2026 Prediction Examination
Level 2 Probability
91267 Apply probability methods in solving problems
Credits: Four
| Achievement | Achievement with Merit | Achievement with Excellence |
|---|---|---|
| Apply probability methods in solving problems. | Apply probability methods, using relational thinking, in solving problems. | Apply probability methods, using extended abstract thinking, in solving problems. |
Instructions
- Suggested time: 60 minutes.
- You should attempt ALL the questions in this paper.
- Show ALL relevant working.
- A scientific calculator is permitted.
- The formula sheet opens from the margin beside any question.
- Use the margin beside each question for its timer and worked answer.
- Start the timer below when you are ready to begin.
Main Exam Timer
Our 2026 prediction paper, written from the 2023 to 2025 papers and this year's assessment specification. Every question has a worked answer.
A Maths Study Hub practice paper. Not an official NZQA examination.
QUESTION ONE
(a)
Kiwifruit are one of New Zealand’s most valuable horticultural exports. Before they are packed, kiwifruit are weighed one at a time, because the weight of each kiwifruit decides where it is sent and how much the grower is paid for it.
At Ridgeline Orchard, the weights of the green kiwifruit picked this season can be modelled by a normal distribution with a mean of 104 g and a standard deviation of 15 g.
Kiwifruit that weigh less than 80 g are too small to export, so they are sent to a factory to be made into juice.
Find the probability that a randomly chosen kiwifruit from Ridgeline Orchard is sent to be made into juice.
What percentage of the kiwifruit from Ridgeline Orchard weigh between 90 g and 115 g?
A supermarket chain buys only the heaviest 15% of the orchard’s kiwifruit.
What is the minimum weight of a kiwifruit that is sent to the supermarket chain?
(b)
The orchard manager, Lucas, wants no more than 3% of next season’s kiwifruit to be sent to be made into juice.
Lucas thinks that thinning the young fruit earlier in the season will make the kiwifruit heavier on average, without changing the standard deviation of their weights.
Calculate the mean weight that next season’s kiwifruit would need to have.
The orchard next door does not record the mean or the standard deviation of its kiwifruit weights. However, its packhouse report for this season shows that 10% of its green kiwifruit weighed less than 85 g, and 20% weighed more than 125 g. Assume that the weights of the neighbouring orchard’s kiwifruit can also be modelled by a normal distribution.
The neighbouring grower claims that, because the mean weight of their kiwifruit is higher than Ridgeline Orchard’s, a smaller proportion of their kiwifruit will be sent to be made into juice.
Calculate the mean and the standard deviation of the weights of the neighbouring orchard’s kiwifruit, and use them to evaluate the neighbouring grower’s claim.
(c)
At the packhouse, Lucas weighs a random sample of 250 kiwifruit from one bin. The bin is labelled as green kiwifruit from Ridgeline Orchard. The weights of the kiwifruit in the sample are shown in the graph below.
Lucas’s calculator gives the sample a mean of 105.0 g and a standard deviation of 15.1 g.
Estimate the median weight of the kiwifruit in the sample.
Lucas says:
“The mean and the standard deviation of the sample are almost exactly the same as our model’s. This shows that the normal model, with a mean of 104 g and a standard deviation of 15 g, describes the kiwifruit in this bin well.”
Do you agree with Lucas?
Compare the distribution of the weights in the sample with the normal model, discussing shape, centre and spread. Use calculations to support your answer, and suggest a reason, in context, for what you notice.
QUESTION TWO
(a)
After picking, kiwifruit are taken in bins to a packhouse, where they are graded and packed for export. At one packhouse, every kiwifruit passes under a camera grader. The grader looks for defects, such as bruises, scars and misshapen fruit, and rejects any kiwifruit that it judges to have a defect. Rejected kiwifruit cannot be exported.
During one week, the packhouse received kiwifruit from three growers. The packhouse found that:
- 45% of the kiwifruit came from Grower A, 35% came from Grower B, and the rest came from Grower C.
- 4% of Grower A’s kiwifruit had a defect, compared with 6% of Grower B’s kiwifruit and 10% of Grower C’s kiwifruit.
- The grader rejected 90% of the kiwifruit that had a defect. However, it also rejected 2% of the kiwifruit that had no defect.
Draw a probability tree to show this information, and use it to answer the parts below.
What is the probability that a randomly chosen kiwifruit came from Grower A, had a defect, and was not rejected by the grader?
The grader checks a batch of 3000 kiwifruit, which came from the three growers in the proportions given above.
How many of the kiwifruit in the batch would you expect to have a defect?
(b)
Grower C’s kiwifruit were 2.5 times as likely to have a defect as Grower A’s kiwifruit.
The manager of Grower C’s orchard, Grace, says that this means Grower C’s kiwifruit must also have been 2.5 times as likely to be rejected by the grader as Grower A’s kiwifruit.
Is Grace correct? Justify your answer with calculations, and explain why this happens.
What proportion of the kiwifruit that were rejected by the grader actually had no defect?
(c)
The packhouse recalibrates the grader, to reduce the number of kiwifruit with no defect that it rejects. The recalibrated grader still rejects 90% of the kiwifruit that have a defect, but now 85% of the kiwifruit that it rejects actually have a defect.
The proportion of kiwifruit from each grower, and the proportion of each grower’s kiwifruit with a defect, have not changed.
What percentage of the kiwifruit with no defect does the recalibrated grader reject?
