Mathematics: Applications & Interpretation HL
Probability distributions — Topic 4 HL
- 1.
Customers arrive at a small bakery counter at an average rate of 5 per 10-minute interval, and the number of arrivals follows a Poisson distribution. (a) State the mean number of arrivals in a 10-minute interval. (b) Calculate the probability that exactly 3 customers arrive in a randomly chosen 10-minute interval. (c) Calculate the probability that at least one customer arrives in a 2-minute interval.
[4 marks]Marking points
- States λ = 5 for a 10-minute interval.
- Uses the Poisson distribution to obtain P(X = 3) ≈ 0.140.
- Scales the mean to λ = 1 for a 2-minute interval.
- Uses P(at least one) = 1 − P(X = 0) to obtain approximately 0.632.
Examiner tip: Always rescale λ to match the time interval given in each part; a common error is reusing the original mean.
- 2.
Marking analysis: A learner attempts the following task: “Customers arrive at a small bakery counter at an average rate of 5 per 10-minute interval, and the number of arrivals follows a Poisson distribution. (a) State the mean number of arrivals in a 10-minute interval. (b) Calculate the probability that exactly 3 customers arrive in a randomly chosen 10-minute interval. (c) Calculate the probability that at least one customer arrives in a 2-minute interval.” Their response addresses only this point: “States λ = 5 for a 10-minute interval.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]Marking points
- Recognises credit for the stated point: States λ = 5 for a 10-minute interval.
- Identifies the missing requirement: Uses the Poisson distribution to obtain P(X = 3) ≈ 0.140.
- Identifies the missing requirement: Scales the mean to λ = 1 for a 2-minute interval.
- Identifies the missing requirement: Uses P(at least one) = 1 − P(X = 0) to obtain approximately 0.632.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
- 3.
The heights of a large population of a plant species are normally distributed with mean 42 cm and standard deviation 6 cm. (a) Calculate the probability that a randomly selected plant is taller than 50 cm. (b) Calculate the probability that a randomly selected plant has a height between 36 cm and 48 cm. (c) The tallest 10% of plants are labelled 'premium'. Find the minimum height for a plant to be labelled 'premium'.
[5 marks]Marking points
- Uses the normal distribution with mean 42 and standard deviation 6.
- Obtains P(X > 50) ≈ 0.0912.
- Obtains P(36 < X < 48) ≈ 0.6827.
- Uses the inverse normal function with the upper 10% tail.
- Obtains a minimum height of approximately 49.7 cm.
Examiner tip: Use the GDC's inverse normal function directly for boundary problems; do not try to invert the standard normal table by hand.
- 4.
Marking analysis: A learner attempts the following task: “The heights of a large population of a plant species are normally distributed with mean 42 cm and standard deviation 6 cm. (a) Calculate the probability that a randomly selected plant is taller than 50 cm. (b) Calculate the probability that a randomly selected plant has a height between 36 cm and 48 cm. (c) The tallest 10% of plants are labelled 'premium'. Find the minimum height for a plant to be labelled 'premium'.” Their response addresses only this point: “Uses the normal distribution with mean 42 and standard deviation 6.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks]Marking points
- Recognises credit for the stated point: Uses the normal distribution with mean 42 and standard deviation 6.
- Identifies the missing requirement: Obtains P(X > 50) ≈ 0.0912.
- Identifies the missing requirement: Obtains P(36 < X < 48) ≈ 0.6827.
- Identifies the missing requirement: Uses the inverse normal function with the upper 10% tail.
- Identifies the missing requirement: Obtains a minimum height of approximately 49.7 cm.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
- 5.
A researcher believes a new fertiliser increases the mean mass of tomatoes. A random sample of 36 tomatoes grown with the fertiliser has a mean mass of 182 g with a sample standard deviation of 24 g. Tomatoes grown without the fertiliser have a known population mean mass of 172 g. Test, at the 5% significance level, whether the sample provides evidence that the fertiliser increases the mean mass. (a) State the null and alternative hypotheses. (b) Calculate the test statistic. (c) State the p-value for this test. (d) State a conclusion in context, referring to the significance level. (e) State one assumption needed for this test to be valid.
[7 marks]Marking points
- States H0: μ = 172 g.
- States H1: μ > 172 g (one-tailed).
