Free worked solutions · 2026 Edition

IB Maths AI HL Chapter 13: Describing data worked solutions

Worked solutions to the practice questions in Chapter 13 of Photon Guidebook: Mathematics Applications and Interpretation HL for the IB Diploma (2026 Edition). Each solution shows every step with the mark it earns, the way an IB mark scheme does.

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Basics

One idea at a time, in the order the chapter teaches them.

Question 1

[2 marks]
A depot orders 1560 tool records by inventory number. An auditor picks a random start from 1 to 26 and then checks every 26th record.
Name the sampling method and state its sampling interval.
[2]
Show solutionHide solution
Systematic sampling.
A1
Interval: 26 records.
A1

Final answer

Systematic sampling; interval 26 records.

Question 2

[4 marks]
A federation has 315 members in group A, 245 in group B and 140 in group C. A proportional stratified random sample of 80 is planned.
Find the required sample size for each group and describe how to select the people within a group.
[4]
Show solutionHide solution
Total =700; use n_i=80N_i/700.
M1
A:36, B:28, C:16.
A2
Select the stated number uniformly at random without replacement from each group list.
R1

Final answer

A 36, B 28, C 16; choose within each group by random numbers from a list of its members.

Question 3

[3 marks]
The table records late deliveries per week.
Late deliveries013
Frequency58k
The mean is exactly 1.4.
Determine whether any non-negative integer frequency k can satisfy this claim. Justify your answer.
[3]
Show solutionHide solution
[8+3k]/[13+k]=1.4
M1
8+3k=18.2+1.4k
M1
k=6.375, so no integer frequency is possible.
A1

Final answer

k=6.375 is not a whole number, so no frequency works.

Question 4

[2 marks]
The masses (g) of six prototype clips are 17,\ 22,\ 24,\ 27,\ 31,\ 35.
Use one-variable statistics to find the population standard deviation, to 3 significant figures.
[2]
Show solutionHide solution
Enter the six masses in one-variable statistics; select \sigma_x.
M1
\sigma_x=5.89 g.
A1

Final answer

\sigma=5.89 g.

Question 5

[3 marks]
An observation is an outlier if it is strictly below Q_1-1.5\mathrm{IQR} or strictly above Q_3+1.5\mathrm{IQR}. A data set has Q_1=29 and Q_3=41.
Determine which of 10, 11 and 60 would be outliers.
[3]
Show solutionHide solution
\mathrm{IQR}=12; fences 29-18=11 and 41+18=59.
M1
10 and 60 are outside the fences and are outliers.
A1
11 is on the boundary, so is not an outlier.
A1

Final answer

Fences 11 and 59; 10 and 60 are outliers, 11 is not.

Question 6

[4 marks]
Two ordinary box plots of battery duration have summaries A: (42,55,63,68,79), B: (39,51,60,73,86), in hours.
Compare typical duration using the medians and compare the widths of the middle-half intervals.
[4]
Show solutionHide solution
Medians: A 63 h, B 60 h; A is 3 h higher.
A1
A middle-half width 68-55=13 h.
A1
B middle-half width 73-51=22 h.
A1
A has the narrower middle-half interval by 9 h.
R1

Final answer

Medians A 63 h, B 60 h; middle halves 13 h and 22 h wide, so A is narrower by 9 h.

Question 7

[3 marks]
The grouped table records packing times.
Time (s)Frequency
12\le t<185
18\le t<249
24\le t<304
Estimate the mean time from class midpoints.
[3]
Show solutionHide solution
Midpoints 15,21,27; N=18.
A1
\bar{t}\approx[15(5)+21(9)+27(4)]/18
M1
\bar{t}\approx20.7 s (unrounded 20.666\dots).
A1

Final answer

20.7 s.

Question 8

[4 marks]
A balance overstates every mass by exactly 2.7 g. Recorded masses have mean 48.6 g and population standard deviation 3.2 g.
(a) Find the corrected mean.
[2]
(b) State the corrected standard deviation and explain.
[2]
Show solutionHide solution
(a) 48.6-2.7
M1
45.9 g.
A1
(b) 3.2 g.
A1
Subtracting the same constant leaves each deviation from the mean unchanged.
R1

Final answer

(a) 45.9 g (b) 3.2 g, unchanged.

