IB · MATH AI SL

Mathematics: Applications & Interpretation SL

Statistics and regression — Topic 4

Name: ____________________Date: October 14, 2026

1.
A survey of 120 students records favourite study time (morning/evening) against year group (Year 12/Year 13). The observed frequencies give a calculated χ² test statistic of 4.83, and the critical value at the 5% significance level is 3.84. (a) State the null hypothesis for this test. (b) State whether favourite study time and year group are independent. (c) Justify your conclusion using both values. (d) State one condition required for a chi-squared test to be valid.
[4 marks]
Marking points
- States H0: favourite study time is independent of year group.
- States that the variables are not independent (they are associated/dependent).
- Justifies this using the fact that the calculated statistic (4.83) exceeds the critical value (3.84).
- States a valid condition, such as all expected frequencies being at least 5.
Examiner tip: A calculated statistic larger than the critical value means the observed pattern is unlikely under independence, so H0 is rejected.
2.
Marking analysis: A learner attempts the following task: “A survey of 120 students records favourite study time (morning/evening) against year group (Year 12/Year 13). The observed frequencies give a calculated χ² test statistic of 4.83, and the critical value at the 5% significance level is 3.84. (a) State the null hypothesis for this test. (b) State whether favourite study time and year group are independent. (c) Justify your conclusion using both values. (d) State one condition required for a chi-squared test to be valid.” Their response addresses only this point: “States H0: favourite study time is independent of year group.” 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 H0: favourite study time is independent of year group.
- Identifies the missing requirement: States that the variables are not independent (they are associated/dependent).
- Identifies the missing requirement: Justifies this using the fact that the calculated statistic (4.83) exceeds the critical value (3.84).
- Identifies the missing requirement: States a valid condition, such as all expected frequencies being at least 5.
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.
A café manager records daily temperature x (°C) and iced-drink sales y for six days: (18,20), (21,29), (24,38), (27,44), (30,55), (33,58). Using your GDC: (a) find the equation of the linear regression line of y on x. (b) State the value of Pearson's correlation coefficient r and interpret its strength. (c) Use your equation to estimate sales on a day with temperature 26°C. (d) Explain why using this equation to estimate sales at 5°C would not be appropriate.
[5 marks]
Marking points
- Uses the GDC regression feature to obtain y ≈ 2.61x − 25.9 (accept equivalent rounding).
- States r ≈ 0.993 (or similarly close to 1).
- Interprets r as showing a very strong positive linear correlation.
- Substitutes x = 26 to estimate y ≈ 42 drinks.
- Explains that 5°C is far outside the range of recorded x-values, so this would be extrapolation and may not be reliable.
Examiner tip: State whether an estimate uses interpolation or extrapolation; extrapolation carries much greater risk of error.
4.
Marking analysis: A learner attempts the following task: “A café manager records daily temperature x (°C) and iced-drink sales y for six days: (18,20), (21,29), (24,38), (27,44), (30,55), (33,58). Using your GDC: (a) find the equation of the linear regression line of y on x. (b) State the value of Pearson's correlation coefficient r and interpret its strength. (c) Use your equation to estimate sales on a day with temperature 26°C. (d) Explain why using this equation to estimate sales at 5°C would not be appropriate.” Their response addresses only this point: “Uses the GDC regression feature to obtain y ≈ 2.61x − 25.9 (accept equivalent rounding).” 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 GDC regression feature to obtain y ≈ 2.61x − 25.9 (accept equivalent rounding).
- Identifies the missing requirement: States r ≈ 0.993 (or similarly close to 1).
- Identifies the missing requirement: Interprets r as showing a very strong positive linear correlation.
- Identifies the missing requirement: Substitutes x = 26 to estimate y ≈ 42 drinks.
- Identifies the missing requirement: Explains that 5°C is far outside the range of recorded x-values, so this would be extrapolation and may not be reliable.
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 factory's machine produces defective components independently with probability 0.04. In a random sample of 50 components, let X be the number of defective components, assumed to follow a binomial distribution. Using your GDC: (a) state the values of n and p. (b) Calculate P(X = 2). (c) Calculate P(X ≤ 1). (d) Calculate the expected number of defective components in the sample.
