Maths Oct Nov 2010

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS

General Certificate of Education

Advanced Subsidiary Level and Advanced Level

MATHEMATICS 9709/61

Paper 6 Probability & Statistics 1 (S1) October/November 2010

1 hour 15 minutes

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At the end of the examination, fasten all your work securely together.

The number of marks is given in brackets [ ] at the end of each question or part question.

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2

1 Anita made observations of the maximum temperature, t◦C, on 50 days. Her results are summarised

by Σ t = 910 and Σ(t − t)2 = 876, where t denotes the mean of the 50 observations. Calculate t and

the standard deviation of the observations. [3]

2 On average, 2 apples out of 15 are classified as being underweight. Find the probability that in a

random sample of 200 apples, the number of apples which are underweight is more than 21 and less

than 35. [5]

3 The times taken by students to get up in the morning can be modelled by a normal distribution with

mean 26.4 minutes and standard deviation 3.7 minutes.

(i) For a random sample of 350 students, find the number who would be expected to take longer

than 20 minutes to get up in the morning. [3]

(ii) ‘Very slow’ students are students whose time to get up is more than 1.645 standard deviations

above the mean. Find the probability that fewer than 3 students from a random sample of 8

students are ‘very slow’. [4]

4 The weights in grams of a number of stones, measured correct to the nearest gram, are represented in

the following table.

Weight (grams) 1 − 10 11 − 20 21 − 25 26 − 30 31 − 50 51 − 70

Frequency 2x 4x 3x 5x 4x x

A histogram is drawn with a scale of 1 cm to 1 unit on the vertical axis, which represents frequency

density. The 1 − 10 rectangle has height 3 cm.

(i) Calculate the value of x and the height of the 51 − 70 rectangle. [4]

(ii) Calculate an estimate of the mean weight of the stones. [3]

5 Three friends, Rick, Brenda and Ali, go to a football match but forget to say which entrance to the

ground they will meet at. There are four entrances, A, B, C and D. Each friend chooses an entrance

independently.

• The probability that Rick chooses entrance A is 13. The probabilities that he chooses entrances

B, C or D are all equal.

• Brenda is equally likely to choose any of the four entrances.

• The probability that Ali chooses entrance C is 27

and the probability that he chooses entrance D

is 35. The probabilities that he chooses the other two entrances are equal.

(i) Find the probability that at least 2 friends will choose entrance B. [4]

(ii) Find the probability that the three friends will all choose the same entrance. [4]

© UCLES 2010 9709/61/O/N/10



3

6

Pegs are to be placed in the four holes shown, one in each hole. The pegs come in different colours

and pegs of the same colour are identical. Calculate how many different arrangements of coloured

pegs in the four holes can be made using

(i) 6 pegs, all of different colours, [1]

(ii) 4 pegs consisting of 2 blue pegs, 1 orange peg and 1 yellow peg. [1]

Beryl has 12 pegs consisting of 2 red, 2 blue, 2 green, 2 orange, 2 yellow and 2 black pegs. Calculate

how many different arrangements of coloured pegs in the 4 holes Beryl can make using

(iii) 4 different colours, [1]

(iv) 3 different colours, [3]

(v) any of her 12 pegs. [3]

7 Sanket plays a game using a biased die which is twice as likely to land on an even number as on an

odd number. The probabilities for the three even numbers are all equal and the probabilities for the

three odd numbers are all equal.

(i) Find the probability of throwing an odd number with this die. [2]

Sanket throws the die once and calculates his score by the following method.

• If the number thrown is 3 or less he multiplies the number thrown by 3 and adds 1.

• If the number thrown is more than 3 he multiplies the number thrown by 2 and subtracts 4.

The random variable X is Sanket’s score.

(ii) Show that P(X = 8) = 29. [2]

The table shows the probability distribution of X.

x 4 6 7 8 10

P(X = x) 39

19

29

29

19

(iii) Given that E(X) = 589

, find Var(X). [2]

Sanket throws the die twice.

