Free worked solutions · 2026 Edition

IB Maths AA SL Chapter 7: Statistics worked solutions

Worked solutions to the practice questions in Chapter 7 of The Ultimate Guide for the IB: Mathematics: Analysis and Approaches SL (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

[4 marks]

An online bookshop has 5000 customers, numbered 0001 to 5000. The manager wants a simple random sample of 25 customers to answer a survey.

(a) Explain what is meant by a simple random sample of 25 customers.

[1]

(b) Describe how the manager could choose the sample using a random number generator.

[2]

(c) The manager instead chooses the 25 customers who most recently placed an order. Explain why this is not a simple random sample.

[1]

Show solutionHide solution

(a) every possible group of 25 customers has the same chance of being chosen (so each customer is equally likely to be chosen)

(b) generate random whole numbers from 1 to 5000

select the customers with those numbers, ignoring any repeated number, until 25 different customers are chosen

(c) customers who did not order recently have no chance of being chosen (the choice depends on when they ordered, not on chance)

Final answer

(a) every set of 25 customers is equally likely to be chosen
(b) 25 distinct random labels from 0001–5000
(c) customers who have not ordered recently cannot be chosen.

Question 2

[5 marks]

An outlier is a data value that is more than 1.5 \times \text{IQR} below the lower quartile or above the upper quartile.

For a set of data, Q_1 = 24 and Q_3 = 36.

(a) Find the interquartile range.

[1]

(b) Find the two values that separate the outliers from the other data.

[2]

(c) The data include the values 5, 54 and 58. Determine which of these values are outliers.

[2]

Show solutionHide solution

(a) \text{IQR} = 36 - 24 = 12

(b) 1.5 \times 12 = 18; \ 24 - 18 and 36 + 18

6 and 54

(c) 5 < 6 and 58 > 54, so 5 and 58 are outliers

54 is exactly 1.5 \times \text{IQR} above Q_3, not more, so 54 is not an outlier

Final answer

(a) 12
(b) 6, 54
(c) 5 and 58 only.

Question 3

[6 marks]

A teacher counts the number of spelling mistakes, x, on each page of a 30-page essay draft.

\begin{array}{c|c}\text{Mistakes} x & \text{Pages} f\\0 & 9\\1 & 10\\2 & 6\\3 & 3\\4 & 2\end{array}

(a) Write down the mode.

[1]

(b) By showing your working, find the mean number of mistakes per page.

[3]

(c) Find the median number of mistakes per page.

[2]

Show solutionHide solution

(a) 1 (highest frequency, 10)

(b) \sum f_i x_i=0(9)+1(10)+2(6)+3(3)+4(2)

=39

\bar{x}=\dfrac{39}{30}=1.3

(c) n=30, so the median is halfway between the 15th and 16th values; cumulative frequencies 9,\ 19,\ \ldots (M1)

the 10th to 19th values are all 1, so median =1

Final answer

(a) 1
(b) 1.3
(c) 1.

Question 4

[6 marks]

An airline weighs the 80 suitcases checked in at one desk. The masses, m kg, are shown in the table.

\begin{array}{c|c}\text{Mass} m \text{(kg)} & \text{Suitcases}\\10\le m<14 & 6\\14\le m<18 & 17\\18\le m<22 & 31\\22\le m<26 & 19\\26\le m<30 & 7\end{array}

(a) Write down the modal class.

[1]

(b) Find an estimate of the mean mass.

[2]

(c) Find an estimate of the standard deviation of the masses.

[2]

Suitcases of mass 26 kg or more are charged a fee.

(d) Write down the percentage of these suitcases that are charged the fee.

[1]

Show solutionHide solution

(a) 18\le m<22

(b) mid-interval values 12,\ 16,\ 20,\ 24,\ 28 with frequencies 6,\ 17,\ 31,\ 19,\ 7 (M1)

estimate of mean =\dfrac{1616}{80}=20.2 kg

(c) using the same mid-interval values and frequencies (population standard deviation) (M1)

\sigma=4.19 kg (4.190465\ldots)

(d) \dfrac{7}{80}\times100=8.75\%

Final answer

(a) 18\le m<22
(b) 20.2 kg
(c) 4.19 kg
(d) 8.75\%.

Question 5

[6 marks]

The raw marks in a mathematics competition have mean 20 and standard deviation 4. Each raw mark x is converted to a scaled mark y=ax+b, where a>0. The scaled marks have mean 47 and standard deviation 10.

(a) Determine a.

[2]

(b) Determine b.

[2]

(c) A student has a raw mark of 26. Find the student's scaled mark.

