Estimating proportions in normal models with appropriate technology | Year 11

01 / LEARN

Understand the idea.

Standardise and select the correct area

A standardised score $z = \frac{x - μ}{σ}$ measures how far a cutoff is from the mean in standard deviations. Negative scores lie below the mean and positive scores above it. Use standard deviation $σ > 0$ , not variance, in this calculation.

The standard normal cumulative function $Φ (z)$ returns area to the left. For an interval, subtract the cumulative area at the lower bound from that at the upper bound. For a right tail, subtract the left cumulative area from $1$ .

P (a < X < b) = Φ (\frac{b - μ}{σ}) - Φ (\frac{a - μ}{σ})

Report a useful probability in its context

Use either a normal-CDF function with the original bounds, mean and standard deviation, or standardise the bounds first. Sketch or describe the intended area before entering values. Use an infinity setting where available; otherwise choose finite bounds with negligible omitted tail area.

Report probabilities as proportions or percentages in context. Multiplying by a group size gives an expected count, not a guarantee. For percentiles, inverse normal uses the cumulative area to the left; check whether the resulting cutoff belongs above or below the mean.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

An interval with unequal distances

Let X have a normal model with mean 100 and sd 15. Find P(85<X<122.5).

Worked solution

1
Standardise both bounds.
$\begin{aligned} z_{lower} & = \frac{85 - 100}{15} = - 1 \\ z_{upper} & = \frac{122.5 - 100}{15} = 1.5 \end{aligned}$
Both bounds must be expressed on the same standard normal scale.
2
Select the interval area.
$P (85 < X < 122.5) = Φ (1.5) - Φ (- 1)$
Subtracting the lower left-tail area removes values below the requested interval.
3
Evaluate the cumulative probabilities.
$\begin{aligned} P (85 < X < 122.5) & \approx 0.9331928 - 0.1586553 \\ = 0.7745375 \end{aligned}$
The normal CDF supplies the two cumulative areas.
4
Report a rounded proportion and percentage.
$P (85 < X < 122.5) \approx 0.7745 \approx 77.45 %$
Multiplying the unrounded proportion by 100 converts it to a percentage.

Answer

Approximately 0.7745, or 77.45%.

EXAMPLE 02

A specification limit

A machine produces lengths with normal mean 50 mm and sd 2 mm. Find the proportion above 53 mm.

Worked solution

1
Standardise the length cutoff.
$z = \frac{53 - 50}{2} = 1.5$
The cutoff is 1.5 standard deviations above the normal mean.
2
Select the upper tail.
$P (X > 53) = 1 - Φ (1.5)$
The cumulative function gives the area below 53 mm, so its complement gives the area above.
3
Evaluate and report the proportion.
$\begin{aligned} P (X > 53) & \approx 1 - 0.9331928 \\ = 0.0668072 \\ \approx 0.0668 = 6.68 % \end{aligned}$
The positive cutoff leaves a small upper tail: approximately 6.68% of lengths exceed 53 mm.

Answer

Approximately 0.0668, or 6.68%.

03 / YOUR TURN

Try it yourself.

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 & reasoning

Near 0.16.

A cutoff at $z = 1$ has about $0.84$ of the normal area below it. The area above is its complement, about $1 - 0.84 = 0.16$ .

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

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
$\frac{66 - 60}{4} = 1.5$ .
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
$z = \frac{105 - 120}{10} = - 1.5$ .
$Φ (- 1.5) = 0.0668072$ .
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
$400 \times 0.023 = 9.2$ .
A particular batch has an integer count and random variation.
Expected failures 9.2, so roughly 9 in the long-run-average sense.

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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 ↗

In this lesson

Understand the idea.

Standardise and select the correct area

Report a useful probability in its context

See the method in action.

An interval with unequal distances

Standardise both bounds.

Select the interval area.

Evaluate the cumulative probabilities.

Report a rounded proportion and percentage.

A specification limit

Standardise the length cutoff.

Select the upper tail.

Evaluate and report the proportion.

Try it yourself.

Check your understanding

Keep practising