S Statistics — 1MA1 revision notes

S1

Infer properties of populations or distributions from a sample, while knowing the limitations of sampling

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

A population is the whole group being studied; a sample is the smaller group from which data are collected.
If the sample is representative, its proportion or mean can be used to estimate a population value.
Scale a sample proportion by the population size when estimating a count.
The result is an estimate, not a certainty.
Small samples, convenience sampling, under-coverage and non-response can make the sample unrepresentative and the inference unreliable.

Worked example

In a random sample of $75$ residents, $27$ support a proposal. Estimate how many of the town's $1250$ residents support it and state the assumption needed.

1.Sample proportion $=\dfrac{27}{75}=0.36$ .
2.Estimated population count $=0.36\times1250=450$ .
3.The estimate assumes that the sample is representative of the town's residents.

Answer: About $450$ residents, provided the sample is representative.

Common mistakes

Don't fall into the trap of using the sample frequency as the population estimate without scaling.
Don't fall into the trap of assuming a large but biased sample must be representative.

Exam tip

State a limitation in context, explaining how it could make the sample unrepresentative.

Tier 1 · Easy

ORIGINAL

1

In a random sample of

50

library users,

30

prefer later opening. Estimate how many of the library's

1000

users prefer later opening.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

A random sample of

80

items from a production run of

800

contains

18

items with a surface mark. Estimate the number in the whole run with a surface mark and state one limitation of the estimate.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

A service invites a random sample of

450

customers to answer a survey. Only

270

reply, and

189

of the replies support a change. Use the replies to estimate the number of supporters among all

12000

customers, then explain a serious limitation.

(4)

(Total for Question 1 is 4 marks)

S2

Interpret and construct frequency tables, bar charts, pie charts and pictograms (categorical data), vertical line charts (ungrouped discrete numerical data), tables and line graphs (time series)

Notes

Worked answers & exam appearances →

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

Match the display to the data. Use separated bars for categories, a vertical line chart for ungrouped discrete numbers, and joined points in time order for a time series.
A pictogram needs a clear key.
For a pie chart, $\text{sector angle}=\dfrac{\text{frequency}}{\text{total}}\times360^\circ$ .
Any graph needs clear labels, an even scale and accurate plotting.
Bar widths and gaps should be consistent, and unrelated categories should not be joined by lines.

Worked example

A $126^\circ$ sector of a pie chart represents $42$ students. Find the total number of students.

1.The sector represents the fraction $\dfrac{126}{360}$ of the total.
2.Total $=42\div\dfrac{126}{360}=42\times\dfrac{360}{126}=120$ .

Answer: $120$ students.

Common mistakes

Don't fall into the trap of using frequency directly as an angle instead of scaling to $360^\circ$ .
Don't fall into the trap of joining categorical bars or plotting a time series out of order.

Exam tip

After calculating pie-chart sectors, check that all sector angles add to $360^\circ$ .

Tier 1 · Easy

ORIGINAL

1

The data are

2,3,2,5,4,3,2,4,5,2

. Construct a frequency table for the values

2

,

3

,

4

and

5

.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

In a survey of

240

journeys,

54

are made by bicycle. Work out the angle of the bicycle sector in a pie chart.

(2)

(Total for Question 1 is 2 marks)

Tier 3 · Hard

ORIGINAL

1

Quarterly sales are Q₁:

320

, Q₂:

410

, Q₃:

390

, Q₄:

520

, and the following Q₁:

350

. State the five points to plot on a time-series line graph. Work out the percentage change from the first Q₁ to the following Q₁ and explain why comparing Q₄ directly with the following Q₁ could mislead.

(4)

(Total for Question 1 is 4 marks)

S3

Construct and interpret diagrams for grouped discrete and continuous data, i.e. histograms with equal and unequal class intervals and cumulative frequency graphs [Higher only]

Notes

Worked answers & exam appearances →

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

Higher tier only. In a histogram, bar area represents frequency and bar height is frequency density: $\text{frequency density}=\dfrac{\text{frequency}}{\text{class width}}$ .
Therefore $\text{frequency}=\text{frequency density}\times\text{class width}$.
Use the class boundaries for each width, draw touching bars, and label both axes.
For a cumulative frequency graph, plot the lower boundary with cumulative frequency $0$ , then each upper class boundary against its running total.
Estimate the median at $n/2$ , the lower quartile at $n/4$ and the upper quartile at $3n/4$ .

