the conditional rate for that group.
P(heavy and lost) = P(heavy) × P(lost | heavy).
Example: 0.252492… × 1/12 ≈ 0.0210.
The question specifies distance from zero?
For 0 < D < 10, enter BOTH bounds.
normalcdf(0,10,26,7) ≈ 0.0110.
The next part counts qualifying items?
Carry the unrounded normal probability into the binomial method already taught in 4.8.1. Do not round it first.
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Check the spread
N(100,15²) means SD 15, variance 225. Enter 15.
Check the region
Below → left. Above → right. invNorm always needs the left area.
Check the answer
Probability: 0 to 1. Measurement: include units. Expected number: group size × probability.
Check the shaded side matches your question.
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Normal values have no ±1 boundary shift
This is a continuous model.
Example: P(X > 120) uses 120, not 119. Binomial count rules do not apply.
How much working?
Write the event and calculator inputs before the result.
Example: P(X > 120) = normalcdf(120,1E99,100,15) ≈ 0.0912.
Very small probability?
Keep significant figures.
Example: 0.000303 = 3.03 × 10⁻⁴, not 0.000.
Variance is not SD
N(100,15²) means SD = 15, variance = 225.
Enter 15 in normalcdf.
Probability or expected number?
A probability is between 0 and 1.
Among 200 items with probability 0.073, expected count = 200 × 0.073 = 14.6.
A percentage requested?
Multiply the probability by 100.
Example: 0.073 → 7.3%.
A sketch requested?
Include a recognisable bell, labelled mean and cutoff, and the correct shaded region.
Additional AA exam practice
IB-style question — read the parameters
The masses of apples are modelled by X ~ N(150, 20²) grams.
State the mean and standard deviation, and the value the distribution is symmetric about.
Step by step
- N(μ, σ²): first is the mean, second is the variance.
- Symmetric about the mean.
Final answer
Mean 150 g, standard deviation 20 g, symmetric about 150 g.
IB-style question — P(X < a)
X ~ N(50, 8²).
Find P(X < 60) and P(X > 60).
Step by step
- P(X < 60) = normalcdf(−∞, 60, 50, 8).
- P(X > 60) is the complement.
Final answer
P(X < 60) ≈ 0.894; P(X > 60) ≈ 0.106.
IB-style question — find σ from the IQR
A normally distributed variable has interquartile range 13.49.
Find its standard deviation σ.
Step by step
- The quartiles sit 0.6745 standard deviations either side of the mean (invNorm(0.75) = 0.6745), so the IQR spans 2 × 0.6745 = 1.349 standard deviations.
- Solve for σ.
Final answer
σ = 10.
IB-style question — conditional on a normal region
X ~ N(20, 4²). Find P(X > 24 | X > 18).
Step by step
- Conditional: keep only X > 18, take the X > 24 slice.
- Each probability from normalcdf on the GDC.
Final answer
≈ 0.230.
IB-style question — find μ and σ from two probabilities
X ~ N(μ, σ²), with P(X < 50) = 0.2 and P(X < 70) = 0.9.
Find μ and σ.
Step by step
- Standardise each probability using invNorm to get the z-values, then write two equations.
- Subtract to eliminate μ.
- Back-substitute for μ.
Final answer
μ ≈ 57.9, σ ≈ 9.42.
IB-style question — empirical rule
Heights are X ~ N(170, 10²) cm.
Use the empirical rule to estimate P(160 < X < 180).
Step by step
- 160 and 180 are μ − σ and μ + σ.
- Within 1σ → empirical rule.
Final answer
About 0.68 (68%).
IB-style question — how many
Bolt lengths are X ~ N(50, 2²) mm.
A bolt is rejected if it is longer than 53 mm.
In a batch of 400 bolts, find the expected number rejected.
Step by step
- Probability one bolt is too long.
- Expected number = probability × 400.
Final answer
About 27 bolts are expected to be rejected.
IB-style question — normalcdf
Reaction times are X ~ N(80, 6²) milliseconds.
Find P(74 < X < 92).
Step by step
- Two bounds, so use normalcdf(lower, upper, μ, σ).
- Read the answer off the GDC (3 s.f.).
Final answer
P(74 < X < 92) ≈ 0.819.