6 Statistics and probability

Exam analysis

Most tested topics

6.1 Graphical representation of data

Choose a display by matching it to the data and the comparison you need. Pictograms and bar charts show discrete categories; pie charts show parts of one whole; two-way tables classify each item by two categorical variables.

Display	Best use	Essential check
pictogram	small category frequencies	read the key, including partial symbols
bar chart	compare discrete categories	equal-width separated bars and a labelled scale
pie chart	compare proportions of one total	sectors total $360^\circ$
two-way table	cross-classify two categories	row and column totals agree with the grand total

For a pie chart, $\text{sector angle}=\dfrac{\text{frequency}}{\text{total}}\times360^\circ$ . Reverse the calculation with $\text{frequency}=\dfrac{\text{angle}}{360^\circ}\times\text{total}$ .

A taller bar may represent a larger frequency, but a wider bar in an ordinary bar chart does not. Do not use bar area unless the graph is a histogram.

A frequency table organises every observation into one category or class, records tallies in groups of five, and converts each tally into a frequency.

Step	Action
1	define categories or non-overlapping class intervals
2	process the raw list once, adding one tally per observation
3	convert each tally to a numerical frequency
4	add the frequencies and compare with the number of observations

Categories must be exhaustive and mutually exclusive: every value belongs somewhere and no value belongs twice. Interval notation such as $10<x\le20$ fixes where a boundary value goes.

Do not recount separately for each category; a single pass with tallies reduces omissions and double-counting.

Before reading a statistical diagram, identify its title, variable, units, category labels or class intervals, and the scale on each axis or in its key.

Task	Reliable method
read a value	trace from the mark or bar to the labelled scale
compare categories	read both values, then subtract or form the requested ratio
find a total	convert every symbol, bar or sector to frequency before adding
interpret a pie sector	use its fraction of $360^\circ$ , not its visual width alone

If one vertical grid step represents 2 games and a bar is 5 steps high, its frequency is $5\times2=10$ games. The scale must be decoded before the height is used.

A diagram supports statements about the displayed data, not an unexplained causal claim. Also check for truncated axes or unequal scales before comparing visual size.

A histogram displays grouped continuous data. Bars touch, class intervals determine bar widths, and the vertical axis is frequency density so that each bar's area represents frequency.

Quantity	Relationship
class width	upper boundary $-$ lower boundary
frequency density	$\dfrac{\text{frequency}}{\text{class width}}$
frequency	class width $\times$ frequency density

Compute every class width and frequency density, mark the continuous class boundaries on the horizontal axis, then draw touching rectangles. To estimate a frequency over part of a class, use the corresponding fraction of that bar's area.

When class widths differ, bar height is not frequency. Compare or add bar areas; treating a histogram as an ordinary bar chart gives the wrong result.

Cumulative frequency is a running total. Each plotted point shows how many observations are at or below an upper class boundary.

Class frequency	Cumulative frequency	Plot at
3	3	first upper boundary
16	$3+16=19$	second upper boundary
24	$19+24=43$	third upper boundary

Add frequencies successively, pair each running total with its upper class boundary, include the lower starting boundary with cumulative frequency 0 when appropriate, plot the points, and join them with a smooth increasing curve or line segments.

Do not plot at class midpoints and do not use the separate class frequencies as vertical coordinates. A cumulative-frequency graph must never decrease.

A cumulative frequency diagram converts between a value and the number at or below that value. Read up from a value to the curve and across to cumulative frequency, or reverse those steps to find a value for a chosen cumulative count.

Required estimate for total $N$	Cumulative position
lower quartile $Q_1$	$N/4$
median	$N/2$
upper quartile $Q_3$	$3N/4$
interquartile range	$Q_3-Q_1$

If the graph gives cumulative frequency $c$ at value $x$ , then about $c$ observations are $\le x$ and about $N-c$ are $>x$ . For an interval $a<x\le b$ , subtract the readings: $CF(b)-CF(a)$ .

Graph readings are estimates, so use sensible precision. When comparing two groups, state what the medians show about typical value and what the interquartile ranges show about spread.

6.2 Statistical measures

An average is a single value used to describe a typical or central feature of a data set. Different averages answer different questions, so the context and data shape determine which is useful.