[4 marks]
Marking points
- States n = 50 and p = 0.04.
- Uses the binomial distribution to obtain P(X = 2) ≈ 0.276.
- Calculates P(X ≤ 1) ≈ 0.400 as the sum of P(X = 0) and P(X = 1).
- Uses E(X) = np to obtain an expected value of 2.
Examiner tip: For binomial problems, n and p come directly from the sample size and the probability given; use the cumulative function for 'at most' and the complement for 'at least'.
6.
Marking analysis: A learner attempts the following task: “A factory's machine produces defective components independently with probability 0.04. In a random sample of 50 components, let X be the number of defective components, assumed to follow a binomial distribution. Using your GDC: (a) state the values of n and p. (b) Calculate P(X = 2). (c) Calculate P(X ≤ 1). (d) Calculate the expected number of defective components in the sample.” Their response addresses only this point: “States n = 50 and p = 0.04.” 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 n = 50 and p = 0.04.
- Identifies the missing requirement: Uses the binomial distribution to obtain P(X = 2) ≈ 0.276.
- Identifies the missing requirement: Calculates P(X ≤ 1) ≈ 0.400 as the sum of P(X = 0) and P(X = 1).
- Identifies the missing requirement: Uses E(X) = np to obtain an expected value of 2.
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 survey of waiting times (minutes) is grouped as: 10 (frequency 4), 20 (frequency 8), 30 (frequency 10), 40 (frequency 5), 50 (frequency 3), where each value is the midpoint of its class. Using your GDC, calculate the mean and standard deviation of the waiting times.
[3 marks]
Marking points
- Enters the midpoints and frequencies into the GDC's one-variable statistics function.
- Obtains a mean of 28.3 minutes (3 s.f.).
- Obtains a standard deviation of 11.6 minutes (3 s.f.).
Examiner tip: For grouped data, the GDC treats each midpoint as if every value in that class equals the midpoint exactly — this is an approximation of the true mean and standard deviation.
8.
Marking analysis: A learner attempts the following task: “A survey of waiting times (minutes) is grouped as: 10 (frequency 4), 20 (frequency 8), 30 (frequency 10), 40 (frequency 5), 50 (frequency 3), where each value is the midpoint of its class. Using your GDC, calculate the mean and standard deviation of the waiting times.” Their response addresses only this point: “Enters the midpoints and frequencies into the GDC's one-variable statistics function.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]
Marking points
- Recognises credit for the stated point: Enters the midpoints and frequencies into the GDC's one-variable statistics function.
- Identifies the missing requirement: Obtains a mean of 28.3 minutes (3 s.f.).
- Identifies the missing requirement: Obtains a standard deviation of 11.6 minutes (3 s.f.).
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.
The following 11 values are listed in ascending order: 12, 15, 18, 21, 23, 25, 27, 29, 33, 36, 40. Find the median, the lower quartile Q1, the upper quartile Q3, and the interquartile range.
[4 marks]
Marking points
- States the median as the 6th value, 25.
- States Q1 as the median of the lower five values (12, 15, 18, 21, 23), giving Q1 = 18.
- States Q3 as the median of the upper five values (27, 29, 33, 36, 40), giving Q3 = 33.
- Calculates the interquartile range as Q3 − Q1 = 15.
Examiner tip: When the data set has an odd number of values, exclude the overall median itself before splitting into lower and upper halves to find Q1 and Q3.
10.
Marking analysis: A learner attempts the following task: “The following 11 values are listed in ascending order: 12, 15, 18, 21, 23, 25, 27, 29, 33, 36, 40. Find the median, the lower quartile Q1, the upper quartile Q3, and the interquartile range.” Their response addresses only this point: “States the median as the 6th value, 25.” 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 median as the 6th value, 25.
- Identifies the missing requirement: States Q1 as the median of the lower five values (12, 15, 18, 21, 23), giving Q1 = 18.
- Identifies the missing requirement: States Q3 as the median of the upper five values (27, 29, 33, 36, 40), giving Q3 = 33.
- Identifies the missing requirement: Calculates the interquartile range as Q3 − Q1 = 15.