Question 9

[3 marks]
A cumulative-frequency polygon for 80 readings joins (0,0), (5,12), (10,44), (15,68), (20,80). Percentile p is read at cumulative frequency pN/100.
Estimate the 60th percentile using straight segments.
[3]
Show solutionHide solution
Target cumulative frequency 0.60(80)=48.
M1
10+(48-44)(5)/(68-44).
M1
10.83 (approximately).
A1

Final answer

10.8 (10.83).

Question 10

[2 marks]
A machine gives almost identical readings of 248.2 g on repeated measurements of a certified 251.0 g mass.
Distinguish reliability from validity in this example.
[2]
Show solutionHide solution
Consistent repeats suggest reliability. R1 The systematic discrepancy from the certified value questions validity/accuracy of the mass measurement.
R1

Final answer

Reliable (consistent repeats) but not valid (readings 2.8 g below the certified mass).

Tutorial

Exam-style questions on each question type, with the marks the IB would give.

Question 11

[6 marks]
A swimming club has 1350 members: 540 juniors, 450 intermediates, 225 seniors and 135 masters. The club secretary takes a stratified sample of 90 members.
(a) Find the number of members of each group in the sample.
[3]
(b) The secretary later needs a stratified sample of 50 members instead. Explain why the exact proportional numbers are not all whole numbers, and give a set of whole-number sample sizes that add up to 50.
[2]
(c) State one advantage of stratified sampling over simple random sampling for this club.
[1]
Show solutionHide solution
(a) Sampling fraction =\dfrac{90}{1350}=\dfrac{1}{15}.
[1]
Juniors \dfrac{540}{15}=36, intermediates \dfrac{450}{15}=30.
[1]
Seniors \dfrac{225}{15}=15, masters \dfrac{135}{15}=9 (check: 36+30+15+9=90).
[1]
(b) For 50: juniors \dfrac{540\times 50}{1350}=20, intermediates 16.\overline{6}, seniors 8.\overline{3}, masters 5.
[1]
Two values are not integers, so round them to 17 and 8: 20+17+8+5=50.
[1]
(c) It guarantees that every group (including the small masters group) is represented in proportion, which a simple random sample might not do.
[1]

Final answer

(a) 36, 30, 15, 9 (b) 20, 16.7, 8.3, 5; for example 20, 17, 8, 5 (c) every group is represented in proportion.

Question 12

[7 marks]
The number of tickets, x, bought in a single transaction at a theatre box office is recorded for 55 transactions.
\begin{array}{|l|c|c|c|c|c|} \hline \text{Tickets} x & \text{1} & \text{2} & \text{3} & \text{4} & \text{5} \\ \hline \text{Frequency} & \text{12} & \text{18} & k & \text{10} & \text{5} \\ \hline \end{array}
The mean number of tickets per transaction is 2.6.
(a) Find the value of k.
[3]
(b) Hence find the median number of tickets.
[2]
(c) Another transaction for m tickets is recorded and the mean of all 56 transactions is 2.625. Find m.
[2]
Show solutionHide solution
(a) \sum f=45+k and \sum fx=12+36+3k+40+25=113+3k.
[1]
\dfrac{113+3k}{45+k}=2.6\ \Rightarrow\ 113+3k=117+2.6k
[1]
0.4k=4, so k=10.
[1]
(b) Total 55, so the median is the 28th value. Cumulative frequencies: 12,30,\ldots
[1]
The 28th value lies in x=2: median =2.
[1]
(c) New total =56(2.625)=147 and the old total was 55(2.6)=143.
[1]
m=147-143=4 tickets.
[1]

Final answer

(a) k=10 (b) 2 tickets (c) m=4.

Question 13

[6 marks]
Two machines fill sachets of instant coffee. A sachet should contain 50 g. An inspector weighs eight sachets from each machine; the masses, in grams, are:
Machine A: 49.2, 50.1, 50.3, 49.8, 50.6, 50.0, 49.7, 50.3
Machine B: 48.1, 51.9, 49.4, 50.8, 47.6, 52.3, 50.2, 49.7
(a) Show that both machines give a mean mass of 50.0 g.
[2]
(b) Use your GDC to find the standard deviation \sigma_x for each machine, to 3 significant figures.
[2]
(c) The inspector must recommend the machine that fills sachets more consistently. State which machine, and justify your choice.
[2]
Show solutionHide solution
(a) Machine A: the masses total 400.0 g, so the mean is \dfrac{400.0}{8}=50.0 g.
[1]
Machine B: the masses also total 400.0 g, so the mean is \dfrac{400.0}{8}=50.0 g. [1]AG
(b) One-Variable Statistics on each list, reading \sigma_x (not s_x):
[1]
Machine A: \sigma_x=0.406 g; machine B: \sigma_x=1.56 g.
[1]
(c) Machine A, because its standard deviation is much smaller (about 0.41 g against 1.56 g), so its sachets are closer to the mean of 50 g.
[2]

Final answer

(a) both totals 400.0 g (b) A 0.406 g, B 1.56 g (c) Machine A.

