Statistics: Reading and Interpreting Given Diagrams (Venn, Tree, Histogram, Scatter, Box Plot) - Worksheets, Questions and Revision

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AS · Statistics

S11 Statistics: Reading and Interpreting Given Diagrams (Venn, Tree, Histogram, Scatter, Box Plot)

EDEXCEL 8MA0 · Calculator allowed · about 120 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A Venn diagram is used to show data about 50 sixth-form students, some of whom study French (F) and some of whom study Spanish (S). The number of students in each region of the diagram is shown, except for one region, which is marked x.

Figure (to be drawn): Venn diagram: an outer rectangle represents the universal set of 50 sixth-form students. Inside the rectangle are two overlapping circles, labelled F (students who study French) and S (students who study Spanish). The overlapping region (F and S) is labelled 6. The part of circle S outside F is labelled 14. The part of circle F outside S is unlabelled, marked x. The region inside the rectangle but outside both circles is labelled 10.

(a)Using the fact that the diagram represents 50 students in total, find the value of x.(2)

(b)Find P(a randomly selected student studies both French and Spanish), giving your answer as a fraction in its simplest form.(2)

(c)Find P(a student studies French, given that they study Spanish).(3)

(d)Determine, showing your working, whether the events 'a student studies French' and 'a student studies Spanish' are independent.(3)

(Total for Question 1 is 10 marks)

The Venn diagram below shows the probabilities associated with two events, A and B, within a sample space. One region of the diagram is unlabelled and marked x.

Figure (to be drawn): Venn diagram: a rectangle represents the sample space. Two overlapping circles inside it are labelled A and B. The region inside both A and B is labelled 0.15. The region inside A only (not B) is labelled 0.25. The region inside B only (not A) is unlabelled, marked x. The region outside both circles, inside the rectangle, is labelled 0.30.

(a)Show that x = 0.30.(2)

(b)Find P(A) and P(B).(2)

(c)Find P(A union B).(2)

(d)State, with a reason, whether A and B are mutually exclusive.(1)

(e)Find P(A | B').(3)

(Total for Question 2 is 10 marks)

A bag contains 5 red counters and 3 blue counters. Two counters are drawn from the bag at random, one after the other, without replacement. The tree diagram below shows the possible outcomes. Two of the second-stage probabilities are not shown, and are marked p and q.

Figure (to be drawn): Tree diagram with two stages, for drawing two counters from a bag without replacement. First stage: two branches from the start, labelled 'Red' with probability 5/8 and 'Blue' with probability 3/8. Second stage, from the 'Red' branch: two further branches labelled 'Red' with probability 4/7 and 'Blue' with probability p (to find). Second stage, from the 'Blue' branch: two further branches labelled 'Red' with probability 5/7 and 'Blue' with probability q (to find).

(a)Write down the value of p and the value of q.(2)

(b)Calculate P(both counters are red).(2)

(c)Calculate the probability that the two counters are different colours.(3)

(d)Given that the two counters drawn are the same colour, find the probability that both are red.(4)

(Total for Question 3 is 11 marks)

The tree diagram below models whether it rains on each of two consecutive days. Two of the second-stage probabilities are not shown, and are marked a and b.

Figure (to be drawn): Tree diagram with two stages, for rain on two consecutive days. First stage: two branches from the start, labelled 'Rain' with probability 0.3 and 'No rain' with probability 0.7. Second stage, from the 'Rain' branch: two further branches labelled 'Rain' with probability 0.5 and 'No rain' with probability a (to find). Second stage, from the 'No rain' branch: two further branches labelled 'Rain' with probability 0.2 and 'No rain' with probability b (to find).

(a)Write down the values of a and b.(2)

(b)Calculate the probability that it rains on exactly one of the two days.(3)

(c)Calculate the probability that it rains on at least one of the two days.(3)

(d)Given that it rains on at least one of the two days, find the probability that it rains on both days.(3)

(Total for Question 4 is 11 marks)

The histogram below shows the times, t minutes, taken by 90 runners to complete a fun run. The height of one of the bars is not shown.