[1]

(d) Write down the variance of the scaled marks.

[1]

Show solutionHide solution

(a) adding b does not change the standard deviation; multiplying by a multiplies it by a, so 4a=10

a=2.5

(b) the mean transforms in the same way: 2.5(20)+b=47

b=-3

(c) 2.5(26)-3=62

(d) 10^2=100

Final answer

(a) 2.5
(b) -3
(c) 62
(d) 100.

Question 6

[5 marks]

Pearson's product-moment correlation coefficient, r, measures the strength of linear correlation, and -1 \le r \le 1. Three studies give these results.

Study 1: hours of daylight in a day and electricity used for lighting in an office, r=-0.87.

Study 2: shoe size and score in a memory test, r=0.06.

Study 3: price of a paperback book and its number of pages, r=0.42.

(a) Describe the correlation in each study.

[3]

(b) State which study shows the strongest linear relationship. Give a reason.

[1]

(c) Describe how the points of a scatter diagram would lie if r=1.

[1]

Show solutionHide solution

(a) Study 1: strong negative correlation

Study 2: no (or almost no) linear correlation

Study 3: weak positive correlation

Note: for Study 3 also accept “moderate positive”.

(b) Study 1, because |-0.87|=0.87 is the closest of the three values to 1

(c) all the points lie exactly on a straight line with positive gradient

Note: in (b) the answer “Study 3 because 0.42>-0.87” earns A0; in (c) both “exactly on a straight line” and “positive gradient” are needed.

Final answer

(a) strong negative; almost none; weak/moderate positive
(b) Study 1
(c) straight line, positive gradient.

Question 7

[5 marks]

A scientist studies how rainfall affects the growth of a pine tree. For 12 years she records the annual rainfall, R mm, and the width, W mm, of the growth ring the tree formed that year. In these years the rainfall was between 380 mm and 700 mm and the ring widths were between 1.8 mm and 3.1 mm. She finds that r = 0.910, the regression line of W on R is W = 0.004R + 0.3 and the regression line of R on W is R = 207W + 27.3.

An old wooden beam from the same forest shows a growth ring of width 2.6 mm.

(a) Use the appropriate line to estimate the rainfall in the year that this ring formed.

[2]

(b) A student instead rearranges the line of W on R to make R the subject and then substitutes W = 2.6. Find the student's estimate.

[2]

(c) Explain why the estimate in part (a) is the better one.

[1]

Show solutionHide solution

(a) R = 207(2.6) + 27.3

R = 565.5 mm (accept 566 mm)

(b) 2.6 = 0.004R + 0.3 \Rightarrow R = \dfrac{2.6 - 0.3}{0.004}

R = 575 mm

(c) the rainfall R is being estimated from a known width W, and the line of R on W is the line found for predicting R; the line of W on R is for predicting W and, since r \ne \pm 1, the two lines are different

Final answer

(a) 566 mm
(b) 575 mm
(c) use R on W.

Tutorial

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

Question 8

[5 marks]

A hospital employs 240 nurses, 90 doctors and 60 administrators. A stratified sample of 65 of the staff is to be taken.

(a) Find the number of nurses, doctors and administrators in the sample.

[3]

(b) Explain why a stratified sample is better here than a simple random sample of the 390 staff.

[2]

Show solutionHide solution

(a) Total staff =240+90+60=390, so the sampling fraction is \dfrac{65}{390}=\dfrac16.

[1]

Nurses: \dfrac16\times240=40.

[1]

Doctors: \dfrac16\times90=15; administrators: \dfrac16\times60=10 (check: 40+15+10=65).

[1]

(b) A simple random sample could, by chance, contain no (or very few) doctors or administrators.

[1]

A stratified sample keeps every group in the same proportion as in the whole staff, so each group is fairly represented.

[1]

Final answer

(a) 40, 15, 10
(b) preserves the group proportions.

Question 9

[6 marks]

A machine makes 2000 phone cases per day. The cases are numbered 1 to 2000 in order of manufacture. An inspector uses systematic sampling to choose 80 cases, starting with case number 7.

(a) Write down the gap between consecutive chosen cases.

[1]

(b) Write down the numbers of the first three cases chosen.

[1]

(c) Find the number of the last case chosen.

[2]

A worn mould produces a flaw in every 25th case (cases 25,\,50,\,75,\dots).

(d) Explain why the inspector's sample would give a misleading picture of the number of flawed cases.

[2]

Show solutionHide solution

(a) \dfrac{2000}{80}=25.

[1]

(b) 7,\ 32,\ 57.

[1]

(c) The 80th case chosen is 7+79\times25.

[1]

=7+1975=1982.