- Uses a one-sample test setup with sample mean 182, sample standard deviation 24, and n = 36.
- Calculates a test statistic of approximately 2.5.
- States a p-value of approximately 0.0062 (accept a consistent equivalent from a t-based calculation).
- Concludes that since the p-value is less than 0.05, H0 is rejected and there is evidence the fertiliser increases mean mass.
- States a valid assumption, such as the sample being random and representative, or the sampling distribution of the mean being approximately normal.
Examiner tip: State the direction of the alternative hypothesis before calculating; a one-tailed claim changes which p-value you compare to the significance level.
- 6.
Marking analysis: A learner attempts the following task: “A researcher believes a new fertiliser increases the mean mass of tomatoes. A random sample of 36 tomatoes grown with the fertiliser has a mean mass of 182 g with a sample standard deviation of 24 g. Tomatoes grown without the fertiliser have a known population mean mass of 172 g. Test, at the 5% significance level, whether the sample provides evidence that the fertiliser increases the mean mass. (a) State the null and alternative hypotheses. (b) Calculate the test statistic. (c) State the p-value for this test. (d) State a conclusion in context, referring to the significance level. (e) State one assumption needed for this test to be valid.” Their response addresses only this point: “States H0: μ = 172 g.” Evaluate the response against the complete 7-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[7 marks]Marking points
- Recognises credit for the stated point: States H0: μ = 172 g.
- Identifies the missing requirement: States H1: μ > 172 g (one-tailed).
- Identifies the missing requirement: Uses a one-sample test setup with sample mean 182, sample standard deviation 24, and n = 36.
- Identifies the missing requirement: Calculates a test statistic of approximately 2.5.
- Identifies the missing requirement: States a p-value of approximately 0.0062 (accept a consistent equivalent from a t-based calculation).
- Identifies the missing requirement: Concludes that since the p-value is less than 0.05, H0 is rejected and there is evidence the fertiliser increases mean mass.
- Identifies the missing requirement: States a valid assumption, such as the sample being random and representative, or the sampling distribution of the mean being approximately normal.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
- 7.
A binomial random variable X has n = 20 trials with probability of success p = 0.3. (a) Find the mean and variance of X. (b) Calculate P(X = 6).
[4 marks]Marking points
- Uses mean = np = 20(0.3) = 6.
- Uses variance = np(1−p) = 20(0.3)(0.7) = 4.2.
- Uses the binomial probability formula with n = 20, p = 0.3, k = 6.
- Obtains P(X = 6) ≈ 0.192.
Examiner tip: Always state the mean and variance formulas np and np(1−p) explicitly before substituting — this is good practice for method marks even when using a GDC for the final probability.
- 8.
Marking analysis: A learner attempts the following task: “A binomial random variable X has n = 20 trials with probability of success p = 0.3. (a) Find the mean and variance of X. (b) Calculate P(X = 6).” Their response addresses only this point: “Uses mean = np = 20(0.3) = 6.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]Marking points
- Recognises credit for the stated point: Uses mean = np = 20(0.3) = 6.
- Identifies the missing requirement: Uses variance = np(1−p) = 20(0.3)(0.7) = 4.2.
- Identifies the missing requirement: Uses the binomial probability formula with n = 20, p = 0.3, k = 6.
- Identifies the missing requirement: Obtains P(X = 6) ≈ 0.192.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
- 9.
A factory produces components with a 2% defect rate. In a batch of 200 components, the number of defective components is modelled using a Poisson approximation to the binomial distribution. (a) State the parameter λ for this approximation. (b) Calculate the probability that exactly 3 components in the batch are defective.
[4 marks]Marking points
- States that the Poisson approximation is valid since n is large and p is small.
- States λ = np = 200(0.02) = 4.
- Uses the Poisson distribution with λ = 4 to calculate P(X = 3).
- Obtains P(X = 3) ≈ 0.195.
Examiner tip: The Poisson approximation to the binomial works well when n is large and p is small, with λ = np kept moderate — always justify why the approximation is appropriate before using it.
- 10.
Marking analysis: A learner attempts the following task: “A factory produces components with a 2% defect rate. In a batch of 200 components, the number of defective components is modelled using a Poisson approximation to the binomial distribution. (a) State the parameter λ for this approximation. (b) Calculate the probability that exactly 3 components in the batch are defective.” Their response addresses only this point: “States that the Poisson approximation is valid since n is large and p is small.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]Marking points
- Recognises credit for the stated point: States that the Poisson approximation is valid since n is large and p is small.