Question 14

[6 marks]
The ages of the 150 runners who completed a charity 10 km run are summarised below.
Age a (years)Number of runners
20\le a<3014
30\le a<4038
40\le a<5045
50\le a<6031
60\le a<7016
70\le a<806
(a) Use the mid-interval values to estimate the mean age and the standard deviation \sigma of the ages.
[3]
(b) Explain why the standard deviation you found is only an estimate.
[1]
(c) The race organisers want the next run to attract a wider range of ages. Say whether they would want the standard deviation to increase or decrease, and why.
[2]
Show solutionHide solution
(a) Midpoints 25,35,45,55,65,75 with frequencies as the frequency list; One-Variable Statistics.
[1]
\bar x\approx46.0 years.
[1]
\sigma_x\approx12.7 years.
[1]
(b) Each age is replaced by the midpoint of its class, so the exact ages (and so the true spread within each class) are unknown.
[1]
(c) They would want it to increase: a larger standard deviation means the ages are more spread out about the mean.
[2]

Final answer

(a) 46.0 years, 12.7 years (b) mid-values replace the ages (c) increase.

Question 15

[7 marks]
Imran, a technician, records the time in seconds that a robotic arm takes to complete each of 12 welds: 41.2, 42.5, 43.1, 43.8, 44.0, 44.6, 45.1, 45.5, 46.3, 47.0, 47.8, 61.5.
(a) Use your GDC to find the lower and upper quartiles.
[1]
(b) Use the 1.5\times\text{IQR} rule to show that 61.5 is the only outlier.
[3]
(c) Find the mean and the median (i) for all 12 times, (ii) after the outlier is removed.
[2]
(d) Imran finds that the 61.5 s weld was interrupted by a power cut. Explain whether the outlier should be removed, and which of the mean and the median changed more.
[1]
Show solutionHide solution
(a) Q_1=43.45 s and Q_3=46.65 s.
[1]
(b) IQR =46.65-43.45=3.20, so 1.5\times\text{IQR}=4.80.
[1]
Fences: 38.65 and 51.45.
[1]
61.5>51.45 is an outlier; the smallest value 41.2>38.65 and every other value is below the upper fence, so no other value is an outlier.
[1]
(c) (i) mean =46.0 s, median =44.9 s.
[1]
(ii) mean =44.6 s, median =44.6 s.
[1]
(d) The mean fell by about 1.41 s while the median fell by only 0.250 s, so the mean is much more affected. The weld was interrupted by a known cause, so the value is not representative of normal operation and may be removed (with the reason stated).
[1]

Final answer

  • (a) 43.45 s, 46.65 s (b) fences 38.65 and 51.45 (c) (i) 46.0 and 44.9
  • (ii) 44.6 and 44.6 (d) remove it; the mean changed more.

Question 16

[6 marks]
A bus company records the trip time, in minutes, of 120 trips on one route. The cumulative frequencies are:
Under (min)Cum. freq.
209
3031
4067
5098
60115
70120
No trip took less than 10 minutes.
(a) Find the number of trips that took between 30 and 40 minutes.
[1]
(b) Draw a cumulative frequency graph, using a scale of 1 cm to 10 minutes on the horizontal axis and 1 cm to 20 trips on the vertical axis.
[2]
(c) Use your graph to estimate the median and the interquartile range.
[3]
Show solutionHide solution
(a) 67-31=36 trips.
[1]
(b) Points (10,0),(20,9),(30,31),(40,67),(50,98),(60\,115),(70\,120) plotted against the upper boundaries and joined by a smooth increasing curve.
[2]
(c) Median: cumulative frequency 60 gives about 38.1\approx38 minutes.
[1]
Q_1 (at 30) \approx29.5 and Q_3 (at 90) \approx47.4 minutes.
[1]
IQR \approx17.9 minutes (accept \pm1.5).
[1]
The curve is steepest between 30 and 40 minutes, where the cumulative frequency rises by 36.

