Mathematics AA: Standard Level
Statistics and probability — Topic 4
- 1.
A discrete random variable X takes the values 0, 1 and 2 with probabilities 0.25, 0.50 and 0.25. Calculate E(X) and Var(X).
[5 marks] - 2.
Marking analysis: A learner attempts the following task: “A discrete random variable X takes the values 0, 1 and 2 with probabilities 0.25, 0.50 and 0.25. Calculate E(X) and Var(X).” Their response addresses only this point: “Calculates E(X) = 0(0.25) + 1(0.50) + 2(0.25).” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks] - 3.
A biased coin has probability 0.60 of landing heads. It is tossed 8 times. Find the probability of exactly 5 heads and state the binomial parameters used.
[4 marks] - 4.
Marking analysis: A learner attempts the following task: “A biased coin has probability 0.60 of landing heads. It is tossed 8 times. Find the probability of exactly 5 heads and state the binomial parameters used.” Their response addresses only this point: “States X ~ B(8, 0.60).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] - 5.
A class test has scores that are normally distributed with mean 68 and standard deviation 8. Find the probability that a randomly chosen student scores above 80.
[3 marks] - 6.
Marking analysis: A learner attempts the following task: “A class test has scores that are normally distributed with mean 68 and standard deviation 8. Find the probability that a randomly chosen student scores above 80.” Their response addresses only this point: “Standardises using z = (80 − 68)/8 = 1.5.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] - 7.
The events A and B satisfy P(A) = 0.4, P(B) = 0.5 and P(A ∩ B) = 0.2. Determine whether A and B are independent, showing your reasoning.
[3 marks] - 8.
Marking analysis: A learner attempts the following task: “The events A and B satisfy P(A) = 0.4, P(B) = 0.5 and P(A ∩ B) = 0.2. Determine whether A and B are independent, showing your reasoning.” Their response addresses only this point: “States the independence condition P(A)×P(B) = P(A ∩ B).” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] - 9.
A survey of 200 students found that 120 study Mathematics and 90 study Physics, with 50 studying both. A student is chosen at random. Find the probability that the student studies Mathematics or Physics (or both).
[4 marks] - 10.
Marking analysis: A learner attempts the following task: “A survey of 200 students found that 120 study Mathematics and 90 study Physics, with 50 studying both. A student is chosen at random. Find the probability that the student studies Mathematics or Physics (or both).” Their response addresses only this point: “Uses the addition rule P(M ∪ P) = P(M) + P(P) − P(M ∩ P).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] - 11.
A discrete random variable X has probability distribution P(X = x) = kx for x = 1, 2, 3, 4, and P(X = x) = 0 otherwise. Find the value of k and calculate E(X).
[5 marks] - 12.
Marking analysis: A learner attempts the following task: “A discrete random variable X has probability distribution P(X = x) = kx for x = 1, 2, 3, 4, and P(X = x) = 0 otherwise. Find the value of k and calculate E(X).” Their response addresses only this point: “Uses the condition that all probabilities sum to 1: k(1+2+3+4) = 1.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks] - 13.
Exam scores for a large cohort are approximately normally distributed with mean 62 and standard deviation 10. Find the score that separates the top 10% of students from the rest.
[4 marks] - 14.
Marking analysis: A learner attempts the following task: “Exam scores for a large cohort are approximately normally distributed with mean 62 and standard deviation 10. Find the score that separates the top 10% of students from the rest.” Their response addresses only this point: “Identifies that the required z-value satisfies P(Z > z) = 0.10.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] - 15.
A fair six-sided die is rolled twice. Find the probability that the sum of the two rolls is greater than 9.
[4 marks] - 16.
Marking analysis: A learner attempts the following task: “A fair six-sided die is rolled twice. Find the probability that the sum of the two rolls is greater than 9.” Their response addresses only this point: “Identifies that there are 36 equally likely outcomes in total.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] - 17.
A test for a disease is 95% accurate for people who have the disease and 90% accurate for people who do not. In a population, 2% of people have the disease. Find the probability that a randomly selected person tests positive.
[5 marks] - 18.
Marking analysis: A learner attempts the following task: “A test for a disease is 95% accurate for people who have the disease and 90% accurate for people who do not. In a population, 2% of people have the disease. Find the probability that a randomly selected person tests positive.” Their response addresses only this point: “Identifies the two paths to a positive test: (disease and correctly positive) or (no disease and incorrectly positive).” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks] - 19.
Each of 10 independent calls to a service desk is resolved on the first call with probability 0.6. Find the probability that exactly 4 calls are resolved on the first call.
[3 marks] - 20.
Marking analysis: A learner attempts the following task: “Each of 10 independent calls to a service desk is resolved on the first call with probability 0.6. Find the probability that exactly 4 calls are resolved on the first call.” Their response addresses only this point: “Models the number resolved as X ~ B(10, 0.6).” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] - 21.
Events A and B satisfy P(A) = 0.6, P(B) = 0.5, and P(A ∪ B) = 0.8. Find P(A ∩ B) and hence P(A | B).
[4 marks] - 22.
