Data Counting And Distributions

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Original, AI-authored English practice with worked guidance. Selected topics only; independent teacher review is pending. This is not a complete syllabus, official past-paper collection or predicted exam score. Check your own exam-board specification: these concepts are not certified against a particular board or exam year.

83 original questions across 16 selected topic areas. No account or payment required to practise.

Cambridge International AS & A Level Mathematics (9709), examination year 2027: AI-assisted conceptual alignment to official sources, not teacher or exam-board certification. Syllabus mapping and remaining coverage.

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Original Cambridge 2027 concept practice for source-backed gaps. Indicative self-marking only, not an official paper or complete syllabus coverage.

Selected subtopics: Grouped frequencies and interpolation · Selections and restricted arrangements · Geometric variables and normal approximations

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Question 1Travel times in minutes are grouped as 0<=t<10 (frequency 4), 10<=t<20 (frequency 8), and 20<=t<40 (frequency 8). State histogram frequency densities and explain why raw frequency is not the correct height for all bars. Estimate mean, population standard deviation, median and interquartile range, stating the assumptions used.6 indicative marks

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Travel times in minutes are grouped as 0<=t<10 (frequency 4), 10<=t<20 (frequency 8), and 20<=t<40 (frequency 8). State histogram frequency densities and explain why raw frequency is not the correct height for all bars. Estimate mean, population standard deviation, median and interquartile range, stating the assumptions used.

Cambridge 2027: Direct conceptual match

Recomputed weighted totals 380 and 9100, variance 94 and cumulative-graph interpolations 11.25,17.5,27.5; histogram area totals 20.

9709: 5.1 Representation of data; candidate bullet 2 (bullet ordinal, not an official subcode): Construct/interpret stem plots, boxes, histograms and cumulative graphs
9709: 5.1 Representation of data; candidate bullet 3 (bullet ordinal, not an official subcode): Compare central tendency and spread
9709: 5.1 Representation of data; candidate bullet 4 (bullet ordinal, not an official subcode): Estimate quartiles, percentiles and proportions from cumulative graphs
9709: 5.1 Representation of data; candidate bullet 5 (bullet ordinal, not an official subcode): Calculate mean/SD from raw, grouped or coded totals

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Your response to question 1

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Worked guidance

Cumulative frequencies are 4,12,20. The median observation index 10 lies in the second class: 10+(10-4)/8*10=17.5.
Indices 5 and 15 give the quartiles. The wide final class contains Q3, so its interpolation uses width 20, not 10. Grouped data do not reveal the actual individual travel times.

Self-marking points

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The class widths are 10,10,20; densities are 0.4,0.8,0.4 observations per minute.
Unequal widths require histogram area, not height, to represent frequency.
Using midpoints 5,15,30 gives sum(f*x)=380 and estimated mean 19 minutes.
sum(f* $x^{2}$ )=9100 gives estimated variance 9100/20-19^2=94, so SD is approximately 9.70 minutes.
Linear interpolation on the cumulative-frequency graph gives median 17.5 minutes.
Quartiles are Q1=11.25 and Q3=27.5 minutes, so IQR=16.25 minutes; midpoint moments and within-class uniform interpolation are estimates, not exact raw-data results.

Examiner tip

State which results depend on class midpoints and which on linear within-class interpolation.

Question 2(a) A committee contains 2 of 5 students and 1 of 4 teachers. How many committees are possible? (b) How many distinct linear arrangements of the letters BANANA exist? (c) Six distinct people, including Ada and Ben, stand in one row. How many rows have Ada and Ben not adjacent?5 indicative marks

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(a) A committee contains 2 of 5 students and 1 of 4 teachers. How many committees are possible? (b) How many distinct linear arrangements of the letters BANANA exist? (c) Six distinct people, including Ada and Ben, stand in one row. How many rows have Ada and Ben not adjacent?

Cambridge 2027: Direct conceptual match

Independently counted committee combinations, multiset permutations and block-complement arrangements as 40,60,480.

9709: 5.2 Permutations and combinations; candidate bullet 1 (bullet ordinal, not an official subcode): Count selections with permutations and combinations
9709: 5.2 Permutations and combinations; candidate bullet 2, repetition subitem (bullet ordinal, not an official subcode): Count linear arrangements with repeated objects
9709: 5.2 Permutations and combinations; candidate bullet 2, restriction subitem (bullet ordinal, not an official subcode): Count linear arrangements under adjacency restrictions

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Your response to question 2

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Worked guidance

Distinguish choosing committee members from arranging people. Changing the order of the selected two students does not make a new committee.
Repeated identical letters do not create distinct arrangements when swapped. For nonadjacency it is simpler to subtract the block count from all unrestricted rows.