Measure	What it represents	Sensitive to extreme values?
mean	equal-share value using every observation	yes
median	middle of ordered data	much less
mode	most frequent value or category	no

For a frequency table, frequencies tell how many times each value occurs; expand conceptually or use cumulative counts to locate the middle.

There is no universally 'best' average. A data set may have no mode or several modes, and a mean can be unrepresentative when a few extreme values pull it away from most observations.

For discrete data, calculate each measure from the actual values and their frequencies, keeping centre and spread distinct.

Measure	Rule
mean	$\dfrac{\sum x}{n}$ , or $\dfrac{\sum fx}{\sum f}$ for a frequency table
median	middle ordered value; average the two middle values when $n$ is even
mode	value with greatest frequency
range	maximum $-$ minimum

In reverse-mean problems, first convert a mean to a total: $\text{total}=\text{mean}\times\text{number}$ . Combine or subtract totals before dividing by the new number of values.

Order the data before finding the median. Do not divide by the number of table rows: for a frequency table the divisor is $\sum f$ .

Grouped data do not reveal the exact observations, so represent every value in a class by its class midpoint and calculate an estimated mean.

Step	Calculation
midpoint	$(\text{lower boundary}+\text{upper boundary})/2$
estimated class total	midpoint $\times$ frequency
estimated mean	$\dfrac{\sum(\text{midpoint}\times f)}{\sum f}$

For $100<h\le110$ with frequency 12, use midpoint 105, contributing $105\times12=1260$ to the estimated total.

The answer is an estimate because actual values need not equal the midpoint. Never multiply frequency by class width when estimating the mean.

The modal class is the class interval containing the greatest frequency. State the complete interval, including its boundary convention.

Class	Frequency
$0<d\le4$	9
$4<d\le8$	15
$8<d\le12$	7

The modal class in the example is $4<d\le8$ because 15 is the largest frequency.

The modal class is an interval, not its midpoint. If information is shown only by a histogram with unequal widths, compare frequency density or bar area according to the task rather than assuming the tallest-looking width gives the class frequency.

For total frequency $N$ , the median lies at cumulative frequency $N/2$ . Read horizontally from $N/2$ to the cumulative-frequency curve, then vertically to the data axis.

Step	Action
1	read the final cumulative total $N$
2	calculate $N/2$
3	move from $N/2$ on the vertical axis to the curve
4	move down to estimate the median value

To compare typical values for two groups, compare their medians in context: the group with the larger median tends to have the larger observed value.

Use the cumulative-frequency axis position, not half the horizontal-axis range. The result is an estimate and should not claim more precision than the graph supports.

A measure of spread describes how variable or dispersed the data are. Larger spread means values are less tightly clustered; smaller spread means greater consistency.

Measure	Uses	Sensitivity
range $=\max-\min$	full data width	strongly affected by extremes
interquartile range $=Q_3-Q_1$	middle 50% width	resistant to extremes

Compare groups with both centre and spread: median describes a typical value, while IQR describes consistency of the middle half. Write comparisons in the units and context of the data.

A higher median does not mean greater spread, and a smaller IQR does not mean a smaller typical value; centre and variation answer different questions.

Order the data, locate the lower quartile $Q_1$ and upper quartile $Q_3$ using the course's median-of-halves convention, then calculate $IQR=Q_3-Q_1$ .

Ordered data count	Split for quartiles
odd	exclude the overall median, then find medians of lower and upper halves
even	split into equal lower and upper halves, then find each half's median

For 11 ordered values, the 6th is the overall median; $Q_1$ is the median of values 1–5 and $Q_3$ is the median of values 7–11.

Quartiles come from ordered positions, not one quarter and three quarters of the numerical range. State and use one consistent convention.

For total frequency $N$ , read the lower quartile at cumulative frequency $N/4$ and the upper quartile at $3N/4$ , then subtract.

Quantity	Cumulative frequency position
$Q_1$	$N/4$
median	$N/2$
$Q_3$	$3N/4$
IQR	$Q_3-Q_1$

A larger estimated IQR means the middle half is more spread out; a smaller estimated IQR means it is more consistent. Support the statement with both estimated IQRs when comparing groups.

Do not subtract cumulative frequencies $3N/4-N/4$ ; those positions locate quartile values on the horizontal axis, and the horizontal readings must be subtracted.

