ZIMSEC A Level Statistics Paper 2 Specimen 2026

Question 101

[3 marks]continuous random variables

A continuous random variable

X

has probability density function

f(x) = 0.1x + k

for

4 \leq x \leq 6

f(x) = 0.3

for

6 \leq x \leq 8

, and

f(x) = 0

otherwise. Find the value of the constant

k

Answer this when you sit the paper.

Question 102

[3 marks]continuous random variables

A continuous random variable

X

has probability density function

f(x) = 0.1x - 0.3

for

4 \leq x \leq 6

f(x) = 0.3

for

6 \leq x \leq 8

, and

f(x) = 0

otherwise. Find

P(5 \leq X \leq 7)

Answer this when you sit the paper.

Question 201

[1 marks]data presentation

Examination marks are displayed as below, with 1 alongside a leaf 3 standing for the mark 13 %.

Stem	Leaves
1	3
2	6
3	1
4	1 3
5	0 2 6 8
6	1 2 2 2 7
7	0 3 4 5 5 8 9
8	0 3 4 4 8
9	2 7 7 8

key: 4|1 means 41 %

What is the name of this type of display?

AA stem and leaf diagram, which splits each value into a stem and a leaf
BA histogram, whose bar areas are proportional to the class frequencies
CA cumulative frequency curve, which plots running totals against class boundaries
DA box and whisker plot, which shows the median, the quartiles and the extremes

Question 202

[3 marks]data presentation

The marks obtained by candidates in a mathematics examination are shown below, where a stem of 1 with a leaf of 3 means 13 %.

Stem	Leaves
1	3
2	6
3	1
4	1 3
5	0 2 6 8
6	1 2 2 2 7
7	0 3 4 5 5 8 9
8	0 3 4 4 8
9	2 7 7 8

key: 4|1 means 41 %

Calculate the range of the marks.

Answer this when you sit the paper.

Question 203

[2 marks]data presentation

For a set of 30 examination marks the mean is 67.0 %, the median is 71.5 %, the lower quartile is 55 % and the upper quartile is 83.25 %. What does this say about the skewness of the distribution?

AIt is symmetrical, because the mean and the median both lie inside the interquartile range, which is what symmetry requires of a distribution
BIt is positively skewed, because the upper quartile is further from zero than the lower quartile is
CIt is negatively skewed, since the mean lies below the median and the lower half of the middle 50 % is the wider one
DIt is positively skewed, since the mean lies below the median and the upper tail must therefore be the longer one

Question 204

[2 marks]data presentation

Examination marks are recorded in a stem and leaf diagram rather than in a grouped frequency table. Which of the following is an advantage of the stem and leaf diagram?

AIt shows the shape of the distribution while hiding the individual values, which keeps the display uncluttered
BIt keeps every original data value, so the median, the range and the mode can all be read straight off it
CIt removes the need to order the data before the median can be found from the display
DIt replaces the raw values by class mid-points, which makes the mean quicker to calculate

Question 301

[2 marks]discrete random variables

A fair spinner lands on one of the six values 1, 2, 3, 4, 5 and 6, each equally likely. Find the probability that the spinner lands on a prime value.

Answer this when you sit the paper.

Question 302

[3 marks]discrete random variables

A fair spinner lands on one of six values, each equally likely, and each value pays the prize shown.

value	1	2	3	4	5	6
prize in $	2	2	6	4	10	6

Find the probability that the spinner lands on a value that gives a prize of not less than $4.

Answer this when you sit the paper.

Question 303

[3 marks]discrete random variables

A fair spinner lands on one of six values, each equally likely, and each value pays the prize shown.

value	1	2	3	4	5	6
prize in $	2	2	6	4	10	6

Calculate the expected prize for a single game.

Answer this when you sit the paper.

Question 401

[2 marks]normal and binomial distributions

The masses of letters are normally distributed with mean 15 g, and 92 % of them lie within 10 g of the mean. Since 92 % lies between the two symmetric limits, 4 % lies in each tail. State the value of

z

for which

P(Z < z) = 0.96

, to 3 decimal places.