Figure (to be drawn): Histogram with horizontal axis 't, time in minutes' from 0 to 80, and vertical axis 'Frequency density' from 0 to 4. Five bars are drawn with no gaps between them, over the class boundaries 0, 10, 20, 30, 50, 80. The bar for 0≤t<10 has height 0.8. The bar for 10≤t<20 has height 2.0. The bar for 20≤t<30 is drawn but its height is not labelled. The bar for 30≤t<50 has height 0.9. The bar for 50≤t<80 has height 0.2.

(a)State the frequency density represented by the bar for 10≤t<20.(1)

(b)Calculate the number of runners with times in the interval 0≤t<10.(2)

(c)Given that 90 runners took part in total, calculate the frequency density of the bar for 20≤t<30 minutes; this bar's height is not labelled on the diagram.(3)

(d)Estimate the number of runners whose time was between 40 and 50 minutes, stating an assumption you make.(3)

(Total for Question 5 is 9 marks)

The histogram below shows the distances, d km, travelled to school by 60 students at a school.

Figure (to be drawn): Histogram with horizontal axis 'd, distance travelled to school in km' from 0 to 20, and vertical axis 'Frequency density' from 0 to 6. Four bars are drawn with no gaps between them, over the class boundaries 0, 2, 6, 10, 20. The bar for 0≤d<2 has height 4. The bar for 2≤d<6 has height 6. The bar for 6≤d<10 has height 5. The bar for 10≤d<20 has height 0.8.

(a)Use the histogram to find the frequency for each of the four class intervals.(2)

(b)Using linear interpolation, estimate the median distance travelled to school.(3)

(c)Assuming distances are uniformly distributed within the class 10≤d<20, estimate the number of students who travelled more than 15 km to school.(3)

(d)Explain why your answer to part (c) is only an estimate.(1)

(Total for Question 6 is 9 marks)

The scatter diagram below shows the age, in years, and the resale value, in thousands of pounds, of 9 cars of the same model, together with a line of best fit.

Figure (to be drawn): Scatter diagram. Horizontal axis: 'Age of car, x (years)', scale 0 to 12 in steps of 2. Vertical axis: 'Resale value, y (GBP thousands)', scale 0 to 14 in steps of 2. Eight points are plotted at (age, value): (1, 11), (2, 10), (3, 9.5), (4, 8), (5, 7.5), (6, 6), (8, 4), (10, 2). One further point is plotted at (9, 9), marked with a cross and labelled 'Car P', lying clearly above the general trend of the other points. A straight line of best fit is drawn from (0, 12) to (10, 2), passing close to the first eight points and well below the point for Car P.

(a)Describe the type of correlation shown between the age of a car and its resale value.(1)

(b)Using the line of best fit, estimate the resale value of a car aged 7 years.(2)

(c)Car P, aged 9 years, has a resale value of GBP 9000, well above the value suggested by the line of best fit. Suggest a reason, in context, why this car's resale value might not fit the general trend.(1)

(d)State whether it would be appropriate to use the line of best fit to estimate the resale value of a car aged 20 years, giving a reason.(2)

(e)The line of best fit has a gradient of -1. Interpret this value in the context of the data.(2)

(Total for Question 7 is 8 marks)

The scatter diagram below shows the number of hours of revision carried out by 12 students before an exam, and the percentage score each student achieved, together with a line of best fit. The product moment correlation coefficient (PMCC) for the 12 students is also shown.

Figure (to be drawn): Scatter diagram. Horizontal axis: 'Hours of revision, x', scale 0 to 10 in steps of 2. Vertical axis: 'Exam score, y (%)', scale 0 to 100 in steps of 20. Eleven points are plotted showing an increasing trend, lying close to the line from (0, 20) to (10, 90). One further point is plotted at (2, 85), marked with a cross and labelled 'Student Q', lying clearly above the general trend of the other points. A straight line of best fit is drawn from (0, 20) to (10, 90). The value 'r = 0.82' is printed below the diagram.