Worked example

The classes $10<x\leq25$ and $25<x\leq35$ are adjacent. The first has frequency $30$ ; the second has histogram height $1.4$ . Find the first bar height, the second frequency and the cumulative-frequency points.

1.First width $=15$ , so its height is $30\div15=2$ .
2.Second width $=10$ , so its frequency is $1.4\times10=14$ .
3.The cumulative-frequency points are $(10,0)$ , $(25,30)$ and $(35,44)$ .

Answer: First height $2$ ; second frequency $14$ ; cumulative points $(10,0)$ , $(25,30)$ and $(35,44)$ .

Common mistakes

Don't fall into the trap of using frequency as the bar height instead of dividing by class width.
Don't fall into the trap of plotting cumulative frequencies at class midpoints instead of upper boundaries.

Exam tip

Write the frequency-density formula beside the histogram before calculating any height or area.

Tier 1 · Easy

ORIGINAL

1

A histogram class is

15<x\leq20

and has frequency

20

. Work out the frequency density.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

Grouped data have classes

0<x\leq10

,

10<x\leq25

and

25<x\leq40

with frequencies

12

,

30

and

18

. Calculate the three frequency densities and identify the tallest histogram bar.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

A cumulative frequency graph for

80

values passes through

(10,8)

,

(20,26)

,

(30,50)

,

(40,70)

and

(50,80)

. Use linear interpolation between the given points to estimate the median and the interquartile range.

(5)

(Total for Question 1 is 5 marks)

S4

Interpret, analyse and compare data-set distributions via graphs (incl. box plots), central tendency (median, mean, mode, modal class) and spread (range, outliers, quartiles, inter-quartile range)

Notes

Worked answers & exam appearances →

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

A measure of central tendency describes a typical value: the mean uses every value, the median is the ordered middle, and the mode or modal class is most frequent.
Spread describes variation: range is maximum minus minimum.
At Higher tier, box plots also show quartiles and the interquartile range $Q_3-Q_1$ .
To compare distributions, make one contextual statement about centre and one about spread.
Outliers can strongly affect the mean and range, so the median and interquartile range may be more representative.

Worked example

Delivery service A has median time $38$ minutes and range $22$ minutes. Service B has median time $43$ minutes and range $12$ minutes. Compare the distributions.

1.A has the lower median, so its deliveries are typically quicker.
2.B has the smaller range, so its delivery times are more consistent.

Answer: Service A is typically quicker, but service B has more consistent delivery times.

Common mistakes

Don't fall into the trap of giving comparison figures without stating what they mean in context.
Don't fall into the trap of finding the median before putting raw data in order.

Exam tip

A full comparison usually needs both a typical-value statement and a spread statement.

Tier 1 · Easy

ORIGINAL

1

For the data

4,6,6,9,10

, work out the mean and the range.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

Data set A has median

24

and range

33

. Data set B has median

25

and range

29

. Compare the two distributions.

(2)

(Total for Question 1 is 2 marks)

Tier 3 · Hard

ORIGINAL

1

The data are

12,13,13,14,15,15,16,52

. Work out the mean, median and range. Decide which of the mean or median better describes a typical value for these data. Give a reason.

(4)

(Total for Question 1 is 4 marks)

S5

Apply statistics to describe a population

Notes

Worked answers & exam appearances →

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

Statistics can summarise a population or estimate its features from representative samples.
When groups have different sizes, combine their means using a weighted mean: add each group size multiplied by its mean, then divide by the total population.
When different subgroups have different sample proportions, estimate each subgroup separately before adding.
A simple average of group means is only valid when the groups are the same size.

Worked example

A town has $900$ northern and $600$ southern residents. In representative samples, $12$ of $40$ northern residents and $20$ of $50$ southern residents cycle to work. Estimate the town total and percentage.

1.North estimate $=900\times\dfrac{12}{40}=270$ .
2.South estimate $=600\times\dfrac{20}{50}=240$ .
3.Total $=270+240=510$ ; percentage $=\dfrac{510}{1500}\times100=34\%$ .