Answer this when you sit the paper.

Question 402

[2 marks]normal and binomial distributions

The masses of letters posted by a school are normally distributed with mean 15 g, and the masses of 92 % of the letters are within 10 g of the mean. Given that

P(Z < 1.751) = 0.96

, find the standard deviation of the masses, in grams to 2 decimal places.

Answer this when you sit the paper.

Question 403

[2 marks]normal and binomial distributions

Each letter posted by a school has probability 0.92 of having a mass within 10 g of the mean, independently of the others. A random sample of 8 letters is taken. Which distribution does the number of letters in the sample whose mass is within 10 g of the mean follow?

AA Poisson distribution with mean $8 \times 0.92 = 7.36$ , since the letters are posted at random and independently
BA binomial distribution with 8 trials and probability of success 0.92
CA normal distribution with mean 8 and standard deviation 0.92
DA binomial distribution with 8 trials and probability of success 0.08

Question 404

[3 marks]normal and binomial distributions

The number of letters within 10 g of the mean, out of a random sample of 8, follows the binomial distribution

B(8, 0.92)

. Find the probability that at least 2 of the 8 letters are within 10 g of the mean, correct to 3 decimal places.

A0.083, which is $8(0.92)(0.08)^7$ doubled to allow for either of the two letters
B0.632, since at least 2 letters out of a sample of 8 is only a little under two thirds likely
C1.000, since the chance of 0 or 1 such letter is about $1.6 \times 10^{-7}$
D0.001, since two or more successes out of only 8 trials is very unlikely

Question 501

[2 marks]sampling

In statistics, what is meant by a random sample?

AA sample chosen by taking every $k$ th member of the population from a list, after a start is fixed
BA sample chosen so that it contains the same proportion of each group as the population does
CA sample chosen without any plan at all, so that the sampler simply takes whichever members of the population happen to come to hand first, in whatever order they appear
DA sample chosen so that every member of the population has an equal chance of being selected and every possible sample of that size is equally likely

Question 502

[1 marks]sampling

State one method by which a random sample can be obtained from a population.

Answer this when you sit the paper.

Question 503

[3 marks]sampling

A school was asked to send 10 students on an exchange programme and to supply their names within 3 days. The head chose 10 students from among those who already had valid passports. Which sampling method did the head use?

ASimple random sampling, since no student was named by the head before the selection was made
BConvenience sampling, since the students were taken from the group that was quickest and easiest to reach
CStratified sampling, since the school roll was divided into groups first and students were then drawn from each group in proportion to its size
DSystematic sampling, since the students were taken at a fixed interval down an ordered list of names

Question 504

[3 marks]sampling

A head of school had to name 10 students for an exchange programme within 3 days, and chose 10 from among those students who already had valid passports. Would this give a random sample?

ANo, because students without a valid passport had no chance at all of being chosen
BYes, because the head did not know in advance which of the passport holders would be picked
CYes, because 10 students is a large enough number for the selection to even out any bias
DNo, because a sample of 10 students is too small for any selection method to count as random

Question 601

[2 marks]Poisson distribution

An insurance company receives claims at random at an average rate of 3 a week. Which feature of the claims makes a Poisson model appropriate here?

AThe claims are received singly, at random and independently, at a constant average rate
BThe claims are received in a fixed number of independent trials, each with the same probability
CThe size of each claim is normally distributed about its own mean value
DThe number of claims received can never exceed the average of 3 a week by very much

Question 602

[3 marks]Poisson distribution

An insurance company receives on average 3 claims in a week, the claims occurring at random and independently. Find the probability that the company receives at least 2 claims in a given week, to 3 decimal places.

Answer this when you sit the paper.

Question 603

[2 marks]Poisson distribution

An insurance company receives on average 3 claims a week and works a 5 day week. What Poisson mean should be used for the number of claims received in one working day?