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.
A multiple-choice test has 15 questions, each with 4 options and only one correct answer. A student guesses randomly on every question. Let X be the number of correct guesses, which follows a binomial distribution. Calculate the probability that the student gets at least 5 questions correct.
[3 marks]
Marking points
- States n = 15 and p = 0.25 (probability of guessing correctly on any one question).
- Uses P(X ≥ 5) = 1 − P(X ≤ 4).
- Obtains P(X ≥ 5) ≈ 0.314.
Examiner tip: ‘At least’ problems are usually easiest using the complement: P(X ≥ k) = 1 − P(X ≤ k − 1), which your GDC's cumulative binomial function calculates directly.
12.
Marking analysis: A learner attempts the following task: “A multiple-choice test has 15 questions, each with 4 options and only one correct answer. A student guesses randomly on every question. Let X be the number of correct guesses, which follows a binomial distribution. Calculate the probability that the student gets at least 5 questions correct.” Their response addresses only this point: “States n = 15 and p = 0.25 (probability of guessing correctly on any one question).” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]
Marking points
- Recognises credit for the stated point: States n = 15 and p = 0.25 (probability of guessing correctly on any one question).
- Identifies the missing requirement: Uses P(X ≥ 5) = 1 − P(X ≤ 4).
- Identifies the missing requirement: Obtains P(X ≥ 5) ≈ 0.314.
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 machine fills bottles with a volume that is normally distributed with mean 60 ml. It is known that 85% of bottles contain less than 70 ml. Find the standard deviation of the filling volume, giving your answer to 3 significant figures.
[4 marks]
Marking points
- Uses the inverse normal function to find the z-value corresponding to a cumulative probability of 0.85, obtaining z ≈ 1.0364.
- Sets up the standardisation equation z = (70 − 60)/σ.
- Rearranges to make σ the subject.
- Obtains σ ≈ 9.65 ml.
Examiner tip: When a standard deviation is unknown, find the z-value from the given probability first, then solve the standardisation formula for σ instead of x.
14.
Marking analysis: A learner attempts the following task: “A machine fills bottles with a volume that is normally distributed with mean 60 ml. It is known that 85% of bottles contain less than 70 ml. Find the standard deviation of the filling volume, giving your answer to 3 significant figures.” Their response addresses only this point: “Uses the inverse normal function to find the z-value corresponding to a cumulative probability of 0.85, obtaining z ≈ 1.0364.” 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 the inverse normal function to find the z-value corresponding to a cumulative probability of 0.85, obtaining z ≈ 1.0364.
- Identifies the missing requirement: Sets up the standardisation equation z = (70 − 60)/σ.
- Identifies the missing requirement: Rearranges to make σ the subject.
- Identifies the missing requirement: Obtains σ ≈ 9.65 ml.
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.
Six judges rank six gymnasts (A to F) in a competition. The ranks given by Judge 1 and Judge 2 are: Judge 1: A=1, B=2, C=3, D=4, E=5, F=6. Judge 2: A=2, B=1, C=4, D=3, E=6, F=5. Calculate Spearman's rank correlation coefficient between the two judges' rankings, and interpret its value.
[4 marks]
Marking points
- Finds the difference d between each pair of ranks: 1, 1, 1, 1, 1, 1 (in absolute value).
- Calculates Σd² = 6.
- Uses the formula rs = 1 − 6Σd²/[n(n² − 1)] with n = 6.
- Obtains rs ≈ 0.829, interpreting this as a fairly strong positive agreement between the two judges' rankings.
Examiner tip: Spearman's coefficient measures agreement in ranking order, not in the exact numerical values — it is the right tool when only ranks, not raw scores, are given.
16.
Marking analysis: A learner attempts the following task: “Six judges rank six gymnasts (A to F) in a competition. The ranks given by Judge 1 and Judge 2 are: Judge 1: A=1, B=2, C=3, D=4, E=5, F=6. Judge 2: A=2, B=1, C=4, D=3, E=6, F=5. Calculate Spearman's rank correlation coefficient between the two judges' rankings, and interpret its value.” Their response addresses only this point: “Finds the difference d between each pair of ranks: 1, 1, 1, 1, 1, 1 (in absolute value).” 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: Finds the difference d between each pair of ranks: 1, 1, 1, 1, 1, 1 (in absolute value).