In a batch of 3000 kiwifruit, how many fewer kiwifruit with no defect would you expect the recalibrated grader to reject than the original grader?
QUESTION THREE
(a)
Psa is a disease, caused by bacteria, that can seriously damage kiwifruit vines. It was first found in New Zealand kiwifruit orchards in 2010. Many growers now spray their vines to protect them from Psa, but spraying costs time and money.
A group of growers ran a trial over one growing season to compare three spray programmes. Each of the 900 vines in the trial was either sprayed every ten days, sprayed only after rain, or not sprayed at all. At the end of the season, every vine was checked for signs of Psa, such as spots on its leaves. The results are shown in Table 1.
Table 1: Signs of Psa on the vines in the trial
| Spray programme | Signs of Psa | No signs of Psa | Total |
|---|---|---|---|
| Sprayed every ten days | 42 | 238 | 280 |
| Sprayed only after rain | 85 | 415 | 500 |
| Not sprayed | 48 | 72 | 120 |
| Total | 175 | 725 | 900 |
What percentage of the vines in the trial showed signs of Psa?
What is the probability that a vine that was not sprayed showed signs of Psa?
Of the vines that showed signs of Psa, what proportion had not been sprayed?
A vine is chosen at random from the vines that were sprayed, on either programme.
What is the probability that it showed signs of Psa?
(b)
Daniel, a grower, looks at Table 1 and says:
“More of the vines that were sprayed only after rain showed signs of Psa than the vines that were not sprayed at all: 85 vines, compared with only 48. So spraying only after rain is worse than not spraying at all.”
Explain the mistake in Daniel’s reasoning.
Use appropriate calculations to correct Daniel’s statement.
(c)
A grower newsletter reported the trial like this:
“Compared with not spraying at all, spraying every ten days made a vine 25% less likely to get Psa, and spraying only after rain made it 23% less likely. So spraying only after rain is almost as good as spraying every ten days, and growers can save money by spraying less often.”
Evaluate the newsletter’s statements.
Use relative risks calculated from Table 1, and interpret the relative risks in context.
How we chose these questions
A prediction paper is our best judgement of what the 2026 examination is likely to ask. It isn’t a copy of any past paper, and nobody can know the real questions in advance. We built this one from NZQA’s 2023, 2024 and 2025 Probability examinations and marking schedules, the examiners’ reports on those years, and the 2026 assessment specification.
Every question is here for at least one of three reasons
- It’s in the 2026 specificationThe specification says candidates may have to draw a probability tree from written information, describe and compare distributions by their shape, centre and spread, and interpret relative risks both above and below 1.
- It comes up every yearEvery paper since 2021 has had a normal distribution question, a probability tree, a two-way table, an expected number, and a claim to check with relative risk. Most papers also keep one theme running through all three questions, as this one does.
- It’s where marks are lostThe examiners’ reports name the mistakes candidates make most often. We built several of them into these questions.
Question by question
| Question | What it tests | Why it’s here |
|---|---|---|
| 1(a) M | Normal distribution probabilities, a percentage, and finding a weight from a percentage | Every paper since 2021 has started its normal distribution question like this. The 2025 report said some candidates found a z-value but couldn’t use it to finish the problem. |
| 1(b) E | Finding a missing mean, then a mean and standard deviation from two percentages | Recent papers have asked for a missing standard deviation or a missing mean, and the 2025 report listed finding a missing standard deviation as a Merit skill. Finding both from two percentages, and then comparing two orchards, is new. |
| 1(c) E | A median from a graph, then real data compared with a normal model | The specification names comparing distributions by shape, centre and spread, and the 2024 and 2025 reports listed estimating a median from a graph. Data with the same mean and standard deviation as the model but a different shape is new. |
| 2(a) M | A probability tree from written information, and an expected number | The specification names drawing a tree from written information, and every paper has one. The 2024 report said some candidates weren’t familiar with very small probabilities shown in standard form. |
| 2(b) E | Comparing rejection rates, and a probability worked backwards through a tree | The 2024 and 2025 reports listed working backwards through a tree as a Merit skill. Explaining why one ratio doesn’t carry through to the next stage of the tree is new. |
| 2(c) E | Finding an unknown branch with algebra | Every paper since 2021 has had an unknown branch on a tree, found with an equation, and the reports list this as Merit and Excellence. Here the equation comes from a ‘given’ probability, which is new. |
| 3(a) M | Probabilities from a two-way table | Every paper has a two-way table. The 2024 report said some candidates were unsure which numbers to divide, so two of these parts share a numerator and need different denominators, and the 2025 report lists combining numbers from different parts of a table as Merit. |
| 3(b) E | Correcting a claim that compares raw numbers | Every paper since 2021 has asked candidates to correct someone who compared raw numbers from groups of different sizes. |
| 3(c) E | Evaluating a claim with relative risks below 1 | The specification names relative risks below 1, and the 2023 and 2024 reports said candidates were weaker at ‘less likely’ conclusions. A claim that confuses percentage points with relative risk is new. |
None of the Excellence questions, 1(b), 1(c), 2(b), 2(c), 3(b) and 3(c), repeats an NZQA question from 2021–2025.
NZQA sources: 2026 assessment specification, 2025 assessment report, 2025 examination, 2025 assessment schedule. The 2023 and 2024 papers, schedules and reports are on the same NZQA pages.
Maths Study Hub is independent and is not affiliated with or endorsed by NZQA. Written October 2026.
More practice: our 2025 Probability prediction paper, with a worked answer for every question.