Self-marking points

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Committee selections are unordered, giving C(5,2)*C(4,1)=10*4=40.
BANANA has three A letters and two N letters, so arrangements number 6!/(3!*2!)=60.
All six-person rows number 6!=720.
Adjacent Ada/Ben rows number 2*5!=240 by treating the pair as one block with two internal orders.
The required nonadjacent count is 720-240=480.

Examiner tip

These are line arrangements; no rotational equivalence or circle convention is involved.

Question 3A device succeeds on each independent attempt with probability 0.25. Let T count attempts up to and including its first success. Find P(T=3), P(T>4) and E(T), and explain the model assumptions. Separately a random variable X takes 0,1,2 with probabilities 0.2,0.5,0.3. Find E(X) and Var(X).5 indicative marks

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A device succeeds on each independent attempt with probability 0.25. Let T count attempts up to and including its first success. Find P(T=3), P(T>4) and E(T), and explain the model assumptions. Separately a random variable X takes 0,1,2 with probabilities 0.2,0.5,0.3. Find E(X) and Var(X).

Cambridge 2027: Direct conceptual match

Checked the support convention and exact powers 9/64 and 81/256; independently summed the table moments to variance 0.49.

9709: 5.4 Discrete random variables; candidate bullet 1 (bullet ordinal, not an official subcode): Build discrete probability tables and compute moments
9709: 5.4 Discrete random variables; candidate bullet 2 (bullet ordinal, not an official subcode): Calculate binomial/geometric probabilities and justify models
9709: 5.4 Discrete random variables; candidate bullet 3 (bullet ordinal, not an official subcode): Use binomial moments and geometric expectation

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Your response to question 3

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Worked guidance

Three attempts up to the first success means two failures followed by one success; no choice coefficient is needed. T>4 means four failures, whatever happens afterwards.
For the separate table compute the first and second raw moments before subtracting the square of the mean. The geometric expectation is a mean waiting count, not a guaranteed stopping time.

Self-marking points

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T has the geometric distribution on 1,2,3,..., with constant success probability and independent trials.
P(T=3)=0.75^2*0.25=0.140625.
P(T>4)=0.75^4=0.31640625 and E(T)=1/0.25=4 attempts.
For X, probabilities sum to one and E(X)=0.5+0.6=1.1.
E( $X^{2}$ )=0.5+1.2=1.7, so Var(X)=1.7-1. $1^{2}$ =0.49.

Examiner tip

Geo(p) here counts the successful attempt; a failures-only convention would give a different support and expectation.

Question 4Let X~Binomial(100,0.4). Explain why a normal approximation is reasonable and estimate P(35<=X<=45), showing the continuity correction and standardisation. Give the probability to 3 significant figures.5 indicative marks

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Let X~Binomial(100,0.4). Explain why a normal approximation is reasonable and estimate P(35<=X<=45), showing the continuity correction and standardisation. Give the probability to 3 significant figures.

Cambridge 2027: Direct conceptual match

Computed the standardised half-unit endpoints and numerically integrated the standard normal independently for the approximately 0.738 probability.

9709: 5.5 The normal distribution; candidate bullet 3 (bullet ordinal, not an official subcode): Approximate binomial probabilities with continuity correction

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Your response to question 4

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Worked guidance

The integer endpoints 35 and 45 include complete probability bars, hence half-unit boundaries outside them.
Use $\sqrt{24}$ , not 24, as the standard deviation in standardisation. The final value is a normal approximation, not the exact binomial probability.

Self-marking points

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np=40 and n(1-p)=60 both exceed 5, supporting the normal approximation.
Use Y~N(40,24), since np(1-p)=24 is the variance.
Continuity correction gives P(34.5<Y<45.5).
Standardised bounds are +/-5.5/ $\sqrt{24}$ , approximately +/-1.12268.
The approximate probability is 2Phi(1.12268)-1, approximately 0.738.

Examiner tip

Keep both half-unit corrections; changing only one endpoint makes the interval asymmetric.

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