AS Paper 3: Statistics Focus
Covers proof and algebraic methods, coordinate geometry, differentiation, sampling and data presentation, and the binomial probability distribution.
Questions
Question 1 [2 marks]
Probability and the Binomial Distribution
Events A and B are independent, with P(A) = 0.4 and P(B) = 0.5.
Find P(A and B).
Question 2 [3 marks]
Coordinate Geometry
Show that the points A(1, 2), B(4, 8) and C(6, 12) are collinear.
Question 3 [3 marks]
Proof and Algebraic Methods
Without using a calculator, solve the inequality x^2 - 2x - 8 <= 0, giving your answer as a single inequality.
Question 4 [3 marks]
Differentiation
Without using a calculator, find dy/dx for y = 6*sqrt(x) + 5x, writing sqrt(x) as x^(1/2) before differentiating.
Question 5 [3 marks]
Sampling and Data Presentation
A researcher wants to select a sample of 50 diners from a large restaurant chain with 40 branches.
Describe how cluster sampling could be used to select the sample, and state one advantage of cluster sampling over stratified sampling in this context.
Question 6 [4 marks]
Differentiation
A curve has equation y = ln(5x^2 + 1).
Find dy/dx using the chain rule, and find the gradient of the curve at the point where x = 2, giving your answer to 3 significant figures.
Question 7 [4 marks]
Coordinate Geometry
The circle C has centre (-1, 2) and radius 5.
Show that the point (3, -1) lies on C, and find the equation of the tangent to C at this point.
Question 8 [4 marks]
Sampling and Data Presentation
A data set has minimum 12, lower quartile 18, median 30, upper quartile 34 and maximum 38.
Describe, with a reason, the skewness of the distribution.
Question 9 [4 marks]
Coordinate Geometry
The points A(-2, 5) and B(6, -1) are the endpoints of a diameter of a circle.
Find the coordinates of the centre of the circle and the radius.
Question 10 [5 marks]
Proof and Algebraic Methods
Without using a calculator, express 3/(x + 1) - 2/(x - 2) as a single fraction in its simplest form.
Question 11 [4 marks]
Coordinate Geometry
A circle has centre C(3, -2) and radius 6.
Determine, showing your working, whether the point P(8, 1) lies inside, on, or outside the circle.
Question 12 [4 marks]
Sampling and Data Presentation
A sample of 12 pairs of data gives a product moment correlation coefficient of r = 0.62. The critical value for a sample of size 12 at the 5% significance level (one-tail) is 0.497.
Test, at the 5% significance level, whether this provides evidence of positive correlation between the variables in the population.
Question 13 [5 marks]
Sampling and Data Presentation
A data set of journey times, in minutes, is coded using y = (x - 30)/5. The coded data has mean 8 and standard deviation 3.
Find the mean and standard deviation of the original journey times x.
Question 14 [6 marks]
Coordinate Geometry
A circle passes through the points A(6, 6) and B(8, 2).
Given that the centre of the circle lies on the line x + y = 5, find the coordinates of the centre.
Question 15 [6 marks]
Sampling and Data Presentation
A college has 900 students: 340 study Sciences, 260 study Humanities and 300 study Arts. A stratified sample of 45 students is to be selected by subject, and within Sciences, systematic sampling is then used to choose the required number from a numbered list of all 340 Science students.
Find the number of Science students needed in the sample, and find the sampling interval that should be used for the systematic sampling.
Model solutions
| Question 1[2 marks] | |
|---|---|
| Answer or working | Marks |
| using P(A and B) = P(A) x P(B) for independent events | M1 |
| 0.2 | A1 |
| Question 2[3 marks] | |
|---|---|
| Answer or working | Marks |
| finding the gradient of AB = 2 | M1 |
| finding the gradient of BC = 2 | M1 |
| concluding AB and BC have the same gradient and share the point B, so A, B and C are collinear | A1 |
| Final answer: Gradient AB = gradient BC = 2, so A, B and C are collinear. | |
| Question 3[3 marks] | |
|---|---|
| Answer or working | Marks |
| factorising x^2 - 2x - 8 as (x - 4)(x + 2) | M1 |
| identifying the critical values x = -2 and x = 4 | M1 |
| the solution -2 <= x <= 4 | A1 |
| Final answer: -2 <= x <= 4 | |
| Question 4[3 marks] | |
|---|---|
| Answer or working | Marks |
| rewriting 6*sqrt(x) as 6x^(1/2) | M1 |
| differentiating each term | M1 |
| dy/dx = 3/sqrt(x) + 5 | A1 |
| Question 5[3 marks] | |
|---|---|
| Answer or working | Marks |
| a correct description of cluster sampling in context, e.g. randomly select a small number of branches (clusters) and sample all, or a proportion of, diners within each selected branch | B1 |
