A Level Paper 3: Statistics and Hypothesis Testing
Covers coordinate geometry, exponentials and logarithms, sampling and data presentation, the binomial probability distribution, and the normal distribution and hypothesis testing.
Questions
Question 1 [2 marks]
Probability and the Binomial Distribution
Events A and B are mutually exclusive, with P(A) = 0.35 and P(B) = 0.22.
Find P(A or B).
Question 2 [2 marks]
Exponentials and Logarithms
Without using a calculator, find the exact value of log_2(32).
Question 3 [2 marks]
The Normal Distribution and Hypothesis Testing
The random variable V ~ N(56, 9^2).
Find P(V > 47).
Question 4 [4 marks]
Sampling and Data Presentation
A company has 200 employees, numbered 1 to 200 on a staff list.
Describe how a simple random sample of 15 employees could be selected using this list and a random number generator, and state one advantage and one disadvantage of simple random sampling compared with stratified sampling.
Question 5 [3 marks]
Coordinate Geometry
Show that the points A(1, 2), B(4, 8) and C(6, 12) are collinear.
Question 6 [4 marks]
Sampling and Data Presentation
The regression line of y on x for a set of data is y = 3.2 + 1.5x, where x is the number of hours of sunshine and y is ice cream sales in hundreds of pounds, valid for 2 <= x <= 10.
Use the regression line to estimate the sales when x = 6, and interpret the gradient of the line in context.
Question 7 [4 marks]
Exponentials and Logarithms
Given that log_a(5) = p and log_a(2) = q, express log_a(20) in terms of p and q.
Question 8 [4 marks]
Probability and the Binomial Distribution
A random variable Y ~ B(15, 0.2).
Find P(Y <= 2), giving your answer to 3 significant figures.
Question 9 [6 marks]
Sampling and Data Presentation
The number of pets owned by each of 8 families is: 0, 1, 1, 2, 2, 3, 4, 7.
Find the mean number of pets, and find the standard deviation, giving your answers to 3 significant figures. State, with a reason, whether the mean or the median would better represent a typical family in this data set.
Question 10 [6 marks]
Coordinate Geometry
A circle has equation x^2 + y^2 + 8x - 2y + 8 = 0.
Find the centre and radius of the circle, and determine whether the point (0, 5) lies inside, on, or outside the circle.
Question 11 [6 marks]
The Normal Distribution and Hypothesis Testing
The random variable X ~ N(mu, sigma^2). Given that P(X < 40) = 0.10 and P(X > 70) = 0.20.
Find the value of mu and the value of sigma.
Question 12 [5 marks]
Exponentials and Logarithms
Without using a calculator, solve the equation (log_2(x))^2 - 5 log_2(x) + 6 = 0, giving both values of x as exact values.
Question 13 [6 marks]
The Normal Distribution and Hypothesis Testing
A seed supplier claims that 40% of a particular type of seed germinate within one week of planting. A gardener believes the true proportion is lower and plants a random sample of 25 of the seeds.
Using the binomial distribution X ~ B(25, 0.4), find the critical region for a test of H0: p = 0.4 against H1: p < 0.4 at the 5% significance level, stating the actual significance level. Given that 6 of the 25 seeds germinate within a week, state the conclusion of the test.
Question 14 [6 marks]
Probability and the Binomial Distribution
A random variable X ~ B(n, 0.25).
Given that P(X = 2) = P(X = 3), find the value of n.
Model solutions
| Question 1[2 marks] | |
|---|---|
| Answer or working | Marks |
| using P(A or B) = P(A) + P(B) for mutually exclusive events | M1 |
| 0.57 | A1 |
| Question 2[2 marks] | |
|---|---|
| Answer or working | Marks |
| recognising 32 = 2^5 | M1 |
| log_2(32) = 5 | A1 |
| Final answer: 5 | |
| Question 3[2 marks] | |
|---|---|
| Answer or working | Marks |
| standardising z = (47 - 56)/9 = -1 | M1 |
| P(V > 47) = 0.841 (3 sf) | A1 |
| Final answer: 0.841 (3 sf) | |
| Question 4[4 marks] | |
|---|---|
| Answer or working | Marks |
| numbering the 200 employees 1 to 200 and using a random number generator to generate numbers in this range | B1 |
| continuing until 15 distinct numbers are generated (discarding repeats or out-of-range numbers) and selecting the matching employees | B1 |
| a valid advantage, e.g. every possible sample of size 15 is equally likely, so the method is free from selection bias | B1 |
| a valid disadvantage, e.g. it does not guarantee proportional representation of subgroups, unlike stratified sampling | B1 |
