The idea, step by step
Area under the curve represents a proportion
A normal model is continuous, symmetric and bell-shaped. Its mean sets the centre and its standard deviation sets the horizontal scale. The mean, median and mode coincide in an ideal normal distribution. Total area under the curve is 1; an interval probability is the area above that interval. A single exact value has probability zero in a continuous model.
Use the empirical rule as an approximation
For a normal model, about 68% lie within one standard deviation of the mean,95% within two and 99.7% within three. Symmetry splits the remaining tails equally. These rounded values are useful for estimates, not substitutes for technology when an accurate normal probability is requested. Check whether a normal shape is plausible before applying them to a strongly skewed data set.
A common trap
Using 68% between the mean and one standard deviation above it.
Instead: 68% covers both sides from mu−sigma to mu+sigma; one half is about 34%.
See the thinking
A central interval
A normal model has mean 70 and standard deviation 8. Estimate the proportion between 54 and 86.
- 54=70−2×8 and 86=70+2×8.
- The interval is within 2 standard deviations of the mean, so use 95%.
An upper tail
Bottle-fill volumes are approximately normal with mean 500 mL and standard deviation 4 mL. Estimate how many of 2,000 bottles exceed 508 mL.
- 508 mL is 2 standard deviations above the mean.
- The upper tail is approximately(100−95)/2=2.5%.
- Expected count=2000×0.025=50; an actual sample count may differ.
Check your understanding
Estimate the proportion below the mean in a normal model.
Show answer
50%.
The distribution is symmetric and continuous.
Your turn
6 questions · 13 marks · Work at your own pace.
Read the questions online
- A normal model has mean 30 and sd 5. Give the interval containing about 68%.2 marks
- For the same model, estimate the proportion above 40.2 marks
- Estimate the proportion between the mean and 1 standard deviation above it.2 marks
- Estimate the proportion more than 3 standard deviations from the mean.2 marks
- A normal model has mean 100 and sd 15. Estimate how many of 800 observations lie from 85 to 115.2 marks
- A distribution of waiting times has a long right tail. Is it justified to assume 68% lie within one sd? Explain.3 marks
Read the worked solutions
Equivalent correct methods earn the corresponding marks. Include units where asked.
- A normal model has mean 30 and sd 5. Give the interval containing about 68%.2 marks
Use mean ± 1 standard deviation.
25 to 35. - For the same model, estimate the proportion above 40.2 marks
40 is 2 standard deviations above 30; split the 5% outside ±2 standard deviations into two tails.
About 2.5%. - Estimate the proportion between the mean and 1 standard deviation above it.2 marks
Take half of the central 68%.
About 34%. - Estimate the proportion more than 3 standard deviations from the mean.2 marks
100%−99.7%=0.3%, including both tails.
About 0.3%. - A normal model has mean 100 and sd 15. Estimate how many of 800 observations lie from 85 to 115.2 marks
The interval is ±1 standard deviation, so 0.68×800=544.
About 544. - A distribution of waiting times has a long right tail. Is it justified to assume 68% lie within one sd? Explain.3 marks
The empirical rule is specific to an approximately normal shape.
A long right tail gives reason to question that model.
No, not from that information.
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 ↗