(a)State, with a reason, whether the value r = 0.82 shown on the diagram indicates a strong or a weak positive correlation.(2)

(b)Using the line of best fit, estimate the exam score of a student who revised for 6 hours.(2)

(c)Student Q revised for only 2 hours but scored 85%, well above the value suggested by the line. State the effect that removing Student Q's data point would be likely to have on the value of r, and explain your reasoning.(2)

(d)A teacher claims 'revising for longer causes a higher exam score.' Using the scatter diagram, comment on whether this conclusion is fully justified.(3)

(Total for Question 8 is 9 marks)

The box plots below show the time spent on homework per week, in hours, by two classes of sixth-form students, Class X and Class Y, drawn on the same scale.

Figure (to be drawn): Two box plots drawn on the same horizontal scale, from 0 to 20 hours, one above the other, both labelled 'Time spent on homework per week (hours)'. The top box plot is labelled 'Class X': minimum 2, lower quartile 5, median 8, upper quartile 11, maximum 14, drawn as a standard box and whisker with no outliers. The bottom box plot is labelled 'Class Y': minimum 3, lower quartile 10, median 12, upper quartile 13, with the upper whisker drawn to 17 (the largest value that is not an outlier), and one further point marked with a cross at 20, representing an outlier.

(a)Write down the median and calculate the interquartile range for Class X.(3)

(b)State the five-figure summary for Class Y, including the value marked as an outlier as the maximum.(2)

(c)Explain why the upper whisker on the box plot for Class Y is drawn to 17 rather than extending to 20.(2)

(d)Using your answers to parts (a) and (b), compare the median times and the interquartile ranges of the two classes, in context.(3)

(e)State one advantage of using box plots, rather than the mean and standard deviation, to compare the two classes' homework times.(1)

(Total for Question 9 is 11 marks)

The box plots below show the length of telephone calls, in minutes, handled by two call centres, P and Q, drawn on the same scale.

Figure (to be drawn): Two box plots drawn on the same horizontal scale, from 0 to 25 minutes, one above the other, both labelled 'Length of phone call (minutes)'. The top box plot is labelled 'Call Centre P': minimum 2, lower quartile 4, median 6, upper quartile 9, with the upper whisker drawn to 15 (the largest value that is not an outlier), and one further point marked with a cross at 24, labelled as an outlier. The bottom box plot is labelled 'Call Centre Q': minimum 5, lower quartile 11, median 15, upper quartile 18, maximum 22, drawn as a standard box and whisker with no outliers.

(a)By calculating the relevant boundary, determine whether the point at 24 minutes for Call Centre P has been correctly identified 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.(3)

(b)State, giving a reason based on the position of the median within the box, the skewness of the distribution of call lengths for Call Centre P (excluding the outlier).(2)

(c)State, giving a reason, the skewness of the distribution of call lengths for Call Centre Q.(2)

(d)A manager claims 'Call Centre Q handles calls more efficiently, since its calls are shorter on average.' Using the box plots, comment on whether the evidence supports this claim.(3)

(Total for Question 10 is 10 marks)

A factory has three machines, A, B and C, that produce identical parts. The tree diagram below shows the proportion of parts produced by each machine, and the probability that a part from each machine is faulty. Two of the probabilities are not shown, and are marked p and q.

Figure (to be drawn): Tree diagram with two stages, for a part randomly selected from a factory's output. First stage: three branches from the start, labelled 'Machine A' with probability 0.5, 'Machine B' with probability 0.3, and 'Machine C' with probability p (to find). Second stage, from 'Machine A': two further branches labelled 'Faulty' with probability 0.02 and 'Not faulty' with probability 0.98. Second stage, from 'Machine B': two further branches labelled 'Faulty' with probability 0.05 and 'Not faulty' with probability 0.95. Second stage, from 'Machine C': two further branches labelled 'Faulty' with probability 0.08 and 'Not faulty' with probability q (to find).

