S Statistics — printable coverage pack

S1 · Infer properties of populations or distributions from a sample, while knowing the limitations of sampling

A population is the whole group of interest and a sample is the smaller group from which data are collected.
Use a representative sample proportion or mean to estimate the corresponding population value, scaling by the population size when needed.
For example, if $18$ of a random sample of $60$ have a feature, estimate that $18/60=30\%$ of the population has it.
An inference is an estimate, not a certainty; small samples, non-response, convenience sampling and under-coverage can all make it unreliable.

Tier 1 · Easy

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 marks]
Answer
- $600$ users
Method: The sample proportion is $30/50=0.6$ . Apply this to the population: $0.6\times1000=600$ .

Tier 2 · Standard

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 marks]
Answer
- $180$ items.
- The sample may differ from the population by chance, so the estimate need not equal the true number.
Method: The sample is one tenth of the production run, so scale $18$ by $10$ to get $180$ . Because only a sample was inspected, sampling variation remains even if the selection was random.

Tier 3 · Hard

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 marks]
Answer
- Estimated supporters $=8400$ .
- Non-response bias may make the replies unrepresentative because the $180$ non-responders may have different views.
Method: Among replies, the support proportion is $189/270=0.7$ , giving $0.7\times12000=8400$ . However, the estimate assumes responders and non-responders have similar opinions; the low response rate may break that assumption.

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)

Choose a display to match the data: separated bars for categories, a vertical line chart for discrete numerical values, and joined time-ordered points for a time series.
For a pie chart, calculate sector angle using $\text{angle}=\dfrac{\text{category frequency}}{\text{total frequency}}\times360^\circ$ .
A correct graph uses a labelled, evenly spaced scale and plots each frequency or value accurately; bar widths and gaps should be consistent.
Do not join unrelated categories with lines, and do not interpret a truncated or uneven axis as though visual lengths were proportional.

Tier 1 · Easy

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 marks]
Answer
- $2:4$ , $3:2$ , $4:2$ , $5:2$ .
Method: Tally each value once. The frequencies $4+2+2+2=10$ match the number of data values.

Tier 2 · Standard

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 marks]
Answer
- $81^\circ$
Method: The bicycle fraction is $54/240$ . Multiply by a full turn: $\dfrac{54}{240}\times360^\circ=81^\circ$ .

Tier 3 · Hard

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 marks]
Answer
- $(\text{Q1},320),(\text{Q2},410),(\text{Q3},390),(\text{Q4},520),(\text{next Q1},350)$ .
- Percentage increase $=9.375\%$ .
- Q₄ and Q₁ may have different seasonal patterns, so that comparison may not represent an underlying fall.
Method: Plot the values in time order and join consecutive points. Like-for-like Q₁ change is $350-320=30$ , so the percentage change is $30/320\times100=9.375\%$ . Comparing different quarters confounds the change with possible seasonality.

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]

In a histogram, bar area represents frequency and frequency density is $\dfrac{\text{frequency}}{\text{class width}}$ .
For unequal class intervals, calculate every class width from its boundaries before finding bar heights; a greater height means greater density, not necessarily greater frequency.
A cumulative frequency graph plots upper class boundaries against running totals and can estimate the median at $n/2$ and quartiles at $n/4$ and $3n/4$ .
Do not use frequency as histogram height unless all class widths are equal, and do not read quartiles from the ordinary frequency axis.

Tier 1 · Easy

1. A histogram class is $15<x\leq20$ and has frequency $20$ . Work out the frequency density.[2 marks]
Answer
- $4$
Method: The class width is $20-15=5$ . Frequency density is $20/5=4$ .

Tier 2 · Standard

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 marks]
Answer
- Frequency densities: $1.2$ , $2$ , $1.2$ .
- The class $10<x\leq25$ has the tallest bar.
Method: Divide each frequency by its class width: $12/10=1.2$ , $30/15=2$ and $18/15=1.2$ . The greatest density, $2$ , gives the tallest bar.

Tier 3 · Hard

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 marks]
Answer
- Median $\approx25.8$ .
- Lower quartile $\approx16.7$ and upper quartile $=35$ .
- Interquartile range $\approx18.3$ .
Method: The median is the $40$ th value, between cumulative frequencies $26$ and $50$ : $20+\dfrac{40-26}{50-26}\times10\approx25.8$ . The lower quartile is the $20$ th value: $10+\dfrac{20-8}{26-8}\times10\approx16.7$ . The upper quartile is the $60$ th value: $30+\dfrac{60-50}{70-50}\times10=35$ . Hence $IQR\approx35-16.7=18.3$ .