A15, since the weekly mean of 3 applies to each of the 5 working days
B1.7, since the mean for a day is the square root of the mean for a week
C3, since the average rate of claims does not change from one period to another
D0.6, since the weekly mean of 3 is shared over the 5 working days

Question 604

[3 marks]Poisson distribution

An insurance company receives on average 3 claims a week and works 5 days in a week, so claims arrive at a mean rate of 0.6 a day. Find the probability that the company receives exactly one claim in a day, to 3 decimal places.

Answer this when you sit the paper.

Question 605

[3 marks]Poisson distribution

An insurance company receives claims at random at an average rate of 3 a week. Find the probability that it receives a total of exactly 2 claims during 3 consecutive weeks, to 3 decimal places.

Answer this when you sit the paper.

Question 606

[3 marks]Poisson distribution

For an insurance company receiving claims at a mean rate of 3 a week, the probability of at least 2 claims in a given week is 0.8009, and the weeks are independent. Find the probability that at least 2 claims are received in exactly one of 3 consecutive weeks, to 3 decimal places.

Answer this when you sit the paper.

Question 701

[2 marks]regression and correlation

For 10 pairs of readings of speed

X

and fuel used

Y

\sum x = 955

\sum x^2 = 97\,025

and

n = 10

. Find

S_{xx}

Answer this when you sit the paper.

Question 702

[2 marks]regression and correlation

For 10 pairs of readings of speed

X

and fuel used

Y

\sum x = 955

\sum y = 106

\sum xy = 10\,810

and

n = 10

. Find

S_{xy}

Answer this when you sit the paper.

Question 703

[3 marks]regression and correlation

For 10 pairs of readings of speed

X

km/hr and fuel used

Y

litres,

\bar x = 95.5

\bar y = 10.6

S_{xx} = 5\,822.5

and

S_{xy} = 687

. Find the equation of the regression line of

Y

X

, giving the coefficients to 3 decimal places.

Answer this when you sit the paper.

Question 704

[3 marks]regression and correlation

The regression line of fuel used

Y

litres on speed

X

km/hr, fitted to speeds between 60 and 140 km/hr, is

Y = -0.668 + 0.118X

. Estimate the amount of fuel likely to be used when travelling at 105 km/hr, in litres to 1 decimal place.

Answer this when you sit the paper.

Question 705

[3 marks]regression and correlation

A regression line of fuel used on speed was fitted to ten readings taken at speeds between 60 and 140 km/hr. Why can the line not be relied on to estimate the fuel used at 50 km/hr?

ABecause the regression line of $Y$ on $X$ can only be used for values of $X$ that lie above the mean speed of the ten readings
BBecause the line gives a negative estimate at 50 km/hr, which cannot be a quantity of fuel
CBecause the product moment correlation coefficient is too far from 1 for any estimate to be made
DBecause 50 km/hr lies outside the range of the speeds observed, so using the line there is extrapolation

Question 706

[3 marks]regression and correlation

For 10 pairs of readings of speed and fuel used,

S_{xx} = 5\,822.5

S_{yy} = 88.4

and

S_{xy} = 687

. Find the product moment correlation coefficient, to 3 decimal places.

Answer this when you sit the paper.

Question 801

[3 marks]confidence intervals

What is meant by a 95 % confidence interval for a population mean?

AAn interval whose width is fixed at 95 % of the standard deviation of the sample that produced it
BAn interval containing 95 % of the individual values in the population from which the sample was drawn, centred on the observed sample mean
CAn interval, calculated from a sample, constructed by a method that captures the population mean in 95 % of repeated samples
DAn interval within which 95 % of all possible sample means would fall if the population mean were zero

Question 802

[2 marks]confidence intervals

A random sample of 50 buses gave a mean of 70 passengers with a standard deviation of 4. Find the standard error of the sample mean, to 4 decimal places.

Answer this when you sit the paper.

Question 803

[3 marks]confidence intervals

A random sample of 50 buses gave a mean of 70 passengers with a standard deviation of 4, so the standard error of the mean is 0.5657. Calculate the 95 % confidence interval for the mean number of passengers in a bus, giving both limits to 1 decimal place.

Answer this when you sit the paper.