[1]

(d) The chosen cases are 7,32,57,\dots, all of which are 7 more than a multiple of 25, so none is a multiple of 25.

[1]

So the sample would contain no flawed cases at all and would suggest the machine is perfect, because the sampling gap matches the pattern of the flaw.

[1]

Final answer

(a) 25
(b) 7,32,57
(c) 1982
(d) no flawed cases selected.

Question 10

[5 marks]

Seven numbers 6,\ 8,\ 9,\ 11,\ 13,\ 17 and k have a mean of 12.

(a) Find the value of k.

[2]

An eighth number, m, is added to the seven numbers. The mean of the eight numbers is 12.5.

(b) Find the value of m.

[2]

(c) Find the median of the eight numbers.

[1]

Show solutionHide solution

(a) \dfrac{6+8+9+11+13+17+k}{7}=12, so 64+k=84.

[1]

k=20.

[1]

(b) The new total is 8\times12.5=100, and the old total was 84.

[1]

m=100-84=16.

[1]

(c) In order: 6,\ 8,\ 9,\ 11,\ 13,\ 16,\ 17,\ 20. The median is the mean of the 4th and 5th values: \dfrac{11+13}{2}=12.

[1]

Final answer

(a) 20
(b) 16
(c) 12.

Question 11

[5 marks]

The number of coffees, x, bought in one day by each of a group of office workers is shown in the table.

\begin{array}{l|ccccc}x & \mathrm{1} & \mathrm{2} & \mathrm{3} & \mathrm{4} & \mathrm{5}\\f & \mathrm{5} & k & \mathrm{8} & \mathrm{4} & \mathrm{3}\end{array}

The mean number of coffees is 2.8.

(a) Write down expressions, in terms of k, for \sum f and for \sum fx.

[2]

(b) Hence find the value of k.

[3]

Show solutionHide solution

(a) \sum f=5+k+8+4+3=20+k.

[1]

\sum fx=5(1)+2k+8(3)+4(4)+3(5)=5+2k+24+16+15=60+2k.

[1]

(b) \dfrac{60+2k}{20+k}=2.8.

[1]

60+2k=56+2.8k.

[1]

4=0.8k, so k=5.

[1]

(Check: \sum fx=70, \sum f=25, and \dfrac{70}{25}=2.8.)

Final answer

(a) 20+k, 60+2k
(b) 5.

Question 12

[6 marks]

Two data sets, A and B, each have eight values.

\begin{array}{l|cccccccc}A & \mathrm{12} & \mathrm{14} & \mathrm{15} & \mathrm{16} & \mathrm{17} & \mathrm{18} & \mathrm{19} & \mathrm{21}\\B & \mathrm{12} & \mathrm{14} & \mathrm{15} & \mathrm{16} & \mathrm{17} & \mathrm{18} & \mathrm{19} & \mathrm{39}\end{array}

(a) Find the median and the interquartile range of set A.

[3]

(b) Write down the median and the interquartile range of set B.

[1]

(c) Find the range of each set.

[1]

(d) State, with a reason, which measure of spread, the range or the interquartile range, gives a better description of the spread of the central values of set B.

[1]

Show solutionHide solution

(a) Median =\dfrac{16+17}{2}=16.5.

[1]

Lower half 12,14,15,16: Q_1=\dfrac{14+15}{2}=14.5; upper half 17,18,19,21: Q_3=\dfrac{18+19}{2}=18.5.

[1]

\text{IQR}=18.5-14.5=4.

[1]

(b) The same: median 16.5 and \text{IQR}=4, because only the largest value has changed and it does not affect the middle values.

[1]

(c) Range of A=21-12=9; range of B=39-12=27.

[1]

(d) The IQR: the range of B is tripled by the one extreme value 39, whereas the IQR is unchanged.

[1]

Final answer

(a),(b) median 16.5, IQR 4
(c) 9,27
(d) IQR.

Question 13

[6 marks]

Eight values in order are

3,\ 6,\ 9,\ p,\ 15,\ q,\ 21,\ 25.

The median of the data is 13 and the interquartile range is 13.

(a) Determine p.

[2]

(b) Determine q.

[3]

(c) Find the mean of the eight values.

[1]

Show solutionHide solution

(a) The median is the mean of the 4th and 5th values: \dfrac{p+15}{2}=13.

[1]

p=11.

[1]

(b) Lower half 3,6,9,11: Q_1=\dfrac{6+9}{2}=7.5.

[1]

Upper half 15,q,21,25: Q_3=\dfrac{q+21}{2}, and Q_3-Q_1=13, so \dfrac{q+21}{2}=20.5.