- Identifies the missing requirement: States λ = np = 200(0.02) = 4.
- Identifies the missing requirement: Uses the Poisson distribution with λ = 4 to calculate P(X = 3).
- Identifies the missing requirement: Obtains P(X = 3) ≈ 0.195.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
- 11.
The number of calls received by a help desk follows a Poisson distribution with mean 3 calls per hour. The number of emails received follows an independent Poisson distribution with mean 5 per hour. Find the probability that exactly 10 contacts (calls and emails combined) are received in one hour.
[4 marks]Marking points
- States that the sum of two independent Poisson random variables is itself Poisson, with parameter equal to the sum of the individual means.
- Obtains the combined parameter λ = 3 + 5 = 8.
- Uses the Poisson distribution with λ = 8 to calculate P(X = 10).
- Obtains P(X = 10) ≈ 0.0993.
Examiner tip: The additive property of independent Poisson variables is a useful shortcut — it avoids having to model two separate distributions and combine them with a convolution.
- 12.
Marking analysis: A learner attempts the following task: “The number of calls received by a help desk follows a Poisson distribution with mean 3 calls per hour. The number of emails received follows an independent Poisson distribution with mean 5 per hour. Find the probability that exactly 10 contacts (calls and emails combined) are received in one hour.” Their response addresses only this point: “States that the sum of two independent Poisson random variables is itself Poisson, with parameter equal to the sum of the individual means.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]Marking points
- Recognises credit for the stated point: States that the sum of two independent Poisson random variables is itself Poisson, with parameter equal to the sum of the individual means.
- Identifies the missing requirement: Obtains the combined parameter λ = 3 + 5 = 8.
- Identifies the missing requirement: Uses the Poisson distribution with λ = 8 to calculate P(X = 10).
- Identifies the missing requirement: Obtains P(X = 10) ≈ 0.0993.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
- 13.
A random sample of 36 batteries has a mean lifespan of 50 hours. The population standard deviation is known to be 8 hours. Construct a 95% confidence interval for the population mean lifespan.
[4 marks]Marking points
- States the confidence interval formula x̄ ± z(σ/√n) with z = 1.96 for 95% confidence.
- Calculates the margin of error as 1.96 × 8/√36 ≈ 2.61.
- Obtains the confidence interval (47.4, 52.6), to 1 decimal place.
- States correctly that we are 95% confident the true population mean lifespan lies within this interval.
Examiner tip: A confidence interval's correct interpretation refers to the method capturing the true parameter in 95% of repeated samples — it does not mean there is a 95% probability the true mean lies in this specific interval.
- 14.
Marking analysis: A learner attempts the following task: “A random sample of 36 batteries has a mean lifespan of 50 hours. The population standard deviation is known to be 8 hours. Construct a 95% confidence interval for the population mean lifespan.” Their response addresses only this point: “States the confidence interval formula x̄ ± z(σ/√n) with z = 1.96 for 95% confidence.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]Marking points
- Recognises credit for the stated point: States the confidence interval formula x̄ ± z(σ/√n) with z = 1.96 for 95% confidence.
- Identifies the missing requirement: Calculates the margin of error as 1.96 × 8/√36 ≈ 2.61.
- Identifies the missing requirement: Obtains the confidence interval (47.4, 52.6), to 1 decimal place.
- Identifies the missing requirement: States correctly that we are 95% confident the true population mean lifespan lies within this interval.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
- 15.
In a survey of 150 randomly selected voters, 45 said they support a new policy. Construct a 95% confidence interval for the true proportion of voters who support the policy.
[5 marks]Marking points
- Calculates the sample proportion p̂ = 45/150 = 0.3.
- States the confidence interval formula p̂ ± z√(p̂(1−p̂)/n) with z = 1.96.
- Calculates the standard error as √(0.3(0.7)/150) ≈ 0.0374.
- Calculates the margin of error as 1.96 × 0.0374 ≈ 0.0733.
- Obtains the confidence interval (0.227, 0.373), to 3 decimal places.