(a)Find the values of p and q.(2)

(b)Calculate the probability that a randomly selected part is faulty.(3)

(c)Given that a randomly selected part is faulty, find the probability that it was produced by Machine C.(3)

(d)The factory manager claims 'Since Machine C has the highest individual fault rate (8%), most faulty parts come from Machine C.' Using your answers to parts (b) and (c), comment on whether this claim is justified.(3)

(Total for Question 11 is 11 marks)

Mark scheme · S11 Statistics: Reading and Interpreting Given Diagrams (Venn, Tree, Histogram, Scatter, Box Plot)

Question 1

(a) M1 correct method: x = 50 - 6 - 14 - 10
(a) A1 cao x = 20
(a) Answer: x = 20
(b) M1 6/50 identified from the diagram
(b) A1 cao 3/25
(b) Answer: 3/25
(c) M1 n(S) = 6 + 14 = 20 identified from the diagram
(c) M1 correct method: 6/20
(c) A1 cao 3/10
(c) Answer: 3/10
(d) M1 P(F) = (20 + 6)/50 = 26/50 (= 13/25 or 0.52) found
(d) M1 P(F) x P(S) = 0.52 x 0.4 = 0.208 attempted, using P(S) = 20/50 = 0.4
(d) A1 cso: compares 0.208 with P(F and S) = 6/50 = 0.12, correctly concludes the values are not equal so F and S are not independent
(d) Answer: Not independent, since P(F) x P(S) = 0.208 does not equal P(F and S) = 0.12.

Question 2

(a) M1 sets up 0.25 + 0.15 + x + 0.30 = 1, using the fact that all probabilities in the diagram sum to 1
(a) A1 cso x = 0.30
(a) Answer: x = 0.30 (given)
(b) B1 P(A) = 0.25 + 0.15 = 0.40
(b) B1 P(B) = 0.30 + 0.15 = 0.45
(b) Answer: P(A) = 0.40, P(B) = 0.45
(c) M1 correct method, e.g. 0.25 + 0.30 + 0.15, or 1 - 0.30
(c) A1 cao 0.70
(c) Answer: 0.70
(d) B1 not mutually exclusive, since P(A and B) = 0.15, which is not equal to 0
(d) Answer: Not mutually exclusive, since P(A and B) = 0.15 is not zero.
(e) M1 P(B') = 1 - 0.45 = 0.55
(e) M1 P(A and B') = 0.25 identified as the 'A only' region
(e) A1 oe 5/11, awrt 0.455
(e) Answer: 5/11 (awrt 0.455)

Question 3

(a) B1 p = 3/7
(a) B1 q = 2/7
(a) Answer: p = 3/7, q = 2/7
(b) M1 correct method: 5/8 x 4/7
(b) A1 oe 5/14
(b) Answer: 5/14
(c) M1 identifies the two relevant outcomes: red-then-blue and blue-then-red
(c) M1 correct method: 5/8 x 3/7 + 3/8 x 5/7
(c) A1 cao 15/28
(c) Answer: 15/28
(d) M1 attempt P(both red) = 5/8 x 4/7 (= 5/14 or 20/56)
(d) M1 attempt P(both blue) = 3/8 x 2/7 (= 3/28 or 6/56)
(d) dM1 correct conditional method: P(both red) / [P(both red) + P(both blue)], dependent on both previous marks
(d) A1 cao 10/13
(d) Answer: 10/13

