S2 Data presentation and interpretation — printable coverage pack

S2.1 · Interpret diagrams for single-variable data, including understanding that area in a histogram represents frequency; connect to probability distributions.

In a histogram, frequency is proportional to bar area and frequency density is $\frac{\text{frequency}}{\text{class width}}$ ; unequal class widths make bar height alone misleading.
Frequency polygons show class patterns, box plots summarise centre and spread, and cumulative frequency diagrams support estimates of medians, quartiles and counts below a value.
For a continuous probability density histogram, total area is $1$ and the area above an interval is the probability of an observation in that interval.
Read class boundaries and axis scales before calculating. A common error is to use frequency density as though it were frequency.

Tier 1 · Easy

1. A histogram class is $12\leq t<17$ and has frequency density $3.6$ . Find the frequency in this class.[2 marks]
Answer
- $18$
Method: The class width is $17-12=5$ . Hence frequency $=$ class width $\times$ frequency density $=5\times3.6=18$ .

Tier 2 · Standard

1. A cumulative frequency diagram represents $80$ observations. The cumulative frequencies at $x=10$ and $x=20$ are $18$ and $54$ respectively. Estimate the number of observations satisfying $10<x\leq20$ .[3 marks]
Answer
- $36$
Method: The cumulative frequency at $20$ counts all $54$ observations up to that value, including the $18$ already counted up to $10$ . Subtract to isolate the interval: $54-18=36$ .

Tier 3 · Hard

1. The probability density histogram for a continuous random variable $X$ has constant heights $0.12$ on $0\leq x<3$ , $0.08$ on $3\leq x<8$ , and $0.12$ on $8\leq x\leq10$ . Verify that it defines a probability distribution and find $P(1<X<6)$ .[5 marks]
Answer
- Total area $=1$ .
- $P(1<X<6)=0.48$ .
Method: The total area is $3(0.12)+5(0.08)+2(0.12)=0.36+0.40+0.24=1$ , so the histogram defines a probability distribution. The area from $1$ to $3$ is $2(0.12)=0.24$ , and the area from $3$ to $6$ is $3(0.08)=0.24$ . Therefore $P(1<X<6)=0.24+0.24=0.48$ .

S2.2 · Interpret scatter diagrams and regression lines for bivariate data, including recognising distinct sections of the population (regression calculations excluded); interpret correlation informally; correlation does not imply causation.

A scatter diagram shows paired bivariate data; describe the direction and strength of its association and note any outliers or clusters.
A regression line estimates the mean response for a given explanatory-variable value and is most defensible within the observed data range.
Distinct clusters may represent different sections of the population, so one overall correlation or regression line can hide different within-group patterns.
Correlation measures association, not causation; a lurking variable, reverse causation or coincidence may explain the observed relationship.

Tier 1 · Easy

1. A scatter diagram of daily ice-cream sales against temperature shows a strong positive correlation. Interpret this and explain why it does not prove that higher temperature causes every increase in sales.[2 marks]
Answer
- Higher temperatures tend to be associated with higher ice-cream sales.
- Correlation alone does not establish causation; other variables such as sunshine, holidays or visitor numbers may affect both quantities.
Method: Translate positive correlation as a tendency for larger values of one variable to occur with larger values of the other. Then distinguish association from a causal conclusion by identifying a plausible lurking variable.

Tier 2 · Standard

1. A scatter diagram of journey distance against journey time contains one cluster for bicycles and a separate cluster for cars. Explain why fitting one regression line to all journeys may be misleading.[3 marks]
Answer
- The two clusters represent distinct sections of the population with different travel speeds.
- A single line may mainly reflect the gap between the clusters rather than either within-group relationship.
- Separate regression lines or separate analyses would be more informative.
Method: Identify vehicle type as a grouping variable. Since the relationship between distance and time is likely to differ between bicycles and cars, pooling the groups can produce a line that fits neither group well. Analyse the sections separately.

Tier 3 · Hard

1. For trees aged between $3$ and $18$ years, a regression line of trunk diameter $d$ cm on age $a$ years is $d=1.8a+4.2$ . A $12$ -year-old tree has diameter $29$ cm. Interpret the difference between the observed and predicted values and critique using the line to predict the diameter of a $40$ -year-old tree.[4 marks]
Answer
- The predicted diameter at $a=12$ is $25.8$ cm, so the difference between the observed and predicted values is $29-25.8=3.2$ cm.
- The tree is $3.2$ cm thicker than the value predicted by the line.
- A prediction at $40$ years is extrapolation far outside the observed age range and the linear relationship may not continue.
Method: Substitute $a=12$ to get the fitted value $d=1.8(12)+4.2=25.8$ . Residual $=$ observed $-$ fitted $=29-25.8=3.2$ cm. Since $40$ is well beyond the data range $3$ to $18$ , using the line there is unsupported extrapolation.

S2.3 · Interpret measures of central tendency and variation, extending to standard deviation; be able to calculate standard deviation, including from summary statistics.