Final answer

(a) 36 (c) median about 38 min, IQR about 17.9 min.

Question 17

[6 marks]
Participants in a cooking class have ages with minimum 18, lower quartile 41, median 49, upper quartile 57 and maximum 96. The largest age that is not an outlier is 74.
(a) Use the 1.5\times\text{IQR} rule to decide whether 18 or 96 is an outlier.
[3]
(b) Draw a box plot of the ages, showing any outlier separately.
[2]
(c) Describe the distribution of the ages in the middle half.
[1]
Show solutionHide solution
(a) IQR =57-41=16, so 1.5\times16=24.
[1]
Fences: 41-24=17 and 57+24=81.
[1]
18>17 is not an outlier; 96>81 is an outlier.
[1]
(b) Whiskers to 18 and to 74; box from 41 to 57 with the median at 49; a cross at 96.
[2]
(c) The median 49 is near the middle of the box (41 to 57), so the middle half is roughly symmetric.
[1]

Final answer

(a) fences 17 and 81: 96 is an outlier, 18 is not (b) whiskers 18 and 74, cross at 96 (c) roughly symmetric.

Question 18

[6 marks]
Two cafes, Brew Bar and The Bean, record the time, in minutes, that customers wait for a coffee. The five-number summaries of 200 waits at each cafe are:
MinutesBrew BarThe Bean
Minimum1.20.8
Q_13.13.6
Median4.54.4
Q_36.05.0
Maximum11.47.9
(a) Find the range and the interquartile range for each cafe.
[2]
(b) Compare the waiting times at the two cafes, referring to both a measure of centre and a measure of spread.
[3]
(c) A customer hates waiting more than 6 minutes. Which cafe should she choose? Support your answer with figures from the table.
[1]
Show solutionHide solution
(a) Brew Bar: range =11.4-1.2=10.2; IQR =6.0-3.1=2.9.
[1]
The Bean: range =7.9-0.8=7.1; IQR =5.0-3.6=1.4.
[1]
(b) Centre: the medians are almost the same (4.5 against 4.4 min), so a typical wait is about the same.
[1]
Spread: The Bean has the smaller IQR (1.4<2.9) and range, so its waits are more consistent.
[1]
Brew Bar has some much longer waits (maximum 11.4 min), which the typical value hides.
[1]
(c) The Bean: Q_3=5.0<6, so at least 75% of its waits are under 5 minutes, while at Brew Bar Q_3=6.0 so a quarter of customers wait 6 minutes or more.
[1]

Final answer

(a) Brew Bar 10.2, 2.9; The Bean 7.1, 1.4 (b) similar medians; The Bean is more consistent (c) The Bean.

Question 19

[6 marks]
A shop sells gadgets priced in US dollars. For 30 gadgets the prices, U, have mean USD 26.20, standard deviation USD 8.40, median USD 24 and interquartile range USD 11. A Singapore customer pays S=1.35U+4.50 dollars, which includes a fixed USD 4.50 delivery charge.
(a) Find the mean price, in Singapore dollars.
[2]
(b) Find the standard deviation of the prices, in Singapore dollars.
[1]
(c) Find the median price, in Singapore dollars.
[1]
(d) Find the interquartile range of the prices, in Singapore dollars, to 3 significant figures.
[2]
Show solutionHide solution
(a) 1.35(26.20)+4.50
[1]
=\text{SGD }39.87.
[1]
(b) 1.35\times8.40=\text{SGD }11.34 (=\text{SGD }11.3 to 3 s.f.).
[1]
(c) 1.35(24)+4.50=\text{SGD }36.90.
[1]
(d) The delivery charge does not change the IQR: 1.35\times11
[1]
=\text{SGD }14.85\approx\text{SGD }14.9.
[1]

Final answer

(a) SGD 39.87 (b) SGD 11.34 (c) SGD 36.90 (d) SGD 14.9.