Question 4

(a) B1 a = 0.5
(a) B1 b = 0.8
(a) Answer: a = 0.5, b = 0.8
(b) M1 identifies the two relevant outcomes: rain-then-no-rain and no-rain-then-rain
(b) M1 correct method: 0.3 x 0.5 + 0.7 x 0.2
(b) A1 cao 0.29
(b) Answer: 0.29
(c) M1 attempts P(no rain on either day) = 0.7 x 0.8
(c) M1 correct method: 1 - their P(no rain on either day)
(c) A1 cao 0.44
(c) Answer: 0.44
(d) M1 P(rain on both days) = 0.3 x 0.5 = 0.15 identified
(d) M1 correct conditional method: their 0.15 divided by their answer to part (c)
(d) A1 oe 15/44, awrt 0.341
(d) Answer: 15/44 (awrt 0.341)

Question 5

(a) B1 2.0
(a) Answer: 2.0
(b) M1 correct method: frequency = frequency density x class width, 0.8 x 10
(b) A1 cao 8
(b) Answer: 8 runners
(c) M1 attempts the total frequency of the four known bars: 8 + 20 + 18 + 6 = 52
(c) M1 correct method: missing frequency = 90 - 52 (= 38), then divides by the class width 10
(c) A1 cao 3.8
(c) Answer: 3.8
(d) B1 states an assumption, e.g. that times are uniformly (evenly) distributed within the class 30≤t<50, oe
(d) M1 correct method: 0.9 x 10, or equivalently (50-40)/(50-30) x 18
(d) A1 cao 9
(d) Answer: 9 runners, assuming times are spread evenly (uniformly) within the class 30≤t<50.

Question 6

(a) M1 attempts frequency = frequency density x class width for at least one class
(a) A1 all four correct: 8, 24, 20, 8
(a) Answer: 0≤d<2: 8; 2≤d<6: 24; 6≤d<10: 20; 10≤d<20: 8
(b) M1 correct cumulative frequencies identified (8, 32, 52, 60) and median class 2≤d<6 identified
(b) M1 correct interpolation: 2 + ((30-8)/24) x 4
(b) A1 awrt 5.67
(b) Answer: awrt 5.67 km
(c) B1 identifies that only the proportion of the class 10≤d<20 above d=15 is needed
(c) M1 correct method: (20-15)/(20-10) x 8
(c) A1 cao 4
(c) Answer: 4 students
(d) B1 the actual distribution of distances within the class 10≤d<20 is not known; only the total frequency for the whole class is given, so a uniform spread has to be assumed, oe
(d) Answer: Because the exact distances within the class 10≤d<20 are unknown, only the total frequency for the class; a uniform spread within the class has to be assumed.

Question 7

(a) B1 negative correlation
(a) Answer: Negative correlation.
(b) M1 reads the line of best fit at x = 7 appropriately
(b) A1 awrt 5 (allow 4.5 to 5.5)
(b) Answer: awrt GBP 5000 (GBP 5 thousand)
(c) B1 valid contextual reason, e.g. Car P may be a rare or collectible model, or may be in exceptionally good condition or have unusually low mileage for its age, oe
(c) Answer: Car P might be a rare or highly sought-after model, or might have very low mileage or be in exceptionally good condition for its age.
(d) B1 no, it would not be appropriate
(d) B1 reason: 20 years is outside the range of the given data (ages 1 to 10), so this would be extrapolation and the linear trend is not known to continue, oe
(d) Answer: No; 20 years is outside the range of the data used to draw the line (ages 1 to 10), so this would be extrapolation, and there is no evidence the linear trend continues beyond the data.
(e) B1 states that resale value decreases by about GBP 1000 for each extra year of age
(e) B1 correctly links the negative sign to a decrease as age increases, oe
(e) Answer: On average, a car in this model loses about GBP 1000 of resale value for each extra year of age.