Examiner tip: The standard error for a proportion uses the sample proportion p̂ itself in place of the usually unknown population proportion — this is a key difference from the confidence interval for a mean.
- 16.
Marking analysis: A learner attempts the following task: “In a survey of 150 randomly selected voters, 45 said they support a new policy. Construct a 95% confidence interval for the true proportion of voters who support the policy.” Their response addresses only this point: “Calculates the sample proportion p̂ = 45/150 = 0.3.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks]Marking points
- Recognises credit for the stated point: Calculates the sample proportion p̂ = 45/150 = 0.3.
- Identifies the missing requirement: States the confidence interval formula p̂ ± z√(p̂(1−p̂)/n) with z = 1.96.
- Identifies the missing requirement: Calculates the standard error as √(0.3(0.7)/150) ≈ 0.0374.
- Identifies the missing requirement: Calculates the margin of error as 1.96 × 0.0374 ≈ 0.0733.
- Identifies the missing requirement: Obtains the confidence interval (0.227, 0.373), to 3 decimal places.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
- 17.
A spinner with five equally sized sectors, numbered 1 to 5, is spun 100 times. The observed frequencies for sectors 1 to 5 are: 18, 22, 25, 15, 20. Test, at the 5% significance level, whether these results are consistent with the spinner being fair (a uniform distribution across the 5 sectors), given a critical value of χ² = 9.49 for 4 degrees of freedom.
[5 marks]Marking points
- States H0: the outcomes are uniformly distributed, so each expected frequency is 100/5 = 20.
- Calculates the chi-squared statistic χ² = Σ(O−E)²/E.
- Obtains χ² = 2.90.
- Compares 2.90 with the critical value 9.49 and notes 2.90 < 9.49.
- Concludes that H0 is not rejected: there is insufficient evidence the distribution is non-uniform.
Examiner tip: For a chi-squared test, a smaller test statistic means the observed data is closer to what was expected — only reject H0 when the statistic exceeds the critical value.
- 18.
Marking analysis: A learner attempts the following task: “A spinner with five equally sized sectors, numbered 1 to 5, is spun 100 times. The observed frequencies for sectors 1 to 5 are: 18, 22, 25, 15, 20. Test, at the 5% significance level, whether these results are consistent with the spinner being fair (a uniform distribution across the 5 sectors), given a critical value of χ² = 9.49 for 4 degrees of freedom.” Their response addresses only this point: “States H0: the outcomes are uniformly distributed, so each expected frequency is 100/5 = 20.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks]Marking points
- Recognises credit for the stated point: States H0: the outcomes are uniformly distributed, so each expected frequency is 100/5 = 20.
- Identifies the missing requirement: Calculates the chi-squared statistic χ² = Σ(O−E)²/E.
- Identifies the missing requirement: Obtains χ² = 2.90.
- Identifies the missing requirement: Compares 2.90 with the critical value 9.49 and notes 2.90 < 9.49.
- Identifies the missing requirement: Concludes that H0 is not rejected: there is insufficient evidence the distribution is non-uniform.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
- 19.
A 2×2 contingency table records 100 people classified by gender and whether they prefer a new product design: (Men who prefer it: 30, Men who do not: 20, Women who prefer it: 15, Women who do not: 35). Test, at the 5% significance level, whether preference is independent of gender, given a critical value of χ² = 3.84 for 1 degree of freedom.
[6 marks]Marking points
- States H0: preference is independent of gender; H1: preference is not independent of gender.
- Calculates row and column totals: Men = 50, Women = 50, Prefer = 45, Do not prefer = 55.
- Calculates each expected frequency using (row total × column total)/100.
- Calculates the chi-squared statistic χ² = Σ(O−E)²/E.
- Obtains χ² ≈ 9.09.
- Compares 9.09 with the critical value 3.84, concludes H0 is rejected, and states that preference and gender are not independent.
Examiner tip: Degrees of freedom for a contingency table are (rows−1)×(columns−1) — for a 2×2 table this is always 1, regardless of the actual frequencies.
- 20.
