Cambridge International AS & A Level Mathematics 9709 (2026-2027) · Paper 5: Probability & Statistics 1 · official syllabus
REPRESENTATION OF DATA LAB
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Four short missions on histogram density, a grouped quartile, coded means and standard deviation. Predict the rule, calculate, then test your answer. Keep exact values and grouped estimates distinct.
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Mission 1 of 4
Make the bar area count
A histogram has classes 0<=x<10, 10<=x<25 and 25<=x<30, with frequencies 20, 30 and 10. The first and last bars have density 2. Set the missing density of the middle bar. Horizontal distance represents x; bar area represents frequency.
With unequal class widths, a histogram bar's frequency is represented by its...
The bars share common x and density scales. Teal is your middle bar; purple bars have their given densities. Area, not height, counts observations.
Your middle bar has density 1.0 and area 15.0; its required frequency is 30.
Study notes
Played the lab above? These notes are the reference version of what you just felt - use them to check exact formulas and the exam wording.
Choose a representation for the question 5.1
A stem-and-leaf diagram preserves individual values and shows the distribution's shape for a small data set. Put leaves in order and give a key with units, such as 3 | 7 means 37 minutes. Repeated values need repeated leaves. Back-to-back diagrams compare two groups using a shared stem and clearly labelled sides.
A box-and-whisker plot shows the median, quartiles and extremes, so it is useful for comparing location and spread. It does not preserve every observation or reveal all features of a distribution. Compare medians for typical level and interquartile ranges for the spread of the middle half; do not claim that equal medians imply identical distributions.
A histogram represents grouped continuous data using touching bars. A cumulative frequency graph answers questions about values below a boundary. A display must have a title or context, labels, units and a scale that fits the data. Choose the display for the information needed, rather than choosing it only because it is easy to draw.
Histogram area, not height, represents frequency 5.1
Frequency density = frequency / class width. The area of each histogram bar represents its class frequency. With unequal class widths, comparing heights alone compares density, not the number of observations.
Worked example: the classes 0<=x<10, 10<=x<25 and 25<=x<30 have frequencies 20, 30 and 10. Their widths are 10, 15 and 5; all three densities are 2. Their equal heights do not mean equal frequencies. Total frequency is 60.
If rounded continuous measurements are grouped as 10-19, 20-29 and so on, identify the true class boundaries before finding widths. For measurements rounded to the nearest whole unit these boundaries are 9.5, 19.5, 29.5, and so on. Follow the recording precision and convention in the question, rather than subtracting printed integer limits blindly.
Cumulative frequency and interpolation 5.1
Plot cumulative frequencies at the upper class boundaries, including zero at the first lower boundary. Read a median at cumulative frequency N/2, quartiles at N/4 and 3N/4, and a pth percentile at pN/100. Values read from a grouped graph are estimates.
Worked example: 80 observations have frequencies 12, 28 and 40 in classes 0<=x<10, 10<=x<20 and 20<=x<40. Cumulative frequencies at 10, 20 and 40 are 12, 40 and 80. The third quartile is estimated at cumulative frequency 60, halfway through the last class, giving 30.
Linear interpolation within a class assumes observations are spread uniformly across that class: estimated value = lower boundary + (required cumulative frequency - cumulative frequency before class) / class frequency x class width. A smoother curve can give a different graphical estimate; report sensible precision rather than invented exactness.
For a proportion between a and b, subtract the cumulative counts below a from those below b and divide by N. For a proportion above b use the remaining count. For continuous models, boundary equality makes no probability difference; when interpreting an actual discrete data set, attend to whether the wording includes the endpoint.
Measures of location and spread 5.1
For individual observations, mean = sum(x)/n. The median is the middle ordered value, or the average of the two middle values when n is even. The mode is a most frequent value and need not be unique. For an equal-width grouped frequency table, the modal class has the greatest frequency. With unequal histogram widths, the tallest bar identifies the greatest density, not necessarily the greatest frequency. Neither identifies a known exact modal observation.
Range = maximum - minimum. Interquartile range = Q3-Q1 and measures spread of the middle half. Extreme values can strongly change the mean, range and standard deviation. The median and interquartile range are less sensitive to isolated extremes, but are not completely independent of the data.
Population standard deviation for the given complete data set is sqrt(sum(x^2)/n - mean^2). Equivalently, use the mean squared distance from the mean. Keep unrounded working values. Variance has squared units; standard deviation has the original units. The sample estimate with denominator n-1 is a different quantity and is not interchangeable with this formula.
Grouped estimates and coded totals 5.1
For a frequency table, n=sum(f), mean=sum(fx)/sum(f), and standard deviation=sqrt(sum(fx^2)/sum(f) - mean^2). If x is a class midpoint, these are estimates because the observations' positions within the classes are unknown.
Coding z=(x-a)/b with b>0 gives mean(x)=a+b mean(z) and standard deviation(x)=b standard deviation(z). Adding a constant changes the mean but not spread; scaling changes standard deviation by the absolute scale factor. If the code uses a negative scale, use its absolute value for standard deviation.
Worked example: z=(x-50)/5, n=8, sum(z)=12 and sum(z^2)=40. Mean(z)=1.5, so mean(x)=57.5. Variance(z)=40/8-1.5^2=2.75, hence standard deviation(x)=5 sqrt(2.75)=sqrt(68.75), approximately 8.29 (2 d.p.). The rounded 8.29 is not the exact standard deviation.
Combine data sets using totals 5.1
To combine two groups, add their counts, sums and sums of squares, then apply the mean and variance formulas to the combined totals. Do not take an unweighted average of group means when group sizes differ, or average group standard deviations.
Worked example: group A has n=2, sum(x)=6, sum(x^2)=20; group B has n=3, sum(x)=15, sum(x^2)=83. Combined n=5, sum(x)=21 and sum(x^2)=103. Mean is 4.2 and variance is 103/5-4.2^2=2.96. Standard deviation is sqrt(2.96), approximately 1.72 (2 d.p.).
If the groups were coded differently, undo or reconcile the coding before adding their totals. Explain comparisons in the data's context, using both a measure of location and a measure of spread rather than reporting bare calculations alone.
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