Question 804

[3 marks]confidence intervals

The number of passengers in a bus is normally distributed with mean 70 and standard deviation 4. Calculate the probability that a randomly chosen bus had fewer than 65 passengers, to 3 decimal places.

Answer this when you sit the paper.

Question 805

[2 marks]confidence intervals

In constructing a symmetric confidence interval for a mean at the 90 % level, state the value of

z

that should be used, to 3 decimal places.

Answer this when you sit the paper.

Question 806

[3 marks]confidence intervals

The number of passengers in a bus has standard deviation 4. Calculate the sample size

n

that should be taken so that one is 90 % confident that the sample mean will be within 0.8 of the true mean, using

z = 1.645

Answer this when you sit the paper.

Question 901

[3 marks]hypothesis testing and chi-squared

What is the difference between a 1 tailed test and a 2 tailed test of a parameter?

AA 1 tailed test is carried out at the 5 % level and a 2 tailed test is always carried out at the 10 % level
BA 1 tailed test can only reject the null hypothesis, while a 2 tailed test can reject either hypothesis
CA 1 tailed test has an alternative naming a direction, so the whole significance level sits in one tail; a 2 tailed test splits it between both tails
DA 1 tailed test uses a sample from a single population, while a 2 tailed test compares samples drawn from two separate populations at the same significance level

Question 902

[2 marks]hypothesis testing and chi-squared

In a survey of newspaper readership the row total for the Northern province is 150, the column total for the newspaper 'current' is 160, and the grand total is 560. Calculate the expected frequency for Northern province readers of 'current', to 2 decimal places.

Answer this when you sit the paper.

Question 903

[2 marks]hypothesis testing and chi-squared

In a survey of newspaper readership the row total for the Southern province is 220, the column total for the newspaper 'News' is 190, and the grand total is 560. Calculate the expected frequency for Southern province readers of 'News', to 2 decimal places.

Answer this when you sit the paper.

Question 904

[2 marks]hypothesis testing and chi-squared

A chi-squared test of association is carried out on a contingency table with 3 rows (three provinces) and 3 columns (three newspapers). State the number of degrees of freedom.

Answer this when you sit the paper.

Question 905

[3 marks]hypothesis testing and chi-squared

A survey of newspaper readership in 3 provinces gave the observed frequencies below, with a grand total of 560.

Province	today	current	News
Northern	55	65	30
Central	80	48	62
Southern	75	47	98

Using expected frequencies of 56.25, 42.86, 50.89 (Northern), 71.25, 54.29, 64.46 (Central) and 82.50, 62.86, 74.64 (Southern), calculate the chi-squared test statistic, to 1 decimal place.

Answer this when you sit the paper.

Question 906

[2 marks]hypothesis testing and chi-squared

A chi-squared test of association between province and newspaper preference gives a test statistic of 33.9 on 4 degrees of freedom. The 5 % critical value of

\chi^2

with 4 degrees of freedom is 9.488. What is the conclusion at the 5 % level of significance?

ANo conclusion can be drawn, because a chi-squared statistic above 30 lies outside the range of the tables
BReject the null hypothesis: there is evidence of an association between province and newspaper preference
CAccept the null hypothesis: the two variables are independent, since 33.9 is well above the critical value
DReject the null hypothesis: the test proves that living in a province causes a reader to choose one newspaper

Question 907

[2 marks]hypothesis testing and chi-squared

For a chi-squared test of association carried out at the 5 % level of significance on 4 degrees of freedom, state the critical value.

Answer this when you sit the paper.

Question 1001

[1 marks]normal distribution

The mass of a randomly chosen key follows a normal distribution with mean 12 g and variance 9. State the standard deviation of the mass of a key, in grams.

Answer this when you sit the paper.

Question 1002

[2 marks]normal distribution

The mass of a key-holder is

N(20, 16)

and the mass of a key is

N(12, 9)

, all items independent. State the mean and the variance of the combined mass of 2 randomly chosen key-holders and 3 randomly chosen keys.

Answer this when you sit the paper.

Question 1003

[3 marks]normal distribution

The combined mass of 2 randomly chosen key-holders and 3 randomly chosen keys is normally distributed with mean 76 g and variance 59. Find the probability that this combined mass is greater than 78 g, to 3 decimal places.

