Statistics: Reading and Interpreting Given Diagrams (Venn, Tree, Histogram, Scatter, Box Plot)
Reading and interpreting given diagrams is the skill of extracting exact values, such as a frequency, a probability, a class boundary, a correlation or a quartile, from a Venn diagram, a tree diagram, a histogram, a scatter diagram or a box plot that a question describes or draws for you, rather than building the diagram yourself from a raw list of data. Exam questions expect specific numbers read off correctly, using what each diagram's regions, branches, bars or axes actually represent.
Before you start
Make sure you're comfortable with these topics first:
Method
- In a Venn diagram, the number written in a closed region is the frequency (or probability, if the diagram is already labelled with probabilities) for exactly that combination of memberships; to find a value that is not directly labelled, use the fact that every region inside the diagram sums to the stated total.
- In a tree diagram, multiply ALONG a branch (following one complete path from start to end) to find the probability of that one sequence of outcomes, and add the results for every branch that leads to the outcome you want, when there is more than one way to reach it.
- A without-replacement tree diagram changes the probabilities on its SECOND set of branches to reflect what the first branch already removed; check whether a diagram is with replacement (later branch probabilities repeat) or without replacement (they do not) before reading off a later probability.
- A histogram plots frequency DENSITY, not frequency, on the vertical axis, so frequency = frequency density x class width, which is the AREA of a bar; find a bar's frequency by multiplying its height by its width, and find its height by dividing a known frequency by its width.
- Because a histogram's area (not its height) represents frequency, estimating the median or a quartile from one means finding the bar in which that proportion of the total frequency falls, then using linear interpolation across the WIDTH of that bar, exactly as for grouped frequency data.
- A scatter diagram shows correlation as a pattern in the scatter of points (positive, negative, or none); the gradient of a line of best fit is the average change in the y-variable per unit change in the x-variable, and its y-intercept is only a genuine prediction if x = 0 is a sensible value for the data.
- Using a line of best fit to predict a value INSIDE the range of the given data (interpolation) is generally reliable; predicting OUTSIDE that range (extrapolation) assumes the same trend continues beyond where it has actually been observed, and a question that asks you to comment on reliability wants exactly this point made.
- A box plot shows a five-figure summary: the minimum, the lower quartile, the median, the upper quartile and the maximum. Read the whisker ends for the minimum and maximum, unless an outlier is plotted as a separate point beyond a whisker, in which case the whisker stops at the last non-outlier value instead.
- Whichever diagram is used, check the axes, key or labels stated in the question before reading off a single number: a histogram's vertical axis, a Venn diagram's regions and a tree diagram's branches only mean what the question says they mean, and misreading which one you are looking at is the most common way marks are lost on this topic.
Worked example
A Venn diagram shows two events, A and B, drawn inside a rectangle representing a total of 60 students. The region where the two circles overlap (A and B) contains 9 students. The part of circle A outside the overlap (A only) contains 15 students. The part of circle B outside the overlap (B only) contains 21 students. Find (a) the number of students outside both circles, (b) P(A), (c) P(A|B).
- Add every given region: overlap (9) + A only (15) + B only (21) = 45 students accounted for inside the two circles.
- Since the rectangle represents all 60 students, the number outside both circles is 60 - 45 = 15.
- n(A) = (A only) + (A and B) = 15 + 9 = 24, so P(A) = 24/60 = 0.4.
- n(B) = (B only) + (A and B) = 21 + 9 = 30, so P(B) = 30/60 = 0.5.
- P(A|B) = P(A and B)/P(B) = (9/60)/(30/60) = 9/30 = 0.3.
Practice questions
Type your answer and press Check to be marked straight away, or reveal the answer and mark yourself.
Q1A Venn diagram shows events C and D inside a rectangle representing 40 items. The overlap (C and D) contains 5 items, C only contains 11 items, and D only contains 9 items. Find the number of items outside both C and D.Show answer
Answer: 15 (40 - (5+11+9) = 40 - 25).
Q2A bag contains 5 red and 3 blue counters. Two counters are drawn without replacement, shown on a tree diagram with branches for each colour at each stage. Use the tree diagram to find the probability that the two counters drawn are DIFFERENT colours.Show answer
Answer: 15/28 (P(red then blue) + P(blue then red) = (5/8)(3/7) + (3/8)(5/7) = 15/56 + 15/56 = 30/56 = 15/28).
Q3A histogram has a bar for the class 10-20 with frequency density 3, and a bar for the class 20-50 with frequency density 1.2. Find the total frequency represented by these two bars.Show answer
Answer: 66 (frequency = density x width: 3x10=30 for the first bar, 1.2x30=36 for the second; 30+36=66).
Q4A scatter diagram of hours revised (x) against test score (y) has a line of best fit with equation y = 4x + 52. Use this line to predict the test score of a student who revised for 6 hours.Show answer
Answer: 76 marks (4(6)+52 = 24+52).
Q5A box plot shows a minimum of 12, a lower quartile of 18, a median of 25, an upper quartile of 33 and a maximum of 40. Find the interquartile range, and state whether a further value of 41 would be plotted as an outlier, using the rule that a value is an outlier if it lies more than 1.5 x IQR beyond the nearer quartile.Show answer
Answer: IQR = 33-18 = 15. The upper outlier boundary is 33 + 1.5(15) = 55.5, and since 41 < 55.5, it would NOT be plotted as an outlier.
Q6A Venn diagram for events E and F shows P(E only) = 0.2, P(F only) = 0.35, and P(neither E nor F) = 0.15. Find P(E and F).Show answer
Answer: 0.3 (every region of a probability Venn diagram sums to 1: 0.2+0.35+0.15+P(E and F) = 1, so P(E and F) = 1 - 0.7).
Exam-style questions
Written in the style of a A Level Maths exam paper, with a full mark scheme.
A doctor's test for a disease is represented by a tree diagram, where the first set of branches shows whether a patient has the disease and the second set shows whether they test positive. In the population being tested, P(has the disease) = 0.02. Given that a patient has the disease, P(tests positive) = 0.95. Given that a patient does NOT have the disease, P(tests positive) = 0.04 (a false positive). (a) State the three missing branch probabilities on the tree diagram. (2) (b) Find the probability that a randomly selected patient tests positive. (2) (c) Given that a patient tests positive, find the probability that they actually have the disease, giving your answer to 3 significant figures. (3)
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A histogram summarises the delivery times, in minutes, of 200 parcels, using four bars over the class intervals 0-10, 10-20, 20-40 and 40-70 minutes, drawn on a frequency density scale. The bar for 0-10 has frequency density 2, the bar for 10-20 has frequency density 9, and the bar for 40-70 has frequency density 0.5. (a) Given that the total number of parcels is 200, find the frequency density of the bar for the class 20-40 minutes. (4) (b) Hence estimate the number of parcels that took longer than 40 minutes to deliver. (1)
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A scatter diagram plots the age, in years, of 12 secondhand cars (x) against their value in pounds (y), for cars aged between 1 and 8 years. The product moment correlation coefficient for the data is r = -0.93, and the equation of the line of best fit is y = -1200x + 14000. (a) Interpret the value r = -0.93 in the context of a car's age and its value. (2) (b) Use the line of best fit to estimate the value of a car that is 4 years old. (2) (c) Comment on the reliability of using this line to estimate the value of a car that is 15 years old. (2)
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Free printable worksheet
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