- Identifies the missing requirement: Calculates Σd² = 6.
- Identifies the missing requirement: Uses the formula rs = 1 − 6Σd²/[n(n² − 1)] with n = 6.
- Identifies the missing requirement: Obtains rs ≈ 0.829, interpreting this as a fairly strong positive agreement between the two judges' rankings.
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 six-sided die is rolled 60 times, with observed frequencies for faces 1 to 6 of: 8, 12, 9, 11, 10, 10. Test, at the 5% significance level, whether the die is fair. (a) State the null hypothesis. (b) Calculate the chi-squared test statistic. (c) Given a critical value of 11.07, state your conclusion.
[5 marks]
Marking points
- States H0: the die is fair (each face has probability 1/6).
- States the expected frequency for each face as 60 ÷ 6 = 10.
- Uses χ² = Σ(O − E)²/E across all six faces.
- Obtains χ² = 1.00.
- States that since 1.00 is less than the critical value 11.07, H0 is not rejected, so there is no evidence the die is unfair.
Examiner tip: A goodness-of-fit test compares observed data to a single theoretical distribution, unlike a test for independence, which compares two categorical variables to each other.
18.
Marking analysis: A learner attempts the following task: “A six-sided die is rolled 60 times, with observed frequencies for faces 1 to 6 of: 8, 12, 9, 11, 10, 10. Test, at the 5% significance level, whether the die is fair. (a) State the null hypothesis. (b) Calculate the chi-squared test statistic. (c) Given a critical value of 11.07, state your conclusion.” Their response addresses only this point: “States H0: the die is fair (each face has probability 1/6).” 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: States H0: the die is fair (each face has probability 1/6).
- Identifies the missing requirement: States the expected frequency for each face as 60 ÷ 6 = 10.
- Identifies the missing requirement: Uses χ² = Σ(O − E)²/E across all six faces.
- Identifies the missing requirement: Obtains χ² = 1.00.
- Identifies the missing requirement: States that since 1.00 is less than the critical value 11.07, H0 is not rejected, so there is no evidence the die is unfair.
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 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. Calculate E(X) and Var(X).
[4 marks]
Marking points
- Confirms the probabilities sum to 1.
- Uses E(X) = Σx·P(X=x) to obtain E(X) = 1.7.
- Uses E(X²) = Σx²·P(X=x) to obtain E(X²) = 3.7.
- Uses Var(X) = E(X²) − [E(X)]² to obtain Var(X) = 0.81.
Examiner tip: Always calculate E(X²) as a separate step before finding the variance — do not square E(X) and call it E(X²).
20.
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. Calculate E(X) and Var(X).” Their response addresses only this point: “Confirms the probabilities sum to 1.” 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: Confirms the probabilities sum to 1.
- Identifies the missing requirement: Uses E(X) = Σx·P(X=x) to obtain E(X) = 1.7.
- Identifies the missing requirement: Uses E(X²) = Σx²·P(X=x) to obtain E(X²) = 3.7.
- Identifies the missing requirement: Uses Var(X) = E(X²) − [E(X)]² 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.
21.
A school has 120 students in Grade 10, 180 in Grade 11, and 300 in Grade 12. A stratified sample of 50 students, proportional to grade size, is to be selected. Calculate the number of students that should be sampled from each grade.
[4 marks]
Marking points
- States the total population as 120 + 180 + 300 = 600.
- Calculates the Grade 10 sample as (120/600) × 50 = 10.
- Calculates the Grade 11 sample as (180/600) × 50 = 15.
- Calculates the Grade 12 sample as (300/600) × 50 = 25, and confirms the three samples total 50.
Examiner tip: Stratified sampling keeps each group's share of the full sample proportional to its share of the whole population.
22.