Answer this when you sit the paper.

Question 1004

[2 marks]normal distribution

The mass of a key-holder is

N(20, 16)

and the mass of a key is

N(12, 9)

, all items independent. Let

D

be the combined mass of 3 key-holders minus the combined mass of 6 keys. State the mean and the variance of

D

Answer this when you sit the paper.

Question 1005

[3 marks]normal distribution

For independent normal masses,

D

= (mass of 3 key-holders) - (mass of 6 keys) has mean -12 and variance 102. Find the probability that the combined mass of 3 key-holders is greater than the combined mass of 6 keys, to 3 decimal places.

Answer this when you sit the paper.

Question 1006

[2 marks]normal distribution

The mass

M

of a key-holder is

N(20, 16)

and the mass

m

of a key is

N(12, 9)

, independently. What is the variance of

M - 2m

A16 - 18 = -2, because subtracting a quantity subtracts its variance as well as its mean
B16 + 18 = 34, because the variance of the key is doubled along with its mean
C16 + 36 = 52, because the multiplier 2 multiplies the key's variance by $2^2$
D16 + 9 = 25, because the multiplier changes the mean of the key but not its variance

Question 1007

[3 marks]normal distribution

For independent normal masses,

D = M - 2m

where

M

N(20, 16)

and

m

N(12, 9)

, so

D

has mean -4 and variance 52. Find the probability that a randomly chosen key-holder is more than twice the mass of a randomly chosen key, to 3 decimal places.

Answer this when you sit the paper.

Question 1101

[2 marks]grouped data and measures of location

The amounts 76 motorists spent on petrol are grouped in the classes

0 \leq x < 50

50 \leq x < 100

100 \leq x < 150

150 \leq x < 200

200 \leq x < 250

and

250 \leq x < 300

dollars. State the mid-points of the six classes.

Answer this when you sit the paper.

Question 1102

[2 marks]grouped data and measures of location

76 motorists recorded the amount they spent on petrol in a month.

petrol purchases ($)	number of motorists
$0 \leq x < 50$	4
$50 \leq x < 100$	11
$100 \leq x < 150$	8
$150 \leq x < 200$	16
$200 \leq x < 250$	22
$250 \leq x < 300$	15

Using the class mid-points 25, 75, 125, 175, 225 and 275, calculate the mean amount spent, in dollars to 2 decimal places.

Answer this when you sit the paper.

Question 1103

[2 marks]grouped data and measures of location

76 motorists recorded the amount they spent on petrol in a month.

petrol purchases ($)	number of motorists
$0 \leq x < 50$	4
$50 \leq x < 100$	11
$100 \leq x < 150$	8
$150 \leq x < 200$	16
$200 \leq x < 250$	22
$250 \leq x < 300$	15

State the number of motorists who spent less than $200.

Answer this when you sit the paper.

Question 1104

[3 marks]grouped data and measures of location

76 motorists recorded the amount they spent on petrol in a month.

petrol purchases ($)	number of motorists
$0 \leq x < 50$	4
$50 \leq x < 100$	11
$100 \leq x < 150$	8
$150 \leq x < 200$	16
$200 \leq x < 250$	22
$250 \leq x < 300$	15

Calculate the median amount spent, in dollars to 2 decimal places.

Answer this when you sit the paper.

Question 1105

[3 marks]grouped data and measures of location

For the petrol spending of 76 motorists, the class mid-points and frequencies give

\sum fx = 13\,800

and

\sum fx^2 = 2\,927\,500

, with mean

181.58

. Calculate the standard deviation, in dollars to 2 decimal places.

Answer this when you sit the paper.

Question 1106

[2 marks]grouped data and measures of location

A histogram is drawn for petrol spending grouped in six classes that are all 50 dollars wide, with frequencies 4, 11, 8, 16, 22 and 15. What should the height of each bar represent?