Question 20

[7 marks]
A city council wants to measure residents' satisfaction with its public parks using a questionnaire with 12 items, scored out of 60.
(a) One item reads: “Do you agree that our beautiful new parks are well maintained and safe?” Give two reasons why this item should be rewritten.
[2]
(b) The council gives the questionnaire to 25 residents, and gives it again to the same residents three weeks later. Name this test and state what it is testing.
[2]
(c) The scores are compared with the number of park visits each resident made in a month, recorded by a ticket app. State the type of validity that this assesses.
[1]
(d) The visit counts of 80 residents are grouped into the categories 0, 1–2, 3–5, 6–9 and 10 or more. Under a proposed model the expected frequencies are 4.4, 17.6, 30.4, 22.8 and 4.8. Combine categories so that every expected frequency is at least 5, and state the number of categories that remain.
[2]
Show solutionHide solution
(a) The word “beautiful” leads the answer (R1). It asks about two things, maintenance and safety, in one item (R1). (b) Test-retest (A1); it tests reliability, whether the questionnaire gives consistent results (R1). (c) Criterion-related validity (A1). (d) Combine 0 with 1–2 to give 4.4+17.6=22.0, and 6–9 with 10 or more to give 22.8+4.8=27.6 (M1). The expected frequencies are 22.0, 30.4 and 27.6, so there are 3 categories (A1).

Final answer

(a) leading word; two ideas in one item (b) test-retest; reliability (c) criterion-related validity (d) 22.0, 30.4, 27.6: 3 categories.

Mastery

Full exam questions that mix question types, at Paper 1 and Paper 2 standard.

Question 21

[8 marks]
Two fitness classes take the same endurance test. Class A has 20 people with mean score 52 and standard deviation 6. Class B has 30 people with mean score 47 and standard deviation 8.
(a) Find the mean score of all 50 people.
[2]
(b) Show that \sum x^2 for class A is 54 800, and find \sum x^2 for class B.
[2]
(c) Hence find the standard deviation of all 50 scores.
[3]
(d) Explain why the combined standard deviation is not simply the average of 6 and 8.
[1]
Show solutionHide solution
(a) \dfrac{20(52)+30(47)}{50}=\dfrac{1040+1410}{50}
[1]
=49.
[1]
(b) \sigma^2=\dfrac{\sum x^2}{n}-\bar x^2, so \sum x^2=n(\sigma^2+\bar x^2).
[1]
Class A: 20(36+2704)=54\,800.
AG
Class B: 30(64+2209)=68\,190.
[1]
(c) Combined \sum x^2=54\,800+68\,190=122\,990.
[1]
\sigma^2=\dfrac{122\,990}{50}-49^2 [1]=2459.8-2401=58.8, so \sigma=7.67.
[1]
(d) The classes have different means (52 and 47); the spread of the class means adds to the variation, so the combined variance (58.8) is larger than the weighted average of the two variances (0.4\times36+0.6\times64=52.8).
[1]

Final answer

(a) 49 (b) 68 190 (c) 7.67 (d) the class means differ.

Question 22

[10 marks]
A researcher measures the reaction times (in milliseconds) of 100 participants. The grouped data are shown, but one frequency is missing.
Reaction time r (ms)Frequency
180\le r<2008
200\le r<22022
220\le r<240k
240\le r<26025
260\le r<28011
(a) Show that k=34.
[2]
(b) Using mid-interval values, estimate the mean reaction time.
[3]
(c) Estimate the standard deviation of the reaction times using your GDC.
[2]
(d) Write down the modal class, and state which class contains the median.
[3]
Show solutionHide solution
(a) Total is 100: 8+22+k+25+11=100\Rightarrow66+k=100\Rightarrow k=34.
[2]
(b) Mid-values 190 210 230 250 270. \sum fx=190(8)+210(22)+230(34)+250(25)+270(11)=1520+4620+7820+6250+2970=23\,180. \bar r=23\,180/100=231.8 ms.
[3]
(c) From GDC (mid-values with frequencies): \sigma\approx22.1 ms.
[2]
(d) Modal class =220\le r<240 (highest frequency 34). Cumulative: 8,30,64,\dots The 50th and 51st values fall in 220\le r<240, so the median lies in that class.
[3]

Final answer

(a) k=34 (b) 231.8 ms (c) 22.1 ms (d) modal class and median class are both 220\le r<240.