Answer: About $510$ residents, or $34\%$ of the town.

Common mistakes

Don't fall into the trap of averaging two group means without using the group sizes.
Don't fall into the trap of applying one subgroup's sample proportion to the whole population.

Exam tip

Turn each mean back into a total first; combine totals, then divide once.

Tier 1 · Easy

ORIGINAL

1

A representative sample of parcels has mean mass

2.4\,\text{kg}

. Estimate the total mass of

250

parcels in the population.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

A population contains

120

junior members with mean attendance

6.5

sessions and

80

senior members with mean attendance

8

sessions. Work out the mean attendance for the whole population.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

A town has

1800

residents in the north and

1200

in the south. In representative samples,

15

of

60

northern residents and

20

of

50

southern residents cycle to work. Estimate the total number and percentage of the town's residents who cycle to work.

(5)

(Total for Question 1 is 5 marks)

S6

Use and interpret scatter graphs of bivariate data; recognise correlation, know it does not indicate causation; draw estimated lines of best fit; make predictions; interpolate/extrapolate with caution

Notes

Worked answers & exam appearances →

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

A scatter graph displays paired values.
An upward pattern shows positive correlation, a downward pattern negative correlation, and no clear pattern no correlation.
Draw a line of best fit through the centre of the points with a roughly balanced spread above and below.
Use it for estimates: interpolation stays within the observed data range, while extrapolation goes beyond it and is less reliable.
Correlation does not prove causation because another variable, reverse causation or coincidence may explain the relationship.

An illustrative scatter graph with an upward trend and a straight line of best fit passing through the centre of the points.

Worked example

A scatter graph compares weekly revision time with test score for revision times from $1$ to $8$ hours. A sensible line of best fit gives a score of about $72$ at $6$ hours and about $84$ when extended to $10$ hours. Interpret both estimates.

1. $6$ hours is inside the observed range, so $72$ is an interpolation and is reasonably reliable.
2. $10$ hours is outside the observed range, so $84$ is an extrapolation and is less reliable because the trend may not continue.
3.The positive correlation does not prove that extra revision alone caused the higher scores; prior attainment could affect both variables.

Answer: $72$ is the more reliable interpolation; $84$ is a less reliable extrapolation, and the graph does not establish causation.

Common mistakes

Don't fall into the trap of saying that correlation proves one variable causes the other.
Don't fall into the trap of using a line of best fit far outside the observed range without warning.

Exam tip

Answer the command precisely: for 'describe the relationship', write 'as $x$ increases, $y$ tends to increase/decrease', not just 'positive/negative'. For a prediction, identify interpolation or extrapolation and comment on reliability.

Tier 1 · Easy

ORIGINAL

1

A scatter graph shows that as daily temperature increases, ice-cream sales usually increase. State the type of correlation and explain why the graph alone does not prove that temperature is the only cause of higher sales.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

For data with observed

x

-values from

5

to

20

, a line of best fit is

y=1.8x+6

. Estimate

y

when

x=14

. Explain why using the line at

x=35

is less reliable.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

A scatter graph compares a puppy's age

a

months with mass

m

kg for ages from

2

to

12

months. Its estimated line of best fit is

m=0.42a+1.8

. Estimate the mass at

9

months and at

16

months. Comment on the reliability of both estimates and on whether the graph proves that age alone causes the change in mass.

(5)

(Total for Question 1 is 5 marks)

Infer properties of populations or distributions from a sample, while knowing the limitations of sampling

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Interpret and construct frequency tables, bar charts, pie charts and pictograms (categorical data), vertical line charts (ungrouped discrete numerical data), tables and line graphs (time series)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Construct and interpret diagrams for grouped discrete and continuous data, i.e. histograms with equal and unequal class intervals and cumulative frequency graphs [Higher only]

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Interpret, analyse and compare data-set distributions via graphs (incl. box plots), central tendency (median, mean, mode, modal class) and spread (range, outliers, quartiles, inter-quartile range)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Apply statistics to describe a population

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Use and interpret scatter graphs of bivariate data; recognise correlation, know it does not indicate causation; draw estimated lines of best fit; make predictions; interpolate/extrapolate with caution

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

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