| Final answer: Number the employees 1-200 and use a random number generator to select 15 distinct numbers; simple random sampling is unbiased, but (unlike stratified sampling) does not guarantee subgroups are proportionally represented. | |
| Question 5[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 6[4 marks] | |
|---|---|
| Answer or working | Marks |
| substituting x = 6 into the regression equation | M1 |
| y = 12.2 (sales of 1220 pounds) | A1 |
| interpreting the gradient: for each extra hour of sunshine, sales increase on average by 1.5 hundred pounds (150 pounds) | B1 |
| noting the estimate is reliable, since x = 6 lies within the given data range 2 to 10 | B1 |
| Final answer: y = 12.2 (sales of 1220 pounds); each extra hour of sunshine is associated with an average increase in sales of 150 pounds | |
| Question 7[4 marks] | |
|---|---|
| Answer or working | Marks |
| writing 20 as 4 x 5 = 2^2 x 5 | M1 |
| using the multiplication law log_a(2^2 x 5) = log_a(2^2) + log_a(5) | M1 |
| using the power law log_a(2^2) = 2 log_a(2) = 2q | A1 |
| combining to give log_a(20) = p + 2q | A1 |
| Final answer: log_a(20) = p + 2q | |
| Question 8[4 marks] | |
|---|---|
| Answer or working | Marks |
| identifying the need to sum P(Y = 0) + P(Y = 1) + P(Y = 2) | M1 |
| setting up the correct binomial terms for each probability | M1 |
| the individual probabilities 0.0352, 0.132 and 0.231 (each to 3 sf) | A1 |
| the total P(Y <= 2) = 0.398 (3 sf) | A1 |
| Final answer: P(Y <= 2) = 0.398 (3 sf) | |
| Question 9[6 marks] | |
|---|---|
| Answer or working | Marks |
| summing the data to give 20 | M1 |
| the mean = 2.5 | A1 |
| finding the sum of squared deviations from the mean = 34 | M1 |
| the variance = 34/8 = 4.25 and the standard deviation = 2.06 (3 sf) | A1 |
| the median = 2 (the average of the two middle values 2 and 2) | B1 |
| a reasoned conclusion: since the value 7 is an outlier that inflates the mean, the median is a better measure of a typical family | B1 |
| Final answer: mean = 2.5, standard deviation = 2.06 (3 sf); median = 2, which better represents a typical family since the mean is inflated by the outlier value 7 | |
| Question 10[6 marks] | |
|---|---|
| Answer or working | Marks |
| completing the square on the x terms, (x + 4)^2 - 16 | M1 |
| completing the square on the y terms, (y - 1)^2 - 1 | M1 |
| centre (-4, 1) | A1 |
| radius 3 | A1 |
| finding the distance from (-4, 1) to (0, 5) as sqrt(4^2 + 4^2) = sqrt(32) | M1 |
| concluding the point lies outside the circle, since sqrt(32) (approx 5.66) > 3 | A1 |
| Final answer: centre (-4, 1), radius 3; (0, 5) lies outside the circle | |
| Question 11[6 marks] | |
|---|---|
| Answer or working | Marks |
| identifying the z-value for P(Z < z) = 0.10, i.e. z = -1.2816 | M1 |
| forming the equation (40 - mu)/sigma = -1.2816 | M1 |
| identifying the z-value for P(Z < z) = 0.80, i.e. z = 0.8416 | M1 |
| forming the equation (70 - mu)/sigma = 0.8416 | M1 |
| solving the simultaneous equations to sigma = 14.1 (3 sf) | A1 |
| mu = 58.1 (3 sf) | A1 |
| Final answer: mu = 58.1 (3 sf), sigma = 14.1 (3 sf) | |
| Question 12[5 marks] | |
|---|---|
| Answer or working | Marks |
| substituting y = log_2(x) to obtain the quadratic y^2 - 5y + 6 = 0 | M1 |
| factorising to (y - 2)(y - 3) = 0 | A1 |
| y = 2 or y = 3 | A1 |
| converting back using x = 2^y | M1 |
| x = 4 or x = 8 | A1 |
| Question 13[6 marks] | |
|---|---|
| Answer or working | Marks |
| identifying X ~ B(25, 0.4) and attempting cumulative probabilities under H0 | M1 |
| P(X <= 5) = 0.0294 (accept | A1awrt |
| P(X <= 6) = 0.0736 (accept awrt), confirming this exceeds 0.05 | A1 |
| the critical region X <= 5, with actual significance level 0.0294 (2.94%) | A1 |
| noting that 6 does not lie in the critical region, since 6 > 5 | A1 |
| the conclusion: insufficient evidence at the 5% level that the true proportion germinating is lower than 40% | A1 |
| Final answer: Critical region X <= 5 (significance level 0.0294); since 6 is not in the critical region, there is insufficient evidence that the true proportion is lower than 40% | |
| Question 14[6 marks] | |
|---|---|
| Answer or working | Marks |
| writing the ratio P(X = 3)/P(X = 2) using the binomial formula | M1 |
| simplifying the combinatorial part of the ratio to C(n,3)/C(n,2) = (n - 2)/3 | M1 |
| simplifying the probability part of the ratio to p/(1 - p) = 1/3 | A1 |
| forming the equation (n - 2)/3 x 1/3 = 1, since P(X = 3) = P(X = 2) | M1 |
| simplifying to (n - 2)/9 = 1 | A1 |
| n = 11 | A1 |