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)

Central tendency describes a typical value: use the mean for all values, the median for the ordered middle, and the mode or modal class for the most frequent outcome.
Spread describes variation: range is maximum minus minimum. At Higher tier, box plots also show quartiles and the interquartile range $Q_3-Q_1$ .
Compare distributions in context using one statement about centre and one about spread, such as a higher median together with a smaller range.
Do not conclude that one set is 'better' from spread alone, and remember that outliers can strongly affect the mean and range.

Tier 1 · Easy

1. For the data $4,6,6,9,10$ , work out the mean and the range.[2 marks]
Answer
- Mean $=7$ ; range $=6$ .
Method: The total is $4+6+6+9+10=35$ , so the mean is $35/5=7$ . The range is $10-4=6$ .

Tier 2 · Standard

1. Data set A has median $24$ and range $33$ . Data set B has median $25$ and range $29$ . Compare the two distributions.[2 marks]
Answer
- B has the slightly higher median: $25$ compared with $24$ .
- B has the smaller range: $29$ compared with $33$ , so its values are less spread out overall.
Method: Compare the typical values using the medians, then compare spread using the ranges. B has both the larger centre and the smaller spread.

Tier 3 · Hard

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 marks]
Answer
- Mean $=18.75$ , median $=14.5$ , range $=40$ .
- The median is more representative because $52$ is an outlier that pulls up the mean.
Method: The total is $150$ , giving mean $150/8=18.75$ . The median is $(14+15)/2=14.5$ and the range is $52-12=40$ . The isolated value $52$ has a strong effect on the mean but not on the median, so the median better represents the main cluster.

S5 · Apply statistics to describe a population

Population statistics summarise a whole group, while sample statistics can be scaled or weighted to estimate population features.
For groups of different sizes, use a weighted mean: add each group size multiplied by its mean, then divide by the total population.
To estimate a population count, apply the appropriate sample proportion to each population subgroup before adding the estimates.
Do not average group means without accounting for group sizes, and do not report an estimate with unjustified precision.

Tier 1 · Easy

1. A representative sample of parcels has mean mass $2.4\,\text{kg}$ . Estimate the total mass of $250$ parcels in the population.[2 marks]
Answer
- $600\,\text{kg}$
Method: Apply the sample mean to all $250$ parcels: $2.4\times250=600\,\text{kg}$ .

Tier 2 · Standard

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 marks]
Answer
- $7.1$ sessions
Method: The total attendances represented are $120\times6.5=780$ and $80\times8=640$ . Divide their sum by all $200$ members: $(780+640)/200=7.1$ .

Tier 3 · Hard

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 marks]
Answer
- Estimated total $=930$ residents.
- Estimated percentage $=31\%$ .
Method: For the north, estimate $1800\times15/60=450$ . For the south, estimate $1200\times20/50=480$ . The total is $450+480=930$ from a population of $3000$ , so the percentage is $930/3000\times100=31\%$ . Calculating by subgroup correctly accounts for their different sizes.

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

A scatter graph shows paired values: upward patterns indicate positive correlation, downward patterns 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, ignoring an outlier only when justified.
Use the line to estimate one variable from the other; interpolation stays within the observed range, while extrapolation extends beyond it and is less reliable.
Correlation does not prove causation because a third variable, reverse causation or coincidence may explain the association.

Tier 1 · Easy

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 marks]
Answer
- Positive correlation.
- Correlation does not establish causation; another factor such as weekends or visitor numbers could affect sales.
Method: An upward association is positive correlation. Then identify a plausible confounding variable to show why the paired data alone cannot isolate a causal effect.

Tier 2 · Standard

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 marks]
Answer
- $y=31.2$ when $x=14$ .
- $x=35$ is outside the observed range, so this would be extrapolation and the trend may not continue.
Method: Substitute $x=14$ : $y=1.8(14)+6=25.2+6=31.2$ . Since $14$ lies inside the data range this is interpolation, whereas $35$ lies outside it.

Tier 3 · Hard

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 marks]
Answer
- At $9$ months, estimated mass $=5.58\,\text{kg}$ .
- At $16$ months, estimated mass $=8.52\,\text{kg}$ .
- The $9$ -month estimate is interpolation and is more reliable; the $16$ -month estimate is extrapolation.
- The association does not prove age alone causes the change because factors such as breed or diet may also affect mass.
Method: Substitute into the line: for $a=9$ , $m=0.42(9)+1.8=5.58$ ; for $a=16$ , $m=0.42(16)+1.8=8.52$ . The first age lies within $2$ to $12$ , while the second lies outside. The graph shows association only and does not control other variables.