AThe frequency density, which here is simply proportional to the frequency because every class has the same width
BThe class mid-point, so that the tallest bar stands over the largest amount of money spent
CThe frequency divided by the total of 76, so that the heights of all six bars add up to 1
DThe cumulative frequency up to the upper boundary of that class, so that the bars always rise steadily from left to right across the display

Question 1107

[2 marks]grouped data and measures of location

In a histogram of petrol spending the modal class is

200 \leq x < 250

with frequency 22; the class below it has frequency 16 and the class above it has frequency 15, and every class is 50 wide. Estimate the mode, in dollars to 2 decimal places.

Answer this when you sit the paper.

Question 1201

[2 marks]time series

What is meant by seasonal variation in a time series?

AThe gradual long term movement of the series in one direction, which is what is left once the short term swings have been removed
BThe unpredictable one-off movement caused by an event outside the pattern, such as a strike
CThe difference between the largest and the smallest value the series takes over the whole period
DThe regular movement that repeats at the same point of every cycle, quarter by quarter or month by month

Question 1202

[2 marks]time series

What is meant by the trend in time series analysis?

AThe pattern that repeats itself in the same quarter of each successive year of the series
BThe underlying long term direction of the series once the seasonal and irregular movements are smoothed away
CThe largest of the seasonal variations in the series, which sets the direction the whole series must then take over time
DThe average of the values within any one year of the series, plotted at the middle of that year

Question 1203

[2 marks]time series

A pharmacy handled 1 700, 3 450, 2 800 and 2 300 customers in the four quarters of 2007, and 2 100 in the first quarter of 2008. Calculate the first 4-point moving average of this series.

Answer this when you sit the paper.

Question 1204

[2 marks]time series

A pharmacy handled these numbers of customers by quarter.

year	quarter	number of customers
2007	1	1 700
2007	2	3 450
2007	3	2 800
2007	4	2 300
2008	1	2 100
2008	2	3 500
2008	3	2 000
2008	4	2 000
2009	1	2 600
2009	2	4 600
2009	3	3 850
2009	4	3 800

Calculate the last of the nine 4-point moving averages of this series.

Answer this when you sit the paper.

Question 1205

[3 marks]time series

The first two 4-point moving averages of a quarterly series are 2 562.5 and 2 662.5. Calculate the first centred moving average, and state the quarter it is plotted at, given that the series starts at the first quarter of 2007.

Answer this when you sit the paper.

Question 1206

[2 marks]time series

The last two 4-point moving averages of a quarterly pharmacy series are 3 262.5 and 3 712.5. Calculate the last centred moving average.

Answer this when you sit the paper.

Question 1207

[3 marks]time series

The centred moving averages of a pharmacy's quarterly customer numbers run 2 612.5, 2 668.75, 2 575, 2 437.5, 2 462.5, 2 662.5, 3 031.25 and 3 487.5, from the third quarter of 2007 to the second quarter of 2009. What do they say about the trend?

AThe trend dips a little through 2008 and then rises steadily, so customer numbers are increasing overall
BThe trend falls steadily throughout, so the pharmacy is handling fewer customers year on year
CThe trend rises and falls with each quarter, which is the seasonal pattern in the data rather than a trend in the series
DThe series has no trend at all, since the values both fall and rise over the period covered

Statistics Paper 2 Specimen 2026

Sit this paper online

The questions

Question 101

Question 102

Question 201

Question 202

Question 203

Question 204

Question 301

Question 302

Question 303

Question 401

Question 402

Question 403

Question 404

Question 501

Question 502

Question 503

Question 504

Question 601

Question 602

Question 603

Question 604

Question 605

Question 606

Question 701

Question 702

Question 703

Question 704

Question 705

Question 706

Question 801

Question 802

Question 803

Question 804

Question 805

Question 806

Question 901

Question 902

Question 903

Question 904

Question 905

Question 906

Question 907

Question 1001

Question 1002

Question 1003

Question 1004

Question 1005

Question 1006

Question 1007

Question 1101

Question 1102

Question 1103

Question 1104

Question 1105

Question 1106

Question 1107

Question 1201

Question 1202

Question 1203

Question 1204

Question 1205

Question 1206

Question 1207

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