Interpreting distribution results | Year 12

01 / LEARN

Understand the idea.

Separate a prediction from a guarantee

Report what a model result measures. An expected count is a long-run average, while a tail probability describes the chance of crossing a specified threshold. Include the event, observation period and units so the result addresses the decision.

A mean is neither a guaranteed count nor a lower bound. If $E (X) = 2$ faults, a fault-free sample can still occur. A question about any fault requires $P (X \geq 1)$ ; a question about exceeding capacity needs the corresponding tail probability.

Keep the model open to challenge

Every probability conclusion depends on its model assumptions. Check whether the trial count, success probability and independence remain plausible. An unusual event can occur under a suitable model; observing it once does not logically prove that the model is false.

Across $m$ comparable repetitions, the expected number of occurrences of an event is $m P (A)$ . The observed count is still an integer that can differ. Retain enough precision to distinguish a very small probability from an impossible event.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Zero is possible even when the mean is positive

The daily count X is modelled by Bin(10, 0.2). A manager claims there must be at least one success because E(X) = 2. Assess the claim.

Worked solution

1
Calculate and identify the expected count.
$E (X) = 10 (0.2) = 2$
This is an average over comparable days, not a compulsory minimum for each day.
2
Test the claimed impossibility of zero.
$P (X = 0) = {(1 - 0.2)}^{10} = {0.8}^{10} = 0.1073741824$
A zero count requires failure on all ten independent trials.
3
Assess the claim using the positive probability.
$P (X = 0) \approx 0.1074 > 0$
Zero successes occurs on about 10.7% of days under the model, so at least one is not guaranteed.

Answer

The mean is not a lower bound. P(X = 0) ≈ 0.1074.

EXAMPLE 02

Expected number of threshold crossings

Each of 50 independent groups has 5 independent trials with success probability 0.2. How many groups are expected to have at least one success?

Worked solution

1
Find the probability that one group qualifies.
$q = 1 - P (no success) = 1 - {0.8}^{5} = 0.67232$
The five independent trials all fail with probability 0.8 to the fifth power.
2
Use one indicator for each qualifying group.
$G = I_{1} + \dots + I_{50}, E (I_{j}) = q$
Each indicator records at least one success, rather than the number of successes inside its group.
3
Add the expectations and interpret.
$E (G) = 50 q = 50 (0.67232) = 33.616$
The expected count is about 33.6 groups on average; an actual count must be an integer and need not equal 34.

Answer

About 33.6 groups on average, not a guarantee of 34 groups.

03 / YOUR TURN

Try it yourself.

Check your understanding

An event has model probability 0.12. In 200 repetitions, how many occurrences are expected?

Show answer & reasoning

24.

The expected count is $200 (0.12) = 24$ . This is an average under the model, so a particular set of repetitions can contain a different count.

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

A sample fault count has mean 1.8. Explain whether any sample can contain exactly 1.8 faults.
2 marks
A model gives probability 0.04 that demand exceeds capacity. Find the expected number of exceedances in 250 comparable days.
2 marks
$X \sim Bin (5, 0.1)$ . Is observing $X = 0$ inconsistent with the model? Quantify your answer.
3 marks
Two systems have equal expected daily cost. Explain why that alone does not establish which one a risk-averse manager should choose.
2 marks
A binomial probability is displayed as 0.0000 after rounding. Must the event be impossible?
2 marks
A return rate estimated last year is used for this year’s changed product range. Give a specific limitation of a binomial forecast using the old rate.
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

A sample fault count has mean 1.8. Explain whether any sample can contain exactly 1.8 faults.
2 marks
Expectation is a weighted average across possible counts.
It is not an individual sample outcome.
No; the count is an integer, while its mean need not be.
A model gives probability 0.04 that demand exceeds capacity. Find the expected number of exceedances in 250 comparable days.
2 marks
Expected count $= 250 \times 0.04 = 10$ .
This is an average over repetitions, not a fixed schedule.
10 days.
$X \sim Bin (5, 0.1)$ . Is observing $X = 0$ inconsistent with the model? Quantify your answer.
3 marks
Zero is in the support.
Its probability is about 59%, so it is not even unusual under the model.
No; $P (X = 0) = {0.9}^{5} = 0.59049$ .
Two systems have equal expected daily cost. Explain why that alone does not establish which one a risk-averse manager should choose.
2 marks
Equal means match only one summary.
The manager needs the risk criteria and further distribution information.
Their variability and high-cost tail probabilities may differ.
A binomial probability is displayed as 0.0000 after rounding. Must the event be impossible?
2 marks
A positive probability below 0.00005 rounds to 0.0000 at four decimal places.
Use more precision or scientific notation when the distinction matters.
No.
A return rate estimated last year is used for this year’s changed product range. Give a specific limitation of a binomial forecast using the old rate.
2 marks
The binomial calculation is conditional on its $p$ value.
Changed products can alter the return probability, so investigate updated data before trusting the forecast.
The success probability may no longer be appropriate or constant.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods: discrete random variables and probability distributions, expected value and variance, transformations and independent sums, Bernoulli and binomial models. Original supporting lessons and exercises follow the approved Ascent chapter structure.

Official curriculum source ↗

In this lesson

Understand the idea.

Separate a prediction from a guarantee

Keep the model open to challenge

See the method in action.

Zero is possible even when the mean is positive

Calculate and identify the expected count.

Test the claimed impossibility of zero.

Assess the claim using the positive probability.

Expected number of threshold crossings

Find the probability that one group qualifies.

Use one indicator for each qualifying group.

Add the expectations and interpret.

Try it yourself.

Check your understanding

Keep practising