IB DP Math AI HL Prediction Paper 2 for 2025 Exams
IB DP Math AI HL Prediction Paper 2 – April/May 2025 Exam
IB DP Math AI HL Prediction Paper 2: Prepare for the IB exams with subject-specific Prediction questions, model answers. All topics covered.
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Question 1
The heights of a certain species of deer are known to have standard deviation 0.35 m. A zoologist takes a random sample of 150 of these deer and finds that the mean height of the deer in the sample is 1.42 m.
(a) Calculate a 96% confidence interval for the population mean height.
(b) Bubay says that 96% of deer of this species are likely to have heights that are within this confidence interval. Explain briefly whether Bubay is correct.
▶️Answer/Explanation
Solution :-
Part (a): Calculating the 96% Confidence Interval
Given:
- Sample mean (\(\bar{x}\)) = 1.42 m
- Population standard deviation (\(\sigma\)) = 0.35 m
- Sample size (n) = 150
- Confidence level = 96%
For a 96% confidence interval, the z-score (z) is found using a standard normal distribution table or calculator. The z-score corresponding to 96% confidence is approximately 2.054 or 2.055.
The formula for the confidence interval is:
\(\bar{x} \pm z \frac{\sigma}{\sqrt{n}}\)
Plugging in the values:
\(1.42 \pm z \frac{0.35}{\sqrt{150}}\)
Lower bound: 1.36 m
Upper bound: 1.48 m
Therefore, the 96% confidence interval is approximately (1.36 m, 1.48 m) (3 sf).
Part (b): Explanation of Bubay’s statement
Bubay is incorrect.
The confidence interval (CI) is about the population mean, not individual values.
Question 2
The masses, in kilograms, of small and large bags of wheat have the independent distributions N(16.0, 0.4) and N(51.0, 0.9) respectively.
Find the probability that the total mass of 3 randomly chosen small bags is greater than the mass of one randomly chosen large bag.
▶️Answer/Explanation
Solution :-
Let \(S_1, S_2, S_3\) be the masses of the three small bags, and \(L\) be the mass of the large bag.
1. Mean of \(S_1 + S_2 + S_3 – L\):
\(E(S_1 + S_2 + S_3 – L) = 16 \times 3 – 51 = 48 – 51 = -3\)
(B1: Oe, using \(L – (S_1 + S_2 + S_3)\))
2. Variance of \(S_1 + S_2 + S_3 – L\):
\(Var(S_1 + S_2 + S_3 – L) = 3 \times 0.4 + 0.9 = 1.2 + 0.9 = 2.1\)
(M1)
3. Standardization:
\(\frac{0 – (-3)}{\sqrt{2.1}} = \frac{3}{\sqrt{2.1}} \approx 2.070\)
(M1: For standardizing with their values)
4. Probability Calculation:
\(1 – \Phi(2.070)\)
(M1: For area consistent with their values)
5. Final Answer:
\(0.0192\) (3 sf)
(A1)
Therefore, the probability that the total mass of 3 randomly chosen small bags is greater than the mass of one randomly chosen large bag is approximately 0.0192.
Question 3
The times, \(T\) minutes, taken by a random sample of 75 students to complete a test were noted. The results were summarised by \(\Sigma t = 230\) and \(\Sigma t^2 = 930\).
(a) Calculate unbiased estimates of the population mean and variance of \(T\).
You should now assume that your estimates from part (a) are the true values of the population mean and variance of \(T\).
(b) The times taken by another random sample of 75 students were noted, and the sample mean, \(\overline{T}\), was found.
Find the value of \(a\) such that \(P(\overline{T} > a) = 0.234\).
▶️Answer/Explanation
Solution :-
Part (a): Calculating the 96% Confidence Interval
Given:
- Sample mean (\(\bar{t}\)) = \(\frac{230}{75} = 3.066… \text{ or } 3.07 \text{ (3 sf) or } \frac{46}{15}\) (B1)
- Population standard deviation (\(S^2\)) = \(\frac{75}{74} \left( \frac{930}{75} – \left( \frac{230}{75} \right)^2 \right) \text{ or } \frac{1}{74} \left( 930 – \frac{230^2}{75} \right)\) (M1: Use of correct formula)
- Sample size (n) = 75
- Confidence level = 96%
For a 96% confidence interval, the z-score (z) is found using a standard normal distribution table or calculator. The z-score corresponding to 96% confidence is approximately 2.054 or 2.055.
The formula for the confidence interval is:
\(\bar{x} \pm z \frac{\sigma}{\sqrt{n}}\)
Plugging in the values:
\(1.42 \pm z \frac{0.35}{\sqrt{150}}\)
Part (b): Finding the value of \(a\)
\([\Phi^{-1} (1 – 0.234)] = 0.726\) (B1)
\(\frac{a – ‘3.0667’}{\sqrt{\frac{‘3.04’}{75}}} = \pm ‘0.726’\) (M1: Ft their 0.726 but must be a z value. Note using 0.766 is M0. Must have sqrt 75.)
\(a = 3.21 \text{ (3 sf) }\)