[1]

q=20.

[1]

(c) \text{Sum}=3+6+9+11+15+20+21+25=110, so the mean is \dfrac{110}{8}=13.75.

[1]

Final answer

(a) 11
(b) 20
(c) 55/4.

Question 14

[5 marks]

The times, t minutes, taken to deliver 60 orders by a restaurant are summarised below.

\begin{array}{c|c}\text{Time} t \text{(min)} & \text{Frequency}\\10<t\le20 & \mathrm{6}\\20<t\le30 & \mathrm{18}\\30<t\le40 & \mathrm{22}\\40<t\le50 & \mathrm{10}\\50<t\le60 & \mathrm{4}\end{array}

(a) Use your GDC to find an estimate of the mean and of the standard deviation of the delivery times.

[3]

A second restaurant has an estimated mean delivery time of 31 minutes with a standard deviation of 6.2 minutes.

(b) Compare the delivery times of the two restaurants.

[2]

Show solutionHide solution

(a) Enter the mid-interval values 15,\ 25,\ 35,\ 45,\ 55 and the frequencies, and run One-Variable Statistics.

[1]

\bar t\approx33.0 minutes.

[1]

\sigma\approx10.5 minutes (10.456\ldots).

[1]

(b) The second restaurant is slightly quicker on average (31<33.0 minutes).

[1]

It is also more consistent: its standard deviation (6.2) is much smaller than 10.5, so its delivery times are less spread out.

[1]

Final answer

(a) 33.0, 10.5 min
(b) second restaurant quicker and more consistent.

Question 15

[5 marks]

The scores of 30 students in a physics quiz have a mean of 48 and a standard deviation of 8. The teacher adjusts every score x to a new score y using y=1.5x-6.

(a) Find the mean of the adjusted scores.

[2]

(b) Find the standard deviation of the adjusted scores.

[2]

(c) Find the variance of the adjusted scores.

[1]

Show solutionHide solution

(a) The mean is transformed in the same way as each value: \bar y=1.5(48)-6.

[1]

\bar y=72-6=66.

[1]

(b) Subtracting 6 does not change the spread; multiplying by 1.5 multiplies the standard deviation by 1.5.

[1]

\sigma_y=1.5\times8=12.

[1]

(c) Variance =12^2=144.

[1]

Final answer

(a) 66
(b) 12
(c) 144.

Question 16

[6 marks]

A candle is lit and its height, h cm, is measured t hours later.

\begin{array}{l|cccccc}t & \mathrm{1} & \mathrm{2} & \mathrm{3} & \mathrm{4} & \mathrm{5} & \mathrm{6}\\h & \mathrm{27.2} & \mathrm{24.1} & \mathrm{21.3} & \mathrm{18.4} & \mathrm{15.2} & \mathrm{12.5}\end{array}

(a) Use your GDC to find the equation of the regression line of h on t, and write down the value of r.

[3]

(b) Use your equation to estimate the height of the candle after 3.5 hours, and comment on the reliability of the estimate.

[2]

(c) Explain why the equation should not be used to predict the height after 12 hours.

[1]

Show solutionHide solution

(a) Linear Regression on the GDC gives h=-2.9457t+30.093\ldots, that is h\approx-2.95t+30.1.

[2]

r=-0.99981\ldots\approx-1.00 (a very strong negative linear correlation).

[1]

(b) h=-2.9457(3.5)+30.093=19.78\ldots\approx19.8 cm.

[1]

t=3.5 lies within the data range (1 to 6) and |r| is very close to 1, so this interpolation is reliable.

[1]

(c) t=12 is far outside the data range; the line would give h\approx-5.3 cm, which is impossible (a candle cannot have a negative height), so the estimate would be unreliable.

[1]

Final answer

(a) h=-2.95t+30.1, r=-0.9998 (-1.00 to 3 s.f.)
(b) 19.8 cm, reliable
(c) extrapolation.

Mastery

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

Question 17

[5 marks]

The waiting times, in seconds, of the callers to a helpline have lower quartile 18 and upper quartile 30. The fences are Q_1-1.5\times\text{IQR} and Q_3+1.5\times\text{IQR}. Flag a value strictly below the lower fence or strictly above the upper fence.

(a) Show that a waiting time of 52 seconds is an outlier.

[3]

(b) State whether a waiting time of 0 seconds is an outlier.

[1]

(c) Write down the largest whole-number waiting time that is not an outlier.

[1]

Show solutionHide solution

(a) \text{IQR}=30-18=12.

[1]

Upper fence =Q_3+1.5\times\text{IQR}=30+1.5(12)=48.

