01 · Heights
For the three classes in the model, calculate the frequency densities.
Hint
Divide each frequency by its class width.
Worked solution
8/10=0.8, 12/10=1.2 and 20/20=1.0 observations per minute.
Understand · explore · practise
Statistics review: displays and regression
Combine histogram area, box plots, percentiles, outlier checks and regression interpretation in original cumulative practice.
Before you startRepresentations of data and correlation.
01 / Read what the diagram actually represents
Use area, position or trend as appropriate.
Histogram frequency is area; box-plot comparisons need context.
A regression line models one response from one explanatory variable. None of these displays alone proves causation. The numerical data in this review are constructed.
Area beyond a thresholdExplore
Constructed grouped times: 0–10 minutes has frequency 8; 10–20 has 12; 20–40 has 20. Assume uniform spread within each class.
02 / Use density when widths differ
The tallest bar need not contain the most observations.
02 · Largest group
Which class contains the most observations, and which has the tallest bar?
Hint
Compare area and height separately.
Worked solution
The 20–40 class contains 20 observations, the largest frequency. The 10–20 class is tallest, with density 1.2.
03 · Partial class
Estimate the number with time above 15 minutes.
Hint
Take the remaining half of the middle class plus the last class.
Worked solution
0.5×12+20=26. Estimated probability is 26/40=0.65, using uniform spread within the middle class.
04 · Another threshold
Estimate the proportion above 30 minutes.
Hint
Only half the final class remains.
Worked solution
10/40=0.25. This is an estimate from grouped data, not an exact count from raw times.
Watch: a threshold cuts through a histogram bar
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / Connect area to cumulative positions
Read the class before interpolating.
Use cumulative frequencies 8, 20 and 40 at upper boundaries 10, 20 and 40.Worked example
Median position = 40/2=20
This is the boundary 20, so the grouped estimate is 20 minutes.
Lower quartile position = 10
Q₁≈10+((10−8)/12)×10=11.6667 minutes.
Upper quartile position = 30
Q₃≈20+((30−20)/20)×20=30 minutes. IQR≈18.3333 minutes.
05 · Cumulative endpoints
Why include a point at the initial lower boundary with cumulative frequency zero?
Hint
No observations have accumulated yet.
Worked solution
It anchors the graph at the start of the covered range. Here use (0,0), followed by (10,8), (20,20) and (40,40).
06 · Interpretation
Does Q₃=30 mean exactly 30 observations equal 30 minutes?
Hint
A percentile is a position, not a repeated value count.
Worked solution
No. It estimates the value below which roughly 75% of observations lie.
04 / Compare centre and spread with context
Use both a location measure and a variability measure.
Two constructed completion-time samples have (Q₁, median, Q₃) of (18,24,31) for group A and (20,27,30) for group B, all in minutes.
07 · Centre comparison
Which group typically completes sooner?
Hint
Compare medians and keep the time context.
Worked solution
Group A has the lower median, 24 rather than 27 minutes, suggesting typically quicker completion.
08 · Spread comparison
Which group has more consistent middle-half times?
Hint
Compare IQRs.
Worked solution
A has IQR 31−18=13 minutes; B has 30−20=10 minutes. Group B has less spread in its middle half.
05 / Use the stated rule and then investigate
An outlier is not automatically an error.
09 · IQR fence
Using group A and the 1.5×IQR rule, find the upper fence and classify a 52-minute result.
Hint
Upper fence = Q₃+1.5IQR.
Worked solution
31+1.5×13=50.5. Since 52>50.5, it is flagged as an upper outlier under this rule.
10 · Retain or remove
Should that 52-minute observation automatically be deleted?
Hint
It may be a valid unusual outcome.
Worked solution
No. Check the record and context. Correct a verified error with documentation; retain a valid observation unless the analysis has a justified exclusion rule.
06 / Interpret units and stay near supported data
A slope is a predicted change per unit of the explanatory variable.
A constructed regression of weekly orders y, measured in hundreds, on advertising spend x, measured in tens of pounds, is y=1.5+0.8x. The observed x values ranged from 2 to 12.
11 · Slope units
Interpret the slope 0.8 in ordinary units.
Hint
One x-unit is £10; one y-unit is 100 orders.
Worked solution
An additional £10 of advertising spend is associated with a predicted increase of 80 weekly orders within this model.
12 · Supported prediction
Predict weekly orders at spend £60.
Hint
x=6, then convert hundreds to orders.
Worked solution
y=1.5+0.8×6=6.3 hundreds, or 630 orders. This is interpolation within the observed spend range.
13 · Extrapolation
Why is predicting at £300 less secure?
Hint
Compare x=30 with the observed range.
Worked solution
It is far beyond x=2 to 12. The fitted linear relationship may not continue there.
14 · Reverse prediction
Can the y-on-x equation automatically serve as the least-squares x-on-y regression?
Hint
The residual directions differ.
Worked solution
No. Solving the equation algebraically does not generally produce the regression fitted in the reverse direction.
07 / Interpret patterns without overclaiming
A useful model still has limitations.
15 · Causal claim
Does a positive fitted slope prove advertising caused extra orders?
Hint
Could another variable affect both?
Worked solution
No. Association alone does not establish causation; seasonality or other changes might influence both variables.
16 · Intercept
Must 1.5 mean that zero advertising causes exactly 150 orders?
Hint
Zero is outside the observed range, and a fitted line is a model.
Worked solution
No. It is the model intercept, but x=0 is outside the observed range and the equation is not a deterministic or causal law.
08 / Check the scale before the calculation
State what is exact, estimated or modelled.
Histogram areas determine frequencies; grouped percentiles assume within-class spread. Compare box plots in context and investigate outliers. Interpret regression units carefully, distinguish interpolation from extrapolation and keep association separate from causation.
Section 1 of 8 · Read what the diagram actually represents