6.3 Probability

An outcome is one possible result of a trial. An event is a specified outcome or set of outcomes. A random trial has an uncertain result, even though its possible outcomes may be known.

Term	Meaning
impossible	cannot occur
unlikely	probability below $\tfrac12$
evens	probability $\tfrac12$
likely	probability above $\tfrac12$
certain	must occur

Equally likely outcomes have the same chance. Do not assume outcomes are equally likely merely because they can all occur.

Probability language compares likelihood, not frequency already observed in a small sample. 'Random' does not mean every result must appear equally often.

Every probability lies from 0 to 1 inclusive. Probability 0 means impossible, $\tfrac12$ means an even chance, and 1 means certain.

Probability	Interpretation
$0$	impossible
between $0$ and $\tfrac12$	unlikely
$\tfrac12$	evens
between $\tfrac12$ and $1$	likely
$1$	certain

Fractions, decimals and percentages can name the same position: $\tfrac34=0.75=75\%$ .

A value below 0 or above 1 cannot be a probability. When marking a scale, use its subdivisions rather than estimating from the page width.

When all elementary outcomes are equally likely, theoretical probability is the number of favourable outcomes divided by the total number of possible outcomes.

Quantity	Rule
probability of event $A$	$P(A)=\dfrac{\text{favourable outcomes}}{\text{all equally likely outcomes}}$
check	$0\le P(A)\le1$

A fair six-sided die has three even faces, so $P(\text{even})=\tfrac36=\tfrac12$ .

Count outcomes, not labels. If sectors, objects or mechanisms are not equally likely, favourable-count over total-count is not justified without weighting them.

A Venn diagram partitions the universal set into disjoint regions. Read the region named by the event, add its frequencies, then divide by the total frequency when one item is chosen at random.

Event	Regions included
$A\cap B$	overlap of $A$ and $B$
$A\cup B$	every region in $A$ or $B$ , overlap once
$A'$	every region outside $A$
neither $A$ nor $B$	outside both circles

For a conditional statement such as 'given $B$ ', restrict the denominator to the total inside $B$ before counting the favourable part.

Do not count an overlap twice when finding a union, and do not omit the region outside all circles from the universal total.

A sample space lists every possible outcome of an experiment. An event is a subset of that sample space.

First coin	Second coin	Ordered outcome
H	H	HH
H	T	HT
T	H	TH
T	T	TT

For fair independent coins, the event 'exactly one head' is $\{HT,TH\}$ , so its probability is $\tfrac24=\tfrac12$ .

Outcomes such as HT and TH are different when order records successive results. A sample space must be exhaustive and must not repeat an outcome.

For two successive choices or trials, use an ordered list, two-way table or tree so every first-stage option is paired with every permitted second-stage option.

Step	Action
1	fix one first-stage outcome
2	pair it with every allowed second-stage outcome
3	repeat for each first-stage outcome
4	check restrictions such as no repetition or order ignored

If there are $m$ first-stage choices and $n$ independent second-stage choices, there are $mn$ ordered outcomes.

Do not use $mn$ when a choice is removed, repeats are forbidden, or AB and BA represent the same unordered pair; adjust the list to the actual rules.

Experimental probability uses observed results: divide the frequency of the event by the total number of trials or observations.

Quantity	Calculation
experimental probability	$\dfrac{\text{event frequency}}{\text{number of trials}}$
estimated event count in $N$ future trials	$N\times\text{experimental probability}$

A larger relevant sample usually gives a more stable estimate, but it does not guarantee the next outcome or make the estimate exact.

Use the total number represented by the data as the denominator. Do not confuse a cumulative frequency, a subgroup total or a graph reading with the whole sample.

An event and its complement cover every outcome without overlap, so their probabilities add to 1.

Required event	Complement method
not $A$	$P(A')=1-P(A)$
at least one success	$1-P(\text{no successes})$
any unlisted category	$1-\text{sum of listed category probabilities}$

If $P(\text{packed lunch})=0.79$ , then $P(\text{no packed lunch})=1-0.79=0.21$ .

Subtract from 1 only when the event being subtracted is exactly the complement of the required event. 'At least one' is complemented by 'none', not by 'exactly one'.