Question 8

(a) B1 strong (positive correlation)
(a) B1 reason: r = 0.82 is close to 1, oe
(a) Answer: Strong positive correlation, since r = 0.82 is close to 1.
(b) M1 reads the line of best fit at x = 6 appropriately
(b) A1 awrt 62 (allow 60 to 64)
(b) Answer: awrt 62%
(c) B1 r would increase (the correlation would become stronger)
(c) B1 reason: Student Q's point does not follow the general upward trend and is currently weakening the correlation, oe
(c) Answer: r would increase, since Student Q's point does not fit the general trend and is currently pulling the correlation coefficient down.
(d) B1 correlation does not necessarily imply causation
(d) B1 identifies a plausible confounding variable, e.g. a student's general ability or motivation, which could affect both revision time and exam score
(d) B1 balanced conclusion: the teacher's claim is not fully justified by the correlation alone, oe
(d) Answer: A strong correlation does not prove causation. A student's ability or motivation could independently cause both more revision and a higher score, so the teacher's causal claim is not fully justified by this data alone.

Question 9

(a) B1 median = 8
(a) M1 correct method: 11 - 5
(a) A1 cao IQR = 6
(a) Answer: median = 8 hours, IQR = 6 hours
(b) B1 minimum = 3, lower quartile = 10, median = 12, upper quartile = 13, all correct
(b) B1 maximum correctly identified as 20 (i.e. including the outlier as the true maximum value)
(b) Answer: minimum = 3, Q1 = 10, median = 12, Q3 = 13, maximum = 20
(c) B1 recognises that 20 is identified as an outlier (it lies more than 1.5 x IQR beyond the upper quartile)
(c) B1 states that the whisker is drawn only as far as the largest value that is not classified as an outlier, which is 17, oe
(c) Answer: Because 20 is classified as an outlier, it is plotted separately rather than as the end of the whisker; the whisker extends only to the largest value that is not an outlier, 17.
(d) B1 Class Y has the higher median (12 compared with 8), so Class Y students typically spent more time on homework than Class X, oe
(d) B1ft compares Class X's IQR (their value from part (a), = 6) with Class Y's IQR (13 - 10 = 3): Class X's IQR is the larger of the two, so homework times were more variable in Class X than in Class Y, oe
(d) B1 valid overall concluding comment in context, e.g. Class Y students spent more time on homework on average and were also more consistent in how long they spent than Class X, oe
(d) Answer: Class Y has the higher median (12 vs 8), so typically spent more time on homework; Class X has the larger IQR (6 vs 3), so homework times were more spread out in Class X. Overall, Class Y students did more homework and more consistently than Class X.
(e) B1 box plots directly show the median, quartiles and any outliers, and are not distorted by extreme values (outliers) the way the mean and standard deviation can be, oe
(e) Answer: Box plots show the median, quartiles and any outliers directly, and are not distorted by extreme values the way the mean and standard deviation can be.

Question 10

(a) M1 IQR = 9 - 4 = 5
(a) M1 upper boundary = 9 + 1.5 x 5
(a) A1 cso: boundary = 16.5, and since 24 > 16.5, the point has been correctly identified as an outlier
(a) Answer: Correctly identified: the boundary is 16.5, and 24 > 16.5.
(b) B1 compares median - Q1 (= 2) with Q3 - median (= 3)
(b) B1 since Q3 - median > median - Q1, the distribution is positively skewed
(b) Answer: Positively skewed: Q3 - median (3) is greater than median - Q1 (2).
(c) B1 compares median - Q1 (= 4) with Q3 - median (= 3)
(c) B1 since median - Q1 > Q3 - median, the distribution is negatively skewed
(c) Answer: Negatively skewed: median - Q1 (4) is greater than Q3 - median (3).
(d) B1 Call Centre Q's median (15) is higher than Call Centre P's median (6), so the typical call is longer at Q, not shorter
(d) B1 the middle 50% of P's calls (IQR 4 to 9 minutes) are much shorter than the middle 50% of Q's calls (IQR 11 to 18 minutes), oe
(d) B1 concludes the evidence does not support the manager's claim, oe
(d) Answer: The evidence does not support the claim: Call Centre Q's median call (15 minutes) is much higher than Call Centre P's (6 minutes), and P's middle 50% of calls (4 to 9 minutes) are far shorter than Q's (11 to 18 minutes). Overall P's calls, aside from the single outlier, tend to be shorter than Q's.