Probability distributions

Paper labels are an unofficial, independently authored grouping, applied only where a question's own format genuinely matches a real paper convention (such as IB Mathematics AA's non-calculator/calculator split). They do not reproduce any exam board's real paper numbering or mark allocation, and uncertain questions are labelled general practice instead.

• 1Customers 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.✓ ReviewedPaper 1 styleMedium4 marksCalculator+
• 2Marking 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.✓ ReviewedMarking analysisMedium4 marksCalculator+
• 3The 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'.✓ ReviewedPaper 2 styleMedium5 marksCalculator+
• 4Marking 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.✓ ReviewedMarking analysisMedium5 marksCalculator+
• 5A 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.✓ ReviewedPaper 3 style (extended)Hard7 marksCalculator+
• 6Marking 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.✓ ReviewedMarking analysisHard7 marksCalculator+
• 7A 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).✓ ReviewedPaper 1 styleMedium4 marksCalculator+
• 8Marking 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.✓ ReviewedMarking analysisMedium4 marksCalculator+
• 9A 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.✓ ReviewedPaper 1 styleMedium4 marksCalculator+
• 10Marking 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.✓ ReviewedMarking analysisMedium4 marksCalculator+
• 11The 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.✓ ReviewedPaper 1 styleMedium4 marksCalculator+
• 12Marking 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.✓ ReviewedMarking analysisMedium4 marksCalculator+
• 13A 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.✓ ReviewedPaper 2 styleMedium4 marksCalculator+
• 14Marking 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.✓ ReviewedMarking analysisMedium4 marksCalculator+
• 15In 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.✓ ReviewedPaper 2 styleMedium5 marksCalculator+
• 16Marking 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.✓ ReviewedMarking analysisMedium5 marksCalculator+
• 17A 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.✓ ReviewedPaper 3 style (extended)Medium5 marksCalculator+
• 18Marking 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.✓ ReviewedMarking analysisMedium5 marksCalculator+
• 19A 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.✓ ReviewedPaper 3 style (extended)Hard6 marksCalculator+
• 20Marking 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.✓ ReviewedMarking analysisHard6 marksCalculator+
• 21In 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.✓ ReviewedPaper 1 styleEasy3 marksNo calculator+
• 22Marking 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.✓ ReviewedMarking analysisEasy3 marksNo calculator+
• 23The 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.✓ ReviewedPaper 2 styleMedium4 marksCalculator+
• 24Marking 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.✓ ReviewedMarking analysisMedium4 marksCalculator+
• 25Two 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.✓ ReviewedPaper 3 style (extended)Hard6 marksCalculator+
• 26Marking 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.✓ ReviewedMarking analysisHard6 marksCalculator+
• 27The 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.✓ ReviewedPaper 3 style (extended)Medium5 marksCalculator+
• 28Marking 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.✓ ReviewedMarking analysisMedium5 marksCalculator+
• 29A 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).✓ ReviewedPaper 1 styleMedium5 marksCalculator+
• 30Marking 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.✓ ReviewedMarking analysisMedium5 marksCalculator+
• 31Ten 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.✓ ReviewedPaper 3 style (extended)Hard7 marksCalculator+
• 32Marking 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.✓ ReviewedMarking analysisHard7 marksCalculator+