[1]

Since 52>48, a waiting time of 52 seconds is an outlier.

[1]

(b) Lower fence =18-1.5(12)=0. A value of 0 is not less than 0, so it is not an outlier.

[1]

(c) 48 (a value must be greater than 48 to be an outlier).

[1]

Final answer

(a) upper fence 48, 52>48
(b) no
(c) 48.

Question 18

[6 marks]

A farmer records the annual rainfall, R cm, and the crop yield, Y tonnes, for a number of years. The regression line of Y on R is

Y=0.7R+7,

and the regression line of R on Y is

R=1.2Y-2.

Both lines pass through the mean point.

(a) Use your GDC to find the mean annual rainfall and the mean yield.

[3]

(b) Estimate the yield in a year when the rainfall is 45 cm.

[1]

(c) Estimate the rainfall needed for a yield of 40 tonnes, stating which line you use and why.

[2]

Show solutionHide solution

(a) Both lines pass through (\bar R,\bar Y), so \bar Y=0.7\bar R+7 and \bar R=1.2\bar Y-2.

[1]

Solving the simultaneous equations on the GDC (or by substitution: \bar R=1.2(0.7\bar R+7)-2=0.84\bar R+6.4, so 0.16\bar R=6.4):

[1]

\bar R=40 and \bar Y=0.7(40)+7=35.

[1]

(b) Use the Y-on-R line: Y=0.7(45)+7=38.5 tonnes.

[1]

(c) To estimate R from a given Y, use the line of R on Y.

[1]

R=1.2(40)-2=46.

[1]

Final answer

(a) 40.0, 35.0
(b) 38.5
(c) 46.0, R on Y.

Question 19

[5 marks]

The temperature of a lake, T^\circC, is measured at depth d metres.

\begin{array}{l|ccccccc}d & \mathrm{2} & \mathrm{5} & \mathrm{8} & \mathrm{12} & \mathrm{15} & \mathrm{20} & \mathrm{25}\\T & \mathrm{21.6} & \mathrm{20.1} & \mathrm{18.5} & \mathrm{16.3} & \mathrm{15.0} & \mathrm{12.2} & \mathrm{10.4}\end{array}

(a) Fit a line predicting temperature from depth. Write it as T=ad+b.

[2]

(b) Write down the value of r.

[1]

(c) Estimate the temperature at a depth of 10 m.

[1]

(d) A student uses the equation to predict the temperature at a depth of 40 m. Comment on the reliability of this prediction.

[1]

Show solutionHide solution

(a) Linear Regression on the GDC: T=-0.497d+22.5, from a=-0.49739\ldots and b=22.4819\ldots.

[2]

(b) r=-0.99848\ldots\approx-0.998.

[1]

(c) T=-0.49739(10)+22.4819=17.5\ldots\approx17.5^\circC.

[1]

(d) 40 m is well outside the data range (2 to 25 m), so this is extrapolation and is unreliable (the line gives \approx2.6^\circC, but the relationship may not stay linear).

[1]

Final answer

(a) T=-0.497d+22.5
(b) -0.998
(c) 17.5\,{}^\circC
(d) extrapolation.

Question 20

[6 marks]

The points scored by a basketball player in ten matches, in order, are

2\quad 11\quad 13\quad 14\quad 16\quad 17\quad 18\quad 20\quad 21\quad 23.

(a) Find the median and the interquartile range.

[3]

(b) Show that the score of 2 is an outlier. (Test against the fences Q_1-1.5\times\text{IQR} and Q_3+1.5\times\text{IQR}, using strict comparisons.)

[2]

(c) Find the mean of the nine scores that remain when the score of 2 is left out.

[1]

Show solutionHide solution

(a) n=10: the median is the mean of the 5th and 6th values: \dfrac{16+17}{2}=16.5.

[1]

Lower half 2,11,13,14,16: Q_1=13; upper half 17,18,20,21,23: Q_3=20.

[1]

\text{IQR}=20-13=7.

[1]

(b) 1.5\times\text{IQR}=10.5, so the lower limit is Q_1-10.5=2.5.

[1]

Since 2<2.5, the score of 2 is an outlier.

[1]

(c) The sum of all ten scores is 155, so the nine remaining scores sum to 153 and their mean is \dfrac{153}{9}=17.

[1]

Final answer

(a) median 33/2, IQR 7
(b) lower fence 5/2, so 2 is an outlier
(c) 17.