Marking analysis: A learner attempts the following task: “A 2×2 contingency table records 100 people classified by gender and whether they prefer a new product design: (Men who prefer it: 30, Men who do not: 20, Women who prefer it: 15, Women who do not: 35). Test, at the 5% significance level, whether preference is independent of gender, given a critical value of χ² = 3.84 for 1 degree of freedom.” Their response addresses only this point: “States H0: preference is independent of gender; H1: preference is not independent of gender.” Evaluate the response against the complete 6-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[6 marks]Marking points
- Recognises credit for the stated point: States H0: preference is independent of gender; H1: preference is not independent of gender.
- Identifies the missing requirement: Calculates row and column totals: Men = 50, Women = 50, Prefer = 45, Do not prefer = 55.
- Identifies the missing requirement: Calculates each expected frequency using (row total × column total)/100.
- Identifies the missing requirement: Calculates the chi-squared statistic χ² = Σ(O−E)²/E.
- Identifies the missing requirement: Obtains χ² ≈ 9.09.
- Identifies the missing requirement: Compares 9.09 with the critical value 3.84, concludes H0 is rejected, and states that preference and gender are not independent.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
- 21.
In a hypothesis test conducted at the 5% significance level, explain what is meant by a Type I error and a Type II error in this specific context, given the test is checking whether a new drug lowers blood pressure compared to a placebo. State the probability of a Type I error.
[3 marks] · no calculatorMarking points
- Explains that a Type I error occurs if the test concludes the drug lowers blood pressure when it actually does not (rejecting a true H0).
- Explains that a Type II error occurs if the test fails to detect that the drug actually does lower blood pressure (failing to reject a false H0).
- States that the probability of a Type I error equals the significance level, 0.05.
Examiner tip: The significance level of a test is always, by definition, the probability of a Type I error — this never needs to be calculated separately from the stated significance level.
- 22.
Marking analysis: A learner attempts the following task: “In a hypothesis test conducted at the 5% significance level, explain what is meant by a Type I error and a Type II error in this specific context, given the test is checking whether a new drug lowers blood pressure compared to a placebo. State the probability of a Type I error.” Their response addresses only this point: “Explains that a Type I error occurs if the test concludes the drug lowers blood pressure when it actually does not (rejecting a true H0).” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculatorMarking points
- Recognises credit for the stated point: Explains that a Type I error occurs if the test concludes the drug lowers blood pressure when it actually does not (rejecting a true H0).
- Identifies the missing requirement: Explains that a Type II error occurs if the test fails to detect that the drug actually does lower blood pressure (failing to reject a false H0).
- Identifies the missing requirement: States that the probability of a Type I error equals the significance level, 0.05.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
- 23.
The weights of bags of flour produced by a factory are normally distributed with mean 80 g and standard deviation 15 g. A random sample of 25 bags is taken. Find the probability that the sample mean weight exceeds 85 g.
[4 marks]Marking points
- States that by the central limit theorem, the sample mean is normally distributed with mean 80 and standard deviation 15/√25 = 3.
- Standardises to obtain z = (85 − 80)/3 ≈ 1.67.
- Uses the standard normal distribution to find P(Z > 1.67).
- Obtains a probability of approximately 0.0478.
Examiner tip: The standard deviation of the sample mean (the standard error) is always the population standard deviation divided by √n, never the population standard deviation itself — using the wrong value is the most common error here.
- 24.
Marking analysis: A learner attempts the following task: “The weights of bags of flour produced by a factory are normally distributed with mean 80 g and standard deviation 15 g. A random sample of 25 bags is taken. Find the probability that the sample mean weight exceeds 85 g.” Their response addresses only this point: “States that by the central limit theorem, the sample mean is normally distributed with mean 80 and standard deviation 15/√25 = 3.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]Marking points
- Recognises credit for the stated point: States that by the central limit theorem, the sample mean is normally distributed with mean 80 and standard deviation 15/√25 = 3.
- Identifies the missing requirement: Standardises to obtain z = (85 − 80)/3 ≈ 1.67.
- Identifies the missing requirement: Uses the standard normal distribution to find P(Z > 1.67).
- Identifies the missing requirement: Obtains a probability of approximately 0.0478.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
- 25.
Two independent samples are collected to compare commute times. Sample A (n = 40) has mean 72 minutes and standard deviation 10 minutes. Sample B (n = 35) has mean 68 minutes and standard deviation 12 minutes. Test, at the 5% significance level, whether there is evidence that the true mean commute time for group A is greater than for group B, given a critical z-value of 1.645.
[6 marks]Marking points
- States H0: μA = μB; H1: μA > μB (one-tailed).
- Calculates the standard error of the difference: √(10²/40 + 12²/35) ≈ 2.572.
- Calculates the test statistic z = (72 − 68)/2.572 ≈ 1.56.
- Compares 1.56 with the critical value 1.645 and notes 1.56 < 1.645.
- Concludes that H0 is not rejected: there is insufficient evidence at the 5% level that group A's mean commute time is greater.
- Notes that the test statistic is close to the critical value, so a larger sample might detect a genuine difference if one exists.
Examiner tip: For a two-sample test with large independent samples, the standard error of the difference combines both sample variances divided by their respective sample sizes, added under a square root — never simply averaged.
- 26.
Marking analysis: A learner attempts the following task: “Two independent samples are collected to compare commute times. Sample A (n = 40) has mean 72 minutes and standard deviation 10 minutes. Sample B (n = 35) has mean 68 minutes and standard deviation 12 minutes. Test, at the 5% significance level, whether there is evidence that the true mean commute time for group A is greater than for group B, given a critical z-value of 1.645.” Their response addresses only this point: “States H0: μA = μB; H1: μA > μB (one-tailed).” Evaluate the response against the complete 6-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[6 marks]Marking points
- Recognises credit for the stated point: States H0: μA = μB; H1: μA > μB (one-tailed).
- Identifies the missing requirement: Calculates the standard error of the difference: √(10²/40 + 12²/35) ≈ 2.572.
- Identifies the missing requirement: Calculates the test statistic z = (72 − 68)/2.572 ≈ 1.56.
- Identifies the missing requirement: Compares 1.56 with the critical value 1.645 and notes 1.56 < 1.645.
- Identifies the missing requirement: Concludes that H0 is not rejected: there is insufficient evidence at the 5% level that group A's mean commute time is greater.
- Identifies the missing requirement: Notes that the test statistic is close to the critical value, so a larger sample might detect a genuine difference if one exists.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
- 27.
The mass of apples from orchard X is normally distributed with mean 50 g and standard deviation 6 g. The mass of apples from orchard Y is independently normally distributed with mean 30 g and standard deviation 4 g. One apple is selected at random from each orchard. Find the probability that the apple from orchard X is more than 25 g heavier than the apple from orchard Y.
[5 marks]Marking points
- States that for independent normal variables, X − Y is normal with mean μX − μY = 50 − 30 = 20.
- States that the variance of X − Y is the sum of the variances: 6² + 4² = 52.
- States the standard deviation of X − Y as √52 ≈ 7.21.
- Standardises to find P(X − Y > 25) using mean 20 and standard deviation 7.21.
- Obtains a probability of approximately 0.244.
Examiner tip: When combining two independent normal variables, variances always add — even when finding the distribution of a difference — because subtracting a random variable does not reduce its uncertainty.
- 28.
Marking analysis: A learner attempts the following task: “The mass of apples from orchard X is normally distributed with mean 50 g and standard deviation 6 g. The mass of apples from orchard Y is independently normally distributed with mean 30 g and standard deviation 4 g. One apple is selected at random from each orchard. Find the probability that the apple from orchard X is more than 25 g heavier than the apple from orchard Y.” Their response addresses only this point: “States that for independent normal variables, X − Y is normal with mean μX − μY = 50 − 30 = 20.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks]Marking points
- Recognises credit for the stated point: States that for independent normal variables, X − Y is normal with mean μX − μY = 50 − 30 = 20.
- Identifies the missing requirement: States that the variance of X − Y is the sum of the variances: 6² + 4² = 52.
- Identifies the missing requirement: States the standard deviation of X − Y as √52 ≈ 7.21.
- Identifies the missing requirement: Standardises to find P(X − Y > 25) using mean 20 and standard deviation 7.21.
- Identifies the missing requirement: Obtains a probability of approximately 0.244.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
- 29.
A discrete random variable X has the probability distribution: P(X=0) = 0.1, P(X=1) = 0.3, P(X=2) = 0.4, P(X=3) = 0.2. Find (a) E(X), (b) E(X²), and (c) Var(X).
[5 marks]Marking points
- Calculates E(X) = Σx·P(X=x) = 0(0.1) + 1(0.3) + 2(0.4) + 3(0.2).