Marking analysis: A learner attempts the following task: “Events A and B satisfy P(A) = 0.6, P(B) = 0.5, and P(A ∪ B) = 0.8. Find P(A ∩ B) and hence P(A | B).” Their response addresses only this point: “Uses P(A ∪ B) = P(A) + P(B) − P(A ∩ B) to find P(A ∩ B) = 0.6 + 0.5 − 0.8.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] - 23.
X ~ B(15, 0.4). Find E(X) and Var(X).
[3 marks] - 24.
Marking analysis: A learner attempts the following task: “X ~ B(15, 0.4). Find E(X) and Var(X).” Their response addresses only this point: “States E(X) = np.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] - 25.
The table shows the number of siblings for 20 students: 0 siblings (5 students), 1 sibling (8 students), 2 siblings (5 students), 3 siblings (2 students). Calculate the mean and standard deviation of the number of siblings.
[5 marks] - 26.
Marking analysis: A learner attempts the following task: “The table shows the number of siblings for 20 students: 0 siblings (5 students), 1 sibling (8 students), 2 siblings (5 students), 3 siblings (2 students). Calculate the mean and standard deviation of the number of siblings.” Their response addresses only this point: “Calculates Σfx = 0(5) + 1(8) + 2(5) + 3(2) = 24.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks] - 27.
A scatter diagram of hours studied against exam score gives a Pearson's product-moment correlation coefficient of r = 0.85. Interpret this value in context, and state one limitation of using r to conclude that studying causes higher scores.
[3 marks] · no calculator - 28.
Marking analysis: A learner attempts the following task: “A scatter diagram of hours studied against exam score gives a Pearson's product-moment correlation coefficient of r = 0.85. Interpret this value in context, and state one limitation of using r to conclude that studying causes higher scores.” Their response addresses only this point: “States that r = 0.85 indicates a strong, positive linear correlation between hours studied and exam score.” 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 calculator - 29.
A field study models afternoon air temperature T (°C) using tree-canopy cover x (%) with T = 31.8 − 0.084x. The correlation coefficient is r = −0.91. (a) Predict T when x = 65. (b) Interpret the gradient in context. (c) Explain why this model alone does not prove that increasing canopy cover causes the temperature change. (d) State the effect on r if every recorded canopy-cover value were increased by 5 percentage points.
[4 marks] - 30.
Marking analysis: A learner attempts the following task: “A field study models afternoon air temperature T (°C) using tree-canopy cover x (%) with T = 31.8 − 0.084x. The correlation coefficient is r = −0.91. (a) Predict T when x = 65. (b) Interpret the gradient in context. (c) Explain why this model alone does not prove that increasing canopy cover causes the temperature change. (d) State the effect on r if every recorded canopy-cover value were increased by 5 percentage points.” Their response addresses only this point: “Substitutes x = 65 to obtain T = 31.8 − 0.084(65) = 26.34 °C (about 26.3 °C).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] - 31.
X takes values 0, 1 and 2 with probabilities 0.2, 0.5 and 0.3 respectively. Find E(X) and Var(X).
[3 marks] · no calculator - 32.
Marking analysis: A learner attempts the following task: “X takes values 0, 1 and 2 with probabilities 0.2, 0.5 and 0.3 respectively. Find E(X) and Var(X).” Their response addresses only this point: “E(X) = 0(0.2) + 1(0.5) + 2(0.3) = 1.1.” 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 calculator - 33.
A fitted model predicts y = 2.4x + 15 from data with 1 <= x <= 5. Predict y at x = 3 and explain why using this model at x = 20 is less reliable.
[3 marks] · no calculator - 34.
Marking analysis: A learner attempts the following task: “A fitted model predicts y = 2.4x + 15 from data with 1 <= x <= 5. Predict y at x = 3 and explain why using this model at x = 20 is less reliable.” Their response addresses only this point: “At x = 3, y = 2.4(3) + 15 = 22.2.” 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 calculator - 35.
Six independent trials each have success probability 0.3. Find the probability of at least two successes and state the distribution used.
[4 marks] - 36.
Marking analysis: A learner attempts the following task: “Six independent trials each have success probability 0.3. Find the probability of at least two successes and state the distribution used.” Their response addresses only this point: “X follows Binomial(6,0.3).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] - 37.
Scores are modelled by a normal distribution with mean 70 and standard deviation 8. Find P(62 < X < 78) to three significant figures and interpret the result as a model prediction.
[3 marks] - 38.
Marking analysis: A learner attempts the following task: “Scores are modelled by a normal distribution with mean 70 and standard deviation 8. Find P(62 < X < 78) to three significant figures and interpret the result as a model prediction.” Their response addresses only this point: “Standardised limits are (62 - 70)/8 = -1 and (78 - 70)/8 = 1.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] - 39.
Factory A makes 60% of a company's items and factory B makes 40%. Defect rates are 2% at A and 5% at B. An item chosen at random is defective. Find the probability it came from B.
[5 marks] - 40.
Marking analysis: A learner attempts the following task: “Factory A makes 60% of a company's items and factory B makes 40%. Defect rates are 2% at A and 5% at B. An item chosen at random is defective. Find the probability it came from B.” Their response addresses only this point: “P(A and defective) = 0.6(0.02) = 0.012.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks]