The idea, step by step
Standardise and select the correct area
For a normal quantity X with mean mu and standard deviation sigma, the standardized score z=(x−mu)/sigma measures distance from the mean in standard deviations. A cumulative normal function returns the area to the left of a cutoff. Subtract two cumulative areas for an interval, or subtract from 1 for a right tail. State the model parameters and use sigma, not variance, in a calculator field asking for standard deviation.
Report a useful probability in its context
Use a normal-CDF feature with lower/upper bounds, mean and standard deviation, or standardize first and use a standard-normal function. Use sufficiently wide practical bounds for an infinite tail if your calculator requires finite input. A probability is not a guaranteed frequency: multiplying it by a group size gives an expected count. For a percentile, use inverse normal with the cumulative area to the left, then check that the cutoff lies on the expected side of the mean.
A common trap
Entering the upper-tail probability directly into inverse normal, which expects a left cumulative area.
Instead: Convert first: the highest 10% begins at left cumulative area 0.90.
See the thinking
An interval with unequal distances
Let X have a normal model with mean 100 and sd 15. Find P(85<X<122.5).
- The standardized bounds are−1 and 1.5.
- Use Φ(1.5)−Φ(−1).
- 0.9331928−0.1586553=0.7745375.
A specification limit
A machine produces lengths with normal mean 50 mm and sd 2 mm. Find the proportion above 53 mm.
- z=(53−50)/2=1.5.
- The right-tail area is 1−Φ(1.5)=0.0668072.
- For 1,000 items the expected number above 53 mm is about 66.8, or 67 items.
Check your understanding
A cutoff is 1 standard deviation above the mean. Is the probability above it near 0.84 or 0.16?
Show answer
Near 0.16.
About 84% lie below that positive z-score; the right tail is the complement.
Your turn
6 questions · 13 marks · Work at your own pace.
Read the questions online
- For normal mean 60 and sd 4, find the z-score of 66.2 marks
- Using Φ(1.25)=0.89435, find P(Z>1.25).2 marks
- Using Φ(0.5)=0.69146 and Φ(−1)=0.15866, find P(−1<Z<0.5).2 marks
- For normal mean 120 and sd 10, estimate P(X<105) using a normal-CDF tool.2 marks
- The inverse standard-normal value for cumulative area 0.90 is 1.28155. Find the 90th percentile of normal mean 80 and sd 6.2 marks
- A normal model predicts a failure probability 0.023 for a batch of 400. Interpret the expected number and explain why it is not a promise.3 marks
Read the worked solutions
Equivalent correct methods earn the corresponding marks. Include units where asked.
- For normal mean 60 and sd 4, find the z-score of 66.2 marks
(66−60)/4=1.5.
1.5. - Using Φ(1.25)=0.89435, find P(Z>1.25).2 marks
1−0.89435=0.10565.
0.10565. - Using Φ(0.5)=0.69146 and Φ(−1)=0.15866, find P(−1<Z<0.5).2 marks
0.69146−0.15866=0.53280.
0.53280. - For normal mean 120 and sd 10, estimate P(X<105) using a normal-CDF tool.2 marks
z=(105−120)/10=−1.5.
Φ(−1.5)=0.0668072.
Approximately 0.0668. - The inverse standard-normal value for cumulative area 0.90 is 1.28155. Find the 90th percentile of normal mean 80 and sd 6.2 marks
x=80+6(1.28155)=87.6893.
Approximately 87.69. - A normal model predicts a failure probability 0.023 for a batch of 400. Interpret the expected number and explain why it is not a promise.3 marks
400×0.023=9.2.
A particular batch has an integer count and random variation.
Expected failures 9.2, so roughly 9 in the long-run-average sense.
Curriculum reference & lesson scope
SACE Board of South Australia
Stage 1 Mathematics: counting, binomial expansion, data, samples, summary measures and introductory normal distributions. These lessons provide the Counting and Statistics strand of the approved Methods preparation route.
Official curriculum source ↗