Question 21

[7 marks]

The times, in minutes, taken by a courier to make deliveries have the following five-number summary:

\begin{array}{c|cccc}\mathrm{minimum} & Q_1 & \mathrm{median} & Q_3 & \mathrm{maximum}\\\mathrm{12} & \mathrm{18} & \mathrm{25} & \mathrm{31} & \mathrm{44}\end{array}

(a) Find the interquartile range, and show that there are no outliers. (Test against the fences Q_1-1.5\times\text{IQR} and Q_3+1.5\times\text{IQR}, using strict comparisons.)

[3]

Each time is converted to seconds, using s=60t.

(b) Write down the five-number summary in seconds and the new interquartile range.

[2]

(c) Show that applying the outlier rule to the times in seconds gives limits that are 60 times the limits in minutes, so the conclusion in part (a) is unchanged.

[2]

Show solutionHide solution

(a) \text{IQR}=31-18=13, so 1.5\times\text{IQR}=19.5.

[1]

Lower limit =18-19.5=-1.5 and upper limit =31+19.5=50.5.

[1]

The minimum 12>-1.5 and the maximum 44<50.5, so there are no outliers.

[1]

(b) In seconds: minimum 720, Q_1=1080, median 1500, Q_3=1860, maximum 2640.

[1]

\text{IQR}=1860-1080=780=60\times13.

[1]

(c) In seconds the limits are 1080-1.5(780)=-90=60(-1.5) and 1860+1.5(780)=3030=60(50.5).

[1]

Every value and both limits are multiplied by the same factor 60, so 720>-90 and 2640<3030 still hold and there are still no outliers.

[1]

Final answer

(a) IQR 13, fences -1.5, 50.5, no outliers
(b) 720,1080,1500,1860,2640, IQR 780
(c) fences -90, 3030.

Question 22

[6 marks]

A population of 500 people consists of three groups: A with 240 people, B with 160 people and C with 100 people. A stratified sample of 50 people is taken and each person is asked how many hours they exercised last week. The mean for group A's sample is 12.4, for group B's sample is 15.0, and the mean for the whole sample is 12.6.

(a) Find the number of people from each group in the sample.

[2]

(b) Find the mean for group C's sample.

[3]

(c) A student calculates the overall mean as \dfrac{12.4+15.0+\bar x_C}{3}. Explain why this method is wrong.

[1]

Show solutionHide solution

(a) The sampling fraction is \dfrac{50}{500}=\dfrac1{10}.

[1]

Group A: 24, group B: 16, group C: 10.

[1]

(b) The total hours for the whole sample is 50\times12.6=630.

[1]

Groups A and B contribute 24(12.4)+16(15.0)=297.6+240=537.6, so group C contributes 630-537.6=92.4.

[1]

\bar x_C=\dfrac{92.4}{10}=9.24 hours.

[1]

(c) The groups have different sizes, so their means must be weighted by the number of people in each group sample (24:16:10); an unweighted average of three means treats the groups as equal in size.

[1]

Final answer

(a) 24,16,10
(b) 9.24 hours
(c) group means need weights.

Question 23

[6 marks]

The table shows the number of hours, x, that six students revised and their marks, y, in a test.

\begin{array}{l|cccccc}x & \mathrm{2} & \mathrm{4} & \mathrm{5} & \mathrm{7} & \mathrm{8} & \mathrm{10}\\y & \mathrm{40} & \mathrm{50} & \mathrm{55} & \mathrm{65} & \mathrm{70} & \mathrm{86}\end{array}

(a) Find the mean point (\bar x,\bar y).

[2]

(b) Draw a scatter diagram for the data, mark the mean point, and draw a line of best fit by eye through the mean point.

[3]

(c) Describe the correlation between x and y.

[1]

Show solutionHide solution

(a) \bar x=\dfrac{2+4+5+7+8+10}{6}=\dfrac{36}{6}=6.

[1]

\bar y=\dfrac{40+50+55+65+70+86}{6}=\dfrac{366}{6}=61. The mean point is (6,61).

[1]

(b) Plot the six points, mark (6,61), and draw a straight line that passes through (6,61) with about as many points above it as below it.

[3]

(A1 for correct scales and labels, A1 for the six points and the mean point, A1 for a sensible line through the mean point.)

(c) Strong positive (linear) correlation.

[1]

Final answer

(a) (6.00,61.0)
(b) six points, line through mean point
(c) strong positive.

Question 24

[16 marks]

A recycling centre weighs 96 bags of sorted paper. Their masses, m kg, are grouped below.

\begin{array}{c|c}\text{Mass (kg)} & \text{Bags}\\0<m\le2 & 8\\2<m\le4 & 20\\4<m\le6 & 32\\6<m\le8 & 24\\8<m\le10 & 12\end{array}

(a) Write down the modal class and the five cumulative frequencies.