| a second valid detail of the method, e.g. the branches selected should themselves be chosen using simple random sampling | B1 |
| a valid advantage over stratified sampling, e.g. it does not require prior knowledge of every diner across all 40 branches, and is more practical to carry out | B1 |
| Final answer: Any valid description of cluster sampling (randomly select branches, then sample diners within them) and a valid advantage (e.g. no need for a full list of every diner). | |
| Question 6[4 marks] | |
|---|---|
| Answer or working | Marks |
| using the chain rule d/dx[ln(u)] = (1/u)(du/dx) | M1 |
| dy/dx = 10x/(5x^2 + 1) | A1 |
| substituting x = 2 | M1 |
| the gradient = 20/21 (or 0.952 to 3 sf) | A1 |
| Final answer: dy/dx = 10x/(5x^2 + 1); gradient at x = 2 is 20/21 (0.952 to 3 sf) | |
| Question 7[4 marks] | |
|---|---|
| Answer or working | Marks |
| the distance from (-1, 2) to (3, -1) equal to sqrt(4^2 + 3^2) = 5, confirming the point lies on C | M1 |
| the radius gradient = -3/4, so the tangent gradient = 4/3 | M1 |
| the tangent as y + 1 = 4/3(x - 3) | A1 |
| the simplified tangent y = 4/3 x - 5 (or equivalent, e.g. 4x - 3y - 15 = 0) | A1 |
| Final answer: y = 4/3 x - 5 (or 4x - 3y - 15 = 0) | |
| Question 8[4 marks] | |
|---|---|
| Answer or working | Marks |
| finding median - lower quartile = 30 - 18 = 12 | M1 |
| finding upper quartile - median = 34 - 30 = 4 | M1 |
| comparing the two values, noting (median - lower quartile) is greater than (upper quartile - median) | A1 |
| concluding the distribution is negatively skewed | A1 |
| Final answer: Negatively skewed, since median - lower quartile (12) is greater than upper quartile - median (4) | |
| Question 9[4 marks] | |
|---|---|
| Answer or working | Marks |
| finding the centre as the midpoint of AB | M1 |
| centre (2, 2) | A1 |
| finding the length AB = sqrt(8^2 + 6^2) = 10 and halving to find the radius | M1 |
| radius = 5 | A1 |
| Final answer: centre (2, 2), radius 5 | |
| Question 10[5 marks] | |
|---|---|
| Answer or working | Marks |
| writing both fractions over the common denominator (x + 1)(x - 2) | M1 |
| expanding the first numerator to 3(x - 2) = 3x - 6 | M1 |
| expanding the second numerator to 2(x + 1) = 2x + 2 | M1 |
| combining to give the numerator x - 8 | A1 |
| the final answer (x - 8)/((x + 1)(x - 2)) | A1 |
| Final answer: (x - 8)/((x + 1)(x - 2)) | |
| Question 11[4 marks] | |
|---|---|
| Answer or working | Marks |
| using the distance formula to find CP | M1 |
| CP = sqrt(34) (or 5.83 to 3 sf) | A1 |
| comparing CP with the radius 6 | M1 |
| concluding P lies inside the circle, since sqrt(34) < 6 | A1 |
| Final answer: P lies inside the circle, since CP = sqrt(34) (approx 5.83) is less than the radius 6 | |
| Question 12[4 marks] | |
|---|---|
| Answer or working | Marks |
| stating hypotheses H0: rho = 0, H1: rho > 0, where rho is the population correlation coefficient | B1 |
| comparing the sample value r = 0.62 with the critical value 0.497 | M1 |
| noting 0.62 > 0.497 | A1 |
| the conclusion: reject H0, there is evidence of positive correlation between the variables in the population | A1 |
| Final answer: 0.62 > 0.497, so reject H0: evidence of positive correlation in the population | |
| Question 13[5 marks] | |
|---|---|
| Answer or working | Marks |
| rearranging the coding to x = 5y + 30 | M1 |
| using mean(x) = 5 x mean(y) + 30 | M1 |
| mean(x) = 70 | A1 |
| using standard deviation(x) = 5 x standard deviation(y), since adding a constant does not change the spread | M1 |
| standard deviation(x) = 15 | A1 |
| Final answer: Mean = 70 minutes, standard deviation = 15 minutes | |
| Question 14[6 marks] | |
|---|---|
| Answer or working | Marks |
| finding the midpoint of AB as (7, 4) | M1 |
| finding the gradient of AB = -2 | M1 |
| the perpendicular gradient 1/2 and the perpendicular bisector y = 0.5x + 0.5 | A1 |
| solving simultaneously with x + y = 5 | M1 |
| x = 3 | A1 |
| the centre (3, 2) | A1 |
| Final answer: Centre = (3, 2) | |
| Question 15[6 marks] | |
|---|---|
| Answer or working | Marks |
| the sampling fraction = 45/900 = 0.05 | M1 |
| multiplying the Science total by the fraction | M1 |
| the Science sample size = 17 | A1 |
| the systematic sampling interval = population size / sample size | M1 |
| the interval = 340/17 = 20 | A1 |
| a correct statement of how the systematic sample would then be chosen, e.g. select a random start between 1 and 20 and then every 20th student thereafter | B1 |
| Final answer: 17 Science students needed; systematic sampling interval = 20 | |