Question 23

[11 marks]
A cycling club records the finishing times (in minutes) of 80 riders in a race. The times are grouped below.
Time t (min)Frequency
40\le t<456
45\le t<5014
50\le t<5526
55\le t<6020
60\le t<6510
65\le t<704
(a) Copy and complete the cumulative frequency values for the upper boundary of each class (45,50,55,60,65,70).
[2]
(b) On graph paper you draw a cumulative frequency curve. Use it (or the table) to estimate:
(i) the median time; (ii) the 90th percentile.
[4]
(c) Riders finishing in under 48 minutes qualify for the regional final. Estimate how many riders qualify.
[3]
(d) Using mid-interval values, estimate the mean finishing time.
[2]
Show solutionHide solution
(a) Cumulative frequencies at 45,50,55,60,65,70: 6,20,46,66,76,80.
[2]
(b) (i) Median at CF =40: lies in 50\le t<55. Interpolating: 50+\dfrac{40-20}{26}\times5=50+3.85=53.8 min. (ii) 90% of 80=72; CF =72 lies in 60\le t<65: 60+\dfrac{72-66}{10}\times5=60+3=63.0 min.
[4]
(c) At 48 min, within 45\le t<50: \text{CF}=6+\dfrac{48-45}{5}\times14=6+8.4=14.4\approx14 riders.
[3]
(d) Mid-values 42.5,47.5,52.5,57.5,62.5,67.5. \sum fx=42.5(6)+47.5(14)+52.5(26)+57.5(20)+62.5(10)+67.5(4)=255+665+1365+1150+625+270=4330. \bar t=4330/80=54.1 min.
[2]

Final answer

(a) 6, 20, 46, 66, 76, 80 (b) 53.8 min; 63.0 min (c) 14 riders (d) 54.1 min.

Question 24

[12 marks]
A phone manufacturer tests the battery life (in hours) of two models, A and B. For a sample of 50 phones of each model the following summary statistics are found.
HoursModel AModel B
Minimum1416
Q_11920
Median2221
Q_32523
Maximum3428
(a) For Model A, find the interquartile range.
[1]
(b) Using the 1.5\times\text{IQR} rule for Model A, determine whether a battery life of 34 hours would be classified as an outlier.
[3]
(c) Draw a box-and-whisker diagram for Model B on a suitable axis.
[3]
(d) A reviewer claims "Model A is the better choice because it can last the longest." Give one statistical reason, using the summary values, why Model B might instead be preferred for reliability.
[2]
(e) For Model A, calculate the range and compare it with the IQR to comment on the spread of the data.
[3]
Show solutionHide solution
(a) \text{IQR}_A=25-19=6 h.
[1]
(b) Upper boundary =Q_3+1.5\times\text{IQR}=25+1.5(6)=25+9=34. Since 34 is not strictly greater than 34, it is on the boundary and is not classified as an outlier.
[3]
(c) Box from Q_1=20 to Q_3=23, median line at 21, whiskers to min 16 and max 28:
[3]
(d) Model B has a smaller IQR (23-20=3 h vs 6 h) and smaller range, so its battery life is more consistent/predictable, which is preferable for reliability.
[2]
(e) Range_A=34-14=20 h, while IQR_A=6 h. The range is much larger than the IQR, indicating that the extreme values (especially the high maximum) stretch the data, but the middle 50% of batteries are quite tightly grouped.
[3]

Final answer

(a) 6 h (b) fence 34 h: on the fence, so not an outlier (c) box 20 to 23, median 21, whiskers 16 and 28 (d) B's IQR 3 h is less than 6 h (e) range 20 h against IQR 6 h.

Question 25

[12 marks]
An environmental group measures the daily water usage (in litres) of 15 households:
210, 185, 240, 305, 198, 225, 260, 215, 190, 250, 235, 470, 205, 220, 245.
(a) (i) Find the median. (ii) Find Q_1 and Q_3, and hence the interquartile range.
[4]
(b) Using the 1.5\times\text{IQR} rule, identify any outliers.
[3]
(c) The group decides to remove any outliers before reporting. Explain the effect that removing the outlier(s) will have on (i) the mean and (ii) the median, compared with the original data.
[3]
(d) Draw a box-and-whisker diagram of the original 15 values, showing any outlier as a separate point.
[2]
Show solutionHide solution
Ordered: 185 190 198 205 210 215 220 225 235 240 245 250 260 305 470.
(a) (i) Median (8th value) =225 L. (ii) Lower 7: 185 190 198 205 210 215 220, Q_1=205. Upper 7: 235 240 245 250 260 305 470, Q_3=250. \text{IQR}=250-205=45 L.
[4]
(b) Upper boundary =250+1.5(45)=250+67.5=317.5; lower boundary =205-67.5=137.5. Since 470>317.5 it is an outlier; all other values lie within the boundaries. So 470 L is the only outlier.
[3]
(c) (i) The mean will decrease noticeably, because the very large value 470 was pulling the mean up. (ii) The median will change only slightly (it is resistant to outliers), moving down by a small amount as one high value is removed.
[3]
(d) Box Q_1=205 to Q_3=250, median 225; lower whisker to 185; upper whisker to the largest non-outlier 305; outlier 470 marked separately:
[2]

Final answer

(a) 225 L; 205 L, 250 L, IQR 45 L (b) 470 L only (c) the mean falls a lot, the median a little (d) whiskers 185 and 305, cross at 470.