Marking analysis: A learner attempts the following task: “A school has 120 students in Grade 10, 180 in Grade 11, and 300 in Grade 12. A stratified sample of 50 students, proportional to grade size, is to be selected. Calculate the number of students that should be sampled from each grade.” Their response addresses only this point: “States the total population as 120 + 180 + 300 = 600.” 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 total population as 120 + 180 + 300 = 600.
- Identifies the missing requirement: Calculates the Grade 10 sample as (120/600) × 50 = 10.
- Identifies the missing requirement: Calculates the Grade 11 sample as (180/600) × 50 = 15.
- Identifies the missing requirement: Calculates the Grade 12 sample as (300/600) × 50 = 25, and confirms the three samples total 50.
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.
Two classes sit the same test. Class A has a mean score of 68 with a standard deviation of 5. Class B has a mean score of 68 with a standard deviation of 14. Explain what the difference in standard deviation tells you about the two classes' results, even though their means are equal.
[2 marks] · no calculator
Marking points
- Explains that Class A's scores are more tightly clustered around the mean of 68, while Class B's scores are much more spread out.
- States that despite having the same average, Class B has much greater variation between individual students' performance.
Examiner tip: Two data sets can share the same mean but tell very different stories — standard deviation is essential for comparing how consistent or spread out results actually are.
24.
Marking analysis: A learner attempts the following task: “Two classes sit the same test. Class A has a mean score of 68 with a standard deviation of 5. Class B has a mean score of 68 with a standard deviation of 14. Explain what the difference in standard deviation tells you about the two classes' results, even though their means are equal.” Their response addresses only this point: “Explains that Class A's scores are more tightly clustered around the mean of 68, while Class B's scores are much more spread out.” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[2 marks] · no calculator
Marking points
- Recognises credit for the stated point: Explains that Class A's scores are more tightly clustered around the mean of 68, while Class B's scores are much more spread out.
- Identifies the missing requirement: States that despite having the same average, Class B has much greater variation between individual students' performance.
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.
A scientist records the growth of a plant, x days after planting, and its height y cm: (2,5), (4,9), (6,13), (8,16), (10,21), (12,24). Using your GDC: (a) find the equation of the linear regression line of y on x. (b) State the value of r and interpret the strength of the correlation. (c) Use your equation to estimate the height at x = 7 days.
[4 marks]
Marking points
- Uses the GDC's regression feature to obtain y ≈ 1.91x + 1.27 (accept equivalent rounding).
- States r ≈ 0.998.
- Interprets this as a very strong positive linear correlation.
- Substitutes x = 7 to estimate y ≈ 14.7 cm.
Examiner tip: Since x = 7 lies within the range of the recorded data (2 to 12), this estimate is interpolation, which is generally more reliable than extrapolation.
26.
Marking analysis: A learner attempts the following task: “A scientist records the growth of a plant, x days after planting, and its height y cm: (2,5), (4,9), (6,13), (8,16), (10,21), (12,24). Using your GDC: (a) find the equation of the linear regression line of y on x. (b) State the value of r and interpret the strength of the correlation. (c) Use your equation to estimate the height at x = 7 days.” Their response addresses only this point: “Uses the GDC's regression feature to obtain y ≈ 1.91x + 1.27 (accept equivalent rounding).” 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 the GDC's regression feature to obtain y ≈ 1.91x + 1.27 (accept equivalent rounding).
- Identifies the missing requirement: States r ≈ 0.998.
- Identifies the missing requirement: Interprets this as a very strong positive linear correlation.
- Identifies the missing requirement: Substitutes x = 7 to estimate y ≈ 14.7 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.
27.
A large box of components contains 40% defective items. A random sample of 40 components is taken, and X is the number of defective components, which follows a binomial distribution. Calculate the mean and variance of X.
[3 marks]
Marking points
- States n = 40 and p = 0.4.
- Uses E(X) = np to obtain a mean of 16.
- Uses Var(X) = np(1 − p) to obtain a variance of 9.6.
Examiner tip: For a binomial distribution, mean and variance both have direct formulas in terms of n and p — there is no need to list every outcome.
28.
Marking analysis: A learner attempts the following task: “A large box of components contains 40% defective items. A random sample of 40 components is taken, and X is the number of defective components, which follows a binomial distribution. Calculate the mean and variance of X.” Their response addresses only this point: “States n = 40 and p = 0.4.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]
Marking points
- Recognises credit for the stated point: States n = 40 and p = 0.4.