- Obtains E(X) = 1.7.
- Calculates E(X²) = Σx²·P(X=x) = 0(0.1) + 1(0.3) + 4(0.4) + 9(0.2).
- Obtains E(X²) = 3.7.
- Uses Var(X) = E(X²) − [E(X)]² = 3.7 − 1.7² to obtain Var(X) = 0.81.
Examiner tip: Always compute E(X²) directly from the distribution table rather than squaring E(X) — these are different quantities, and Var(X) requires both.
- 30.
Marking analysis: A learner attempts the following task: “A discrete random variable X has the probability distribution: P(X=0) = 0.1, P(X=1) = 0.3, P(X=2) = 0.4, P(X=3) = 0.2. Find (a) E(X), (b) E(X²), and (c) Var(X).” Their response addresses only this point: “Calculates E(X) = Σx·P(X=x) = 0(0.1) + 1(0.3) + 2(0.4) + 3(0.2).” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks]Marking points
- Recognises credit for the stated point: Calculates E(X) = Σx·P(X=x) = 0(0.1) + 1(0.3) + 2(0.4) + 3(0.2).
- Identifies the missing requirement: Obtains E(X) = 1.7.
- Identifies the missing requirement: Calculates E(X²) = Σx²·P(X=x) = 0(0.1) + 1(0.3) + 4(0.4) + 9(0.2).
- Identifies the missing requirement: Obtains E(X²) = 3.7.
- Identifies the missing requirement: Uses Var(X) = E(X²) − [E(X)]² = 3.7 − 1.7² to obtain Var(X) = 0.81.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
- 31.
Ten participants' reaction times (in hundredths of a second) were measured before and after a training programme. The differences (before − after) were: 2, −1, 3, 4, 0, 2, 1, −2, 3, 2. Test, at the 5% significance level, whether the training programme reduced reaction times, given a critical t-value of 1.833 for 9 degrees of freedom. (a) State the null and alternative hypotheses. (b) Calculate the mean and standard deviation of the differences. (c) Calculate the test statistic. (d) State a conclusion in context.
[7 marks]Marking points
- States H0: μd = 0 (no change); H1: μd > 0 (reaction time reduced, since before − after is positive if faster).
- Calculates the mean difference d̄ = 1.4.
- Calculates the sample standard deviation of the differences sd ≈ 1.90.
- Calculates the test statistic t = d̄/(sd/√n) = 1.4/(1.90/√10).
- Obtains t ≈ 2.33.
- Compares 2.33 with the critical value 1.833 and notes 2.33 > 1.833.
- Concludes that H0 is rejected: there is evidence at the 5% level that the training programme reduced reaction times.
Examiner tip: A paired t-test reduces two related samples to a single sample of differences — always compute the mean and standard deviation of the differences themselves, not of the original before/after values separately.
- 32.
Marking analysis: A learner attempts the following task: “Ten participants' reaction times (in hundredths of a second) were measured before and after a training programme. The differences (before − after) were: 2, −1, 3, 4, 0, 2, 1, −2, 3, 2. Test, at the 5% significance level, whether the training programme reduced reaction times, given a critical t-value of 1.833 for 9 degrees of freedom. (a) State the null and alternative hypotheses. (b) Calculate the mean and standard deviation of the differences. (c) Calculate the test statistic. (d) State a conclusion in context.” Their response addresses only this point: “States H0: μd = 0 (no change); H1: μd > 0 (reaction time reduced, since before − after is positive if faster).” Evaluate the response against the complete 7-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[7 marks]Marking points
- Recognises credit for the stated point: States H0: μd = 0 (no change); H1: μd > 0 (reaction time reduced, since before − after is positive if faster).
- Identifies the missing requirement: Calculates the mean difference d̄ = 1.4.
- Identifies the missing requirement: Calculates the sample standard deviation of the differences sd ≈ 1.90.
- Identifies the missing requirement: Calculates the test statistic t = d̄/(sd/√n) = 1.4/(1.90/√10).
- Identifies the missing requirement: Obtains t ≈ 2.33.
- Identifies the missing requirement: Compares 2.33 with the critical value 1.833 and notes 2.33 > 1.833.
- Identifies the missing requirement: Concludes that H0 is rejected: there is evidence at the 5% level that the training programme reduced reaction times.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.