Question 26

[18 marks]
A bus company in Wales has 2400 regular passengers on one route: 1080 commuters, 840 students and 480 retired passengers.
(a) The company takes a stratified sample of 120 regular passengers. Find the number of students in the sample.
[2]
The trip times, t minutes, of the 120 passengers in the sample are summarised below.
Time t (min)Passengers
10\le t<209
20\le t<3027
30\le t<4041
40\le t<5028
50\le t<6011
60\le t<704
(b) Use mid-interval values to estimate the mean and the standard deviation of the trip times.
[3]
(c) Write down the cumulative frequencies for the upper class boundaries.
[1]
A cumulative frequency graph drawn from these values gives a lower quartile of 28 minutes, a median of 36 minutes and an upper quartile of 45 minutes. The shortest trip in the sample took 12 minutes and the longest took 67 minutes.
(d) Determine whether the longest trip is an outlier.
[3]
(e) Draw a box plot of the trip times.
[2]
The fare, in GBP, for a trip of t minutes is 1.80+0.12t.
(f) Estimate the mean and the standard deviation of the fares paid by the passengers in the sample.
[3]
(g) Under a model for the trip times, the expected frequencies of the six classes are 6.0, 24.6, 40.8, 31.2, 13.2 and 4.2. Explain why classes must be combined before a chi-squared test, and write down the expected frequencies after combining.
[2]
(h) The sample was taken only on weekday mornings. Explain how this may affect the estimate of the mean trip time for all regular passengers.
[2]
Show solutionHide solution
(a) \frac{840}{2400}\times120 (M1) =42 (A1). (b) Mid-values 15,25,\ldots,65 (M1); mean \frac{4370}{120}=36.4 min (A1); \sigma=11.9 min (A1). (c) 9,36,77\,105\,116\,120 (A1). (d) \mathrm{IQR}=17 (M1); upper fence 45+25.5=70.5 (A1); 67<70.5, so it is not an outlier (R1). (e) Box from 28 to 45 with median 36; whiskers to 12 and 67 (A1 A1); the lower fence is 2.5, below 12, so there is no low outlier. (f) Mean fare 1.80+0.12(36.4167)=\text{GBP }6.17 (M1 A1); standard deviation 0.12\times11.853=\text{GBP }1.42 (A1). (g) 4.2<5 (R1); combine the last two classes: 6.0,24.6,40.8,31.2,17.4 (A1). (h) Weekday-morning trips may be busier and slower than trips at other times (R1), so the sample may not represent all trips and the mean may be biased (R1).

Final answer

(a) 42 (b) 36.4 min, 11.9 min (c) 9, 36, 77, 105, 116, 120 (d) fence 70.5: 67 is not an outlier (e) box 28 to 45, median 36, whiskers 12 and 67 (f) GBP 6.17, GBP 1.42 (g) combine the last two classes: 6.0, 24.6, 40.8, 31.2, 17.4 (h) weekday mornings may not represent all trips.

Going further

Beyond the syllabus: the worked examples and investigations of the chapter's Going further pages.

G1 A grouped mean sits between two bounds

G1 Worked

[4 marks]
Lucía records the time, t minutes, that 60 customers spend in her bookshop in Seville. The times are grouped as follows: 0\le t<10: 7 customers; 10\le t<20: 15 customers; 20\le t<30: 19 customers; 30\le t<40: 12 customers; 40\le t<50: 7 customers.
(a) Find the mean L obtained by putting every time at the lower end of its class, and the mean U obtained by putting every time at the upper end.
[2]
(b) Show that the mid-interval estimate of the mean, 24.5 minutes, equals \frac{L+U}{2}.
[1]
(c) Explain why the true mean must lie between L and U.
[1]
Show solutionHide solution
(a) L=\frac{0(7)+10(15)+20(19)+30(12)+40(7)}{60} (M1) =\frac{1170}{60}=19.5 minutes, and each class is 10 wide, so U=29.5 minutes (A1). (b) \frac{19.5+29.5}{2}=24.5 minutes (A1). (c) Every time is at least the lower end of its class and less than the upper end, so the mean lies between L and U (R1).