- Identifies the missing requirement: Uses E(X) = np to obtain a mean of 16.
- Identifies the missing requirement: Uses Var(X) = np(1 − p) to obtain a variance of 9.6.
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.
The lifetime of a particular brand of light bulb is normally distributed with mean 500 hours and standard deviation 20 hours. (a) Calculate the probability that a randomly selected bulb lasts between 480 and 530 hours. (b) Find the lifetime x such that only 20% of bulbs last longer than x hours.
[4 marks]
Marking points
- Uses the normal distribution with mean 500 and standard deviation 20.
- Obtains P(480 < X < 530) ≈ 0.775.
- Recognises that P(X > x) = 0.2 is equivalent to P(X < x) = 0.8 for the inverse normal function.
- Uses the inverse normal function to obtain x ≈ 516.83 hours.
Examiner tip: Your GDC's inverse normal function always expects the area to the left — convert an 'at least' or 'more than' probability to its complement before using it.
30.
Marking analysis: A learner attempts the following task: “The lifetime of a particular brand of light bulb is normally distributed with mean 500 hours and standard deviation 20 hours. (a) Calculate the probability that a randomly selected bulb lasts between 480 and 530 hours. (b) Find the lifetime x such that only 20% of bulbs last longer than x hours.” Their response addresses only this point: “Uses the normal distribution with mean 500 and standard deviation 20.” 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 the normal distribution with mean 500 and standard deviation 20.
- Identifies the missing requirement: Obtains P(480 < X < 530) ≈ 0.775.
- Identifies the missing requirement: Recognises that P(X > x) = 0.2 is equivalent to P(X < x) = 0.8 for the inverse normal function.
- Identifies the missing requirement: Uses the inverse normal function to obtain x ≈ 516.83 hours.
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.
A survey records smoking status (smoker/non-smoker) across three age groups, with observed frequencies: Age group 1: 30 smokers, 20 non-smokers. Age group 2: 25 smokers, 35 non-smokers. Age group 3: 15 smokers, 25 non-smokers. Test, at the 5% significance level with a critical value of 5.991, whether smoking status is independent of age group. (a) State the null hypothesis. (b) Calculate the chi-squared test statistic. (c) State your conclusion.
[6 marks]
Marking points
- States H0: smoking status is independent of age group.
- Calculates row totals (50, 60, 40), column totals (70, 80), and a grand total of 150.
- Calculates each expected frequency as (row total × column total) ÷ grand total.
- Uses χ² = Σ(O − E)²/E across all six cells.
- Obtains χ² ≈ 5.52.
- States that since 5.52 is less than the critical value 5.991, H0 is not rejected, so there is insufficient evidence of an association between smoking status and age group.
Examiner tip: Calculate every expected frequency carefully before computing chi-squared; a single arithmetic slip in the expected table changes the whole result.
32.
Marking analysis: A learner attempts the following task: “A survey records smoking status (smoker/non-smoker) across three age groups, with observed frequencies: Age group 1: 30 smokers, 20 non-smokers. Age group 2: 25 smokers, 35 non-smokers. Age group 3: 15 smokers, 25 non-smokers. Test, at the 5% significance level with a critical value of 5.991, whether smoking status is independent of age group. (a) State the null hypothesis. (b) Calculate the chi-squared test statistic. (c) State your conclusion.” Their response addresses only this point: “States H0: smoking status is independent of age group.” 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: smoking status is independent of age group.
- Identifies the missing requirement: Calculates row totals (50, 60, 40), column totals (70, 80), and a grand total of 150.
- Identifies the missing requirement: Calculates each expected frequency as (row total × column total) ÷ grand total.
- Identifies the missing requirement: Uses χ² = Σ(O − E)²/E across all six cells.
- Identifies the missing requirement: Obtains χ² ≈ 5.52.
- Identifies the missing requirement: States that since 5.52 is less than the critical value 5.991, H0 is not rejected, so there is insufficient evidence of an association between smoking status and age group.
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.