The mean uses every value, while the median is resistant to extremes; choose a measure that suits the distribution and context.
Range and interquartile range measure spread using endpoints or quartiles; standard deviation measures typical spread about the mean using all observations.
For $n$ data values, use $\sigma=\sqrt{\frac{\sum x^2}{n}-\left(\frac{\sum x}{n}\right)^2}$ unless a different convention is stated.
When groups are combined, add $n$ , $\sum x$ and $\sum x^2$ before recalculating; averaging separate standard deviations is not valid.

Tier 1 · Easy

1. Calculate the mean and population standard deviation of $4,7,7,8,9$ . Give the standard deviation to $3$ significant figures.[3 marks]
Answer
- Mean $=7$ .
- Standard deviation $=1.67$ to $3$ significant figures.
Method: $\sum x=35$ and $\sum x^2=259$ , so $\bar{x}=35/5=7$ . Then $\sigma=\sqrt{259/5-7^2}=\sqrt{2.8}=1.673\ldots$ , giving $1.67$ .

Tier 2 · Standard

1. For $20$ observations, $\sum x=310$ and $\sum x^2=5020$ . Calculate the population standard deviation to $3$ significant figures.[3 marks]
Answer
- $3.28$
Method: The mean is $310/20=15.5$ . Hence $\sigma=\sqrt{5020/20-15.5^2}=\sqrt{251-240.25}=\sqrt{10.75}=3.278\ldots$ , so the standard deviation is $3.28$ .

Tier 3 · Hard

1. Group A has $12$ values with mean $18$ and $\sum x^2=3996$ . Group B has $8$ values with mean $24$ and $\sum x^2=4736$ . Find the mean and population standard deviation of all $20$ values.[5 marks]
Answer
- Combined mean $=20.4$ .
- Combined standard deviation $=4.52$ to $3$ significant figures.
Method: The combined sum is $12(18)+8(24)=408$ , so the combined mean is $408/20=20.4$ . The combined sum of squares is $3996+4736=8732$ . Therefore $\sigma=\sqrt{8732/20-20.4^2}=\sqrt{436.6-416.16}=\sqrt{20.44}=4.521\ldots$ , giving $4.52$ .

S2.4 · Recognise and interpret possible outliers in data sets and statistical diagrams; select or critique data presentation techniques in context; clean data, including dealing with missing data, errors and outliers.

A common outlier rule flags values below $Q_1-1.5\operatorname{IQR}$ or above $Q_3+1.5\operatorname{IQR}$ , but context should guide the final decision.
Investigate a suspicious value against the original record before correcting or removing it; an unusual valid observation is not automatically an error.
Handle missing data transparently: record how many values are missing, avoid inventing unsupported values and consider whether missingness could bias conclusions.
Choose displays to suit the data and purpose: histograms for continuous grouped data, box plots for comparing distributions, and scatter diagrams for paired variables.

Tier 1 · Easy

1. For a data set, $Q_1=14$ and $Q_3=22$ . Use the $1.5\operatorname{IQR}$ rule to determine whether the value $36$ is a possible outlier.[2 marks]
Answer
- The upper boundary is $34$ , so $36$ is a possible outlier.
Method: The interquartile range is $22-14=8$ . The upper outlier boundary is $22+1.5(8)=34$ . Since $36>34$ , it is flagged as a possible outlier.

Tier 2 · Standard

1. A table of package masses contains one blank entry and one value $482$ among values near $48.2$ grams. Describe a defensible way to clean these two entries before analysis.[4 marks]
Answer
- Check the original measurement record for both entries.
- Correct $482$ to $48.2$ only if the source confirms a decimal-point error; otherwise retain and flag it or exclude it with a stated reason.
- Treat the blank as missing rather than replacing it without evidence, and report the reduced sample size or justified imputation method.
Method: First distinguish a data-entry error from a genuine extreme value by consulting the source. Do not silently divide $482$ by $10$ . The blank contains no observed value, so omit it from calculations unless a justified imputation rule has been chosen, and document the decision so its possible bias is visible.

Tier 3 · Hard

1. A data set of $40$ readings was summarised as $\sum x=504$ and $\sum x^2=6856.9$ . One reading was entered as $31$ but the source record confirms it should be $13$ . Calculate the corrected mean and population standard deviation, and state the likely effect of the error on the original spread.[5 marks]
Answer
- Corrected mean $=12.15$ .
- Corrected standard deviation $=2.00$ .
- The error inflated the original spread.
Method: Replace the contribution of $31$ by $13$ : the corrected sum is $504-31+13=486$ , and the corrected sum of squares is $6856.9-31^2+13^2=6064.9$ . Thus $\bar{x}=486/40=12.15$ and $\sigma=\sqrt{6064.9/40-12.15^2}=\sqrt{4}=2.00$ . The erroneous value lay much farther from the centre, so it made the original spread larger.