Final answer

L=19.5 min, U=29.5 min; \frac{L+U}{2}=24.5 min.

G2 The variance identity and combining groups

G2 Investigation

[20 marks]
Mateo records the times, in minutes, of 24 deliveries: 12, 15, 17, 18, 21, 22, 23, 24, 25, 26, 27, 27, 28, 29, 30, 31, 32, 33, 35, 36, 38, 41, 44, 49.
(a) Find the mean and the standard deviation of the 24 times.
[2]
(b) Group the times in classes of width 10 starting at 10 (10\le t<20, and so on). Write down the frequencies and use mid-interval values to estimate the mean and standard deviation.
[2]
(c) Repeat (b) with classes of width 5 starting at 10.
[2]
(d) Find the percentage error of each estimate of the mean. Comment on whether the narrower classes gave a better estimate.
[2]
(e) For the width-10 table, find the mean L obtained by putting every time at the lower end of its class, and the mean U obtained by putting every time at the upper end. Show that the estimate in (b) equals \frac{L+U}{2}.
[3]
(f) Show that \sum(x-\bar{x})^2=\sum x^2-n\bar{x}^2, and hence verify your standard deviation in (a) using \sum x=683 and \sum x^2=21\,353.
[3]
(g) A second courier, Ines, made 16 deliveries with mean 31.5 minutes and standard deviation 6.2 minutes. Find the mean and the standard deviation of all 40 deliveries.
[4]
(h) Mateo wants to use his 24 times in a chi-squared goodness-of-fit test. Suggest, with a reason, whether classes of width 5 or width 10 are more suitable.
[2]
Show solutionHide solution
(a) \bar{x}=\frac{683}{24}=28.5 minutes (28.458…) (A1) and \sigma=8.93 minutes from the GDC (A1). (b) Frequencies 4, 10, 7, 3 (A1); mid-values 15, 25, 35, 45 give \bar{x}=28.75\approx28.8 minutes and \sigma=9.04 minutes (A1). (c) Frequencies 1, 3, 4, 6, 4, 3, 2, 1 (A1); mean 29.0 minutes (28.958…) and \sigma=8.72 minutes (A1). (d) Errors \frac{28.75-28.458}{28.458}=1.02\% and \frac{28.958-28.458}{28.458}=1.76\% (A1); the narrower classes gave a worse mean here, because the times sit unevenly inside the classes, so narrower classes do not guarantee a better estimate (R1). (e) L=\frac{10(4)+20(10)+30(7)+40(3)}{24}=23.75 (M1) and U=33.75 (A1); \frac{L+U}{2}=28.75, the estimate in (b) (R1). (f) \sum(x-\bar{x})^2=\sum x^2-2\bar{x}\sum x+n\bar{x}^2 (M1), and \sum x=n\bar{x} gives \sum x^2-n\bar{x}^2 (A1). Then \sigma^2=\frac{21353}{24}-28.4583^2=79.83 and \sigma=8.93 minutes (A1). (g) Combined mean \frac{683+16(31.5)}{40}=\frac{1187}{40}=29.7 minutes (M1 A1). Ines has \sum x^2=16(6.2^2+31.5^2)=16\,491.04 (M1); \sigma^2=\frac{21353+16491.04}{40}-29.675^2=65.50, so \sigma=8.09 minutes (A1). (h) Width 10 is more suitable (R1): with 24 values, eight classes of width 5 give expected frequencies of about 3, below 5, so classes would have to be combined anyway (R1).

Final answer

  • (a) 28.5 min, 8.93 min
  • (b) 28.8 min, 9.04 min
  • (c) 29.0 min, 8.72 min
  • (d) 1.02% and 1.76%
  • (e) L=23.75, U=33.75
  • (f) 79.83
  • (g) 29.7 min, 8.09 min
  • (h) width 10.

From the book

Photon Guidebook: Mathematics Applications and Interpretation HL for the IB Diploma

Chapter 13 teaches describing data from the basics to full exam questions: how the IB asks it, every question type with a marked example, and these practice questions in three steps.

About the book

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