[2]

(b) Estimate the mean and the standard deviation of the masses.

[3]

(c) Draw a cumulative frequency graph. Use it to estimate the median and the interquartile range.

[4]

(d) One of these bags is selected at random. Find the probability that its mass exceeds 6 kg.

[2]

(e) Given that its mass exceeds 4 kg, find the probability that it exceeds 8 kg.

[2]

(f) A normal model is fitted using the estimates in (b). Find the model probability of a mass above 6 kg, and explain why it may differ from (d).

[3]

Show solutionHide solution

(a) 4<m\le6 [A1]; cumulative frequencies 8,28,60,84,96 [A1].

(b) Enter midpoints 1,3,5,7,9 and frequencies 8,20,32,24,12 [M1]. Mean 5.25 kg [A1]; population standard deviation 2.26 kg [A1].

(c) Plot (0,0),(2,8),(4,28),(6,60),(8,84),(10,96) [A1]; labelled axes and suitable curve [A1]. Median at CF 48: about 5.25 kg [A1]. Quartiles at CF 24 and 72 give about 3.60 and 7.00, hence IQR about 3.40 kg [A1]. Smooth-curve readings close to these are accepted.

(d) (24+12)/96=0.375 [M1 A1].

(e) 12/(32+24+12)=12/68=0.176 [M1 A1].

(f) Normal upper tail with \mu=5.25, unrounded \sigma: 0.370 [M1 A1]. The observed proportion is empirical; the normal curve is a model based on grouped estimates [R1].

Final answer

(a) 4<m\le6; 8,28,60,84,96
(b) 5.25, 2.26 kg
(c) median 5.25, IQR 3.40 kg (approx.)
(d) 0.375
(e) 0.176
(f) 0.370, model vs counts.

Going further

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

G1 Which centre is best?

G1 Worked example

[8 marks]

Four villages lie on a straight road at 0, 3, 5 and 12 km from a junction. The numbers of commuters who live in them are 40, 25, 20 and 15. A bus stop is to be placed on the road, t km from the junction.

(a) Show that the sum of the squared distances from the 100 commuters' homes to the stop is T(t)=100t^2-710t+2885.

[2]

(b) Find the value of t that minimises T(t), and the least value of T.

[2]

(c) Find the value of t that minimises D(t)=\sum f|x-t|, and the least value of D.

[3]

(d) Another 15 commuters move into the village at 12 km. State, with a reason, which of the two best positions moves.

[1]

Show solutionHide solution

(a) T(t)=\sum f(x-t)^2=t^2\sum f-2t\sum fx+\sum fx^2

\sum f=100, \sum fx=75+100+180=355, \sum fx^2=225+500+2160=2885, so T(t)=100t^2-710t+2885

A1AG

(b) vertex at t=\dfrac{710}{200}=3.55, which is the mean \bar x

T(3.55)=2885-100\times3.55^2=1624.75

(c) moving the stop a small distance to the right adds that distance for every commuter to its left and saves it for every commuter to its right; for 0<t<3 the gradient of D is 40-60=-20, for 3<t<5 it is 65-35=30

so D is least at t=3, the median village

D(3)=40\times3+0+20\times2+15\times9=295 (km)

(d) the median stays at 3 km (now 115 commuters, and the 58th lives at 3 km) but the mean moves to \frac{535}{115}\approx4.65 km: only the position from (b) moves

Final answer

(b) t=3.55, least value 1624.75
(c) t=3, 295 km
(d) the mean moves, the median does not

G1 Investigation

[7 marks]

For data x_1<x_2<\cdots<x_n and a number t, let S(t)=\sum(x_i-t)^2, D(t)=\sum|x_i-t| and M(t) be the largest of the distances |x_i-t|.

(a) For the data 1,2,4,10, find the least value of S(t), and show that D(t)=11 for 2\le t\le4.

[3]

(b) The value 100 is added to the data in (a). Find the new value of t that minimises S, and the new value of t that minimises D. Comment on your answers.

[2]

(c) Show that M(t) is least when t=\frac12(x_1+x_n), the mid-range.

[2]

Show solutionHide solution

(a) \bar x=4.25, and the least value is \sum x^2-n\bar x^2=121-4\times4.25^2=48.75

M1A1

for 2\le t\le4: D(t)=(t-1)+(t-2)+(4-t)+(10-t)=11

A1AG

(b) S is least at the mean, \frac{117}{5}=23.4; D is least at the median, 4

one large value drags the mean (and the minimiser of S) a long way; the median, and so the minimiser of D, hardly moves

(c) M(t) is the larger of t-x_1 (increasing) and x_n-t (decreasing)

the larger of the two is least where they are equal, t=\frac12(x_1+x_n), giving M=\frac12(x_n-x_1)

Final answer

(a) 48.75 at t=4.25; D=11
(b) 23.4, 4
(c) mid-range.