Mutually exclusive events cannot happen on the same trial. For such events, the probability of one or the other is the sum of their probabilities.

Condition	Addition rule
$A$ and $B$ mutually exclusive	$P(A\cup B)=P(A)+P(B)$
$A$ and $B$ may overlap	$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

If a spinner lands on exactly one colour, red and yellow are mutually exclusive, so $P(\text{red or yellow})=P(\text{red})+P(\text{yellow})$ .

The word 'or' does not by itself prove mutual exclusivity. Check whether both events can occur together before adding without an overlap correction.

Expected frequency is the long-run estimate of how often an event occurs: multiply its probability by the number of trials.

Known information	Calculation
probability $p$ , trials $n$	expected frequency $=np$
event frequency $f$ , probability $p$	estimated trials $=f/p$
expected frequency $f$ , trials $n$	probability $=f/n$

If $P(\text{blue})=0.4$ over 280 spins, the expected frequency is $280\times0.4=112$ .

Expected frequency is an estimate, not a guaranteed result. First find any missing probability, and keep trials and event counts in consistent units.

A probability tree shows successive stages. Branches leaving one node list all possible next outcomes and must sum to 1.

Step	Action
1	label every stage and branch outcome
2	complete each sibling pair or group so it sums to 1
3	multiply probabilities along one path
4	add probabilities of the distinct paths that satisfy the event

Decide whether later branch probabilities stay the same or change according to replacement, dependence and earlier outcomes.

Do not add probabilities along a path or multiply alternative paths. Multiplication means successive events on one path; addition combines disjoint completed paths.

Events are independent when knowing one occurred does not change the probability of the other. Then the probability of both is the product of their probabilities.

Relationship	Rule
independent $A$ and $B$	$P(A\cap B)=P(A)P(B)$
repeated independent event	use the same branch probabilities at each repetition
exactly one of two events	add the two orders $A B'$ and $A' B$

For at least one success in independent repetitions, the complement is often shorter: $1-P(\text{all failures})$ .

Repeated-looking trials are not automatically independent. Replacement or a separate mechanism may preserve probabilities; removing an item usually changes them.

Without replacement, the first selection changes both the total number of objects and possibly the favourable number. Later probabilities are conditional on the earlier path.

After one object is removed	Update
denominator	subtract 1 from the total
favourable numerator	subtract 1 only if the removed object was favourable
one order	multiply its successive conditional probabilities
several valid orders	calculate each order and add

From 7 red and 5 blue counters, $P(\text{two red without replacement})=\tfrac7{12}\times\tfrac6{11}$ .

Do not reuse the original fraction on the second draw. For 'one of each', include both red-then-blue and blue-then-red unless the order is fixed.

A multi-step probability problem is solved by defining the required event, choosing a representation, applying the correct rule at each stage, and checking that the answer is plausible.

Feature in the problem	Useful move
all outcomes can be listed	sample space or systematic table
successive stages	probability tree
'not' or 'at least one'	test a complement
disjoint alternatives	add path probabilities
without replacement	update conditional branches
repeated trials estimate	expected frequency $=np$

State assumptions, keep exact fractions until the end when practical, and verify $0\le P\le1$ and that sibling branches sum to 1.

Do not select a rule from a keyword alone. Translate the event first, then check for overlap, independence, replacement and whether the question asks for probability or expected count.

6 Statistics and probability

Exam analysis

Most tested topics

6.1 Graphical representation of data

Choose and use a suitable data display

Turn raw data into a frequency table

Read statistical diagrams accurately

Construct and interpret histograms

Build a cumulative frequency diagram

Use a cumulative frequency diagram

6.2 Statistical measures

Understand what an average represents

Calculate mean, median, mode and range

Estimate the mean from grouped data

Identify the modal class

Estimate the median from cumulative frequency

Interpret measures of spread

Find the interquartile range of discrete data

Estimate interquartile range from cumulative frequency

6.3 Probability

Use precise probability language

Read and use the probability scale

Calculate theoretical probability

Find probabilities from a Venn diagram

Build a sample space and identify events

List successive outcomes systematically

Estimate probability from collected data

Use complementary events

Add mutually exclusive probabilities

Calculate expected frequency

Construct and use probability trees

Calculate with independent events

Calculate conditional probability without replacement

Model and solve mixed probability problems