G2 Where the regression line comes from

G2 Worked example

[8 marks]

For data points (x_i,y_i), the line y=ax+b is chosen to make E=\sum(y_i-ax_i-b)^2 as small as possible. Write S_{xx}=\sum(x_i-\bar x)^2, S_{xy}=\sum(x_i-\bar x)(y_i-\bar y) and S_{yy}=\sum(y_i-\bar y)^2. The data are (1,2), (2,3), (3,5), (4,6).

(a) Explain why, for a fixed a, E is least when b=\bar y-a\bar x.

[2]

(b) Find S_{xx}, S_{xy} and S_{yy}. Show that, with b=\bar y-a\bar x, E=5a^2-14a+10.

[3]

(c) Find the values of a and b that minimise E, and the least value of E.

[2]

(d) Compare your line with the regression line of y on x from your GDC.

[1]

Show solutionHide solution

(a) for fixed a, E=\sum\bigl((y_i-ax_i)-b\bigr)^2 is a sum of squared distances from b to the numbers y_i-ax_i

that sum is least when b is their mean, \bar y-a\bar x

(b) \bar x=2.5, \bar y=4; S_{xx}=5, S_{xy}=3+0.5+0.5+3=7, S_{yy}=10

with b=\bar y-a\bar x, each bracket is (y_i-\bar y)-a(x_i-\bar x), so E=S_{yy}-2aS_{xy}+a^2S_{xx}

E=10-14a+5a^2

A1AG

(c) vertex at a=\dfrac{14}{10}=1.4; b=4-1.4\times2.5=0.5

least E=10-14(1.4)+5(1.4)^2=0.2

(d) the GDC gives y=1.40x+0.500: the same line

Final answer

(b) S_{xx}=5, S_{xy}=7, S_{yy}=10
(c) a=1.4, b=0.5, E=0.2

G2 Investigation

[8 marks]

Use the notation of Worked example G2.

(a) For the data (0,1), (1,1), (2,4), (3,4), (4,5), find S_{xx}, S_{xy} and S_{yy}, and the regression line of y on x from a=\dfrac{S_{xy}}{S_{xx}}. Check it with your GDC.

[2]

(b) The regression line of x on y makes the sum of the squared horizontal distances as small as possible. Explain why it is x-\bar x=\dfrac{S_{xy}}{S_{yy}}(y-\bar y), and find it for these data.

[2]

(c) Show that the least value of E is S_{yy}-\dfrac{S_{xy}^2}{S_{xx}}. Deduce that S_{xy}^2\le S_{xx}S_{yy}, with equality only when all the points lie on one straight line.

[3]

(d) Your GDC computes r=\dfrac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}. Show that -1\le r\le1.

[1]

Show solutionHide solution

(a) \bar x=2, \bar y=3; S_{xx}=10, S_{xy}=11, S_{yy}=14

a=1.1, b=3-2.2=0.8: y=1.10x+0.800, as the GDC gives

(b) swap the roles of x and y in Worked example G2: the line passes through (\bar x,\bar y) with \dfrac{\mathrm dx}{\mathrm dy}=\dfrac{S_{xy}}{S_{yy}}

x-2=\dfrac{11}{14}(y-3), that is x=0.786y-0.357 (3 s.f.)

(c) E=S_{xx}a^2-2S_{xy}a+S_{yy} is least at a=\dfrac{S_{xy}}{S_{xx}}, with value S_{yy}-\dfrac{S_{xy}^2}{S_{xx}}

M1A1

E is a sum of squares, so this is \ge0, giving S_{xy}^2\le S_{xx}S_{yy}; it is 0 only if every y_i=ax_i+b, that is, all the points lie on the line

(d) By (c), |S_{xy}|\le\sqrt{S_{xx}S_{yy}}. Divide by the positive square root to obtain |r|\le1, so -1\le r\le1.

Final answer

(a) y=1.10x+0.800
(b) x-2=\frac{11}{14}(y-3)

TI-84 Plus CE keystrokes

The GDC panels in the book give keystrokes for the TI-Nspire CX II and the Casio fx-CG50. The TI-84 Plus CE keystrokes for the common calculator jobs are listed in Chapter 12.

From the book

The Ultimate Guide for the IB: Mathematics: Analysis and Approaches SL

Chapter 7 teaches statistics 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