(iv) Find the probability that the total of the scores on the two throws is 16. [2]

(v) Given that the total of the scores on the two throws is 16, find the probability that the score on

the first throw was 6. [3]

© UCLES 2010 9709/61/O/N/10



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9709/61/O/N/10






MATHEMATICS 9709/62


1 hour 15 minutes


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2

1 The discrete random variable X takes the values 1, 4, 5, 7 and 9 only. The probability distribution of

X is shown in the table.

x 1 4 5 7 9

P(X = x) 4p 5p2 1.5p 2.5p 1.5p

Find p. [3]

2 Esme noted the test marks, x, of 16 people in a class. She found that Σ x = 824 and that the standard

deviation of x was 6.5.

(i) Calculate Σ(x − 50) and Σ(x − 50)2. [3]

(ii) One person did the test later and her mark was 72. Calculate the new mean and standard deviation

of the marks of all 17 people. [3]

3 A fair five-sided spinner has sides numbered 1, 2, 3, 4, 5. Raj spins the spinner and throws two fair

dice. He calculates his score as follows.

• If the spinner lands on an even-numbered side, Raj multiplies the two numbers showing on

the dice to get his score.

• If the spinner lands on an odd-numbered side, Raj adds the numbers showing on the dice to

get his score.

Given that Raj’s score is 12, find the probability that the spinner landed on an even-numbered side.

[6]

4 The weights in kilograms of 11 bags of sugar and 7 bags of flour are as follows.

Sugar: 1.961 1.983 2.008 2.014 1.968 1.994 2.011 2.017 1.977 1.984 1.989

Flour: 1.945 1.962 1.949 1.977 1.964 1.941 1.953

(i) Represent this information on a back-to-back stem-and-leaf diagram with sugar on the left-hand

side. [4]

(ii) Find the median and interquartile range of the weights of the bags of sugar. [3]

5 The distance the Zotoc car can travel on 20 litres of fuel is normally distributed with mean 320 km and

standard deviation 21.6 km. The distance the Ganmor car can travel on 20 litres of fuel is normally

distributed with mean 350 km and standard deviation 7.5 km. Both cars are filled with 20 litres of fuel

and are driven towards a place 367 km away.

(i) For each car, find the probability that it runs out of fuel before it has travelled 367 km. [3]

(ii) The probability that a Zotoc car can travel at least (320 + d) km on 20 litres of fuel is 0.409. Find

the value of d. [4]

© UCLES 2010 9709/62/O/N/10



3

6 (i) State three conditions that must be satisfied for a situation to be modelled by a binomial

distribution. [2]

On any day, there is a probability of 0.3 that Julie’s train is late.

(ii) Nine days are chosen at random. Find the probability that Julie’s train is late on more than 7 days

or fewer than 2 days. [3]

(iii) 90 days are chosen at random. Find the probability that Julie’s train is late on more than 35 days

or fewer than 27 days. [5]

7 A committee of 6 people, which must contain at least 4 men and at least 1 woman, is to be chosen

from 10 men and 9 women.

(i) Find the number of possible committees that can be chosen. [3]

(ii) Find the probability that one particular man, Albert, and one particular woman, Tracey, are both

on the committee. [2]

(iii) Find the number of possible committees that include either Albert or Tracey but not both. [3]

(iv) The committee that is chosen consists of 4 men and 2 women. They queue up randomly in a line

for refreshments. Find the probability that the women are not next to each other in the queue.

[3]

© UCLES 2010 9709/62/O/N/10



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9709/62/O/N/10






MATHEMATICS 9709/63


1 hour 15 minutes


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2

1 Name the distribution and suggest suitable numerical parameters that you could use to model the

weights in kilograms of female 18-year-old students. [2]

2 In a probability distribution the random variable X takes the value x with probability kx, where x takes

values 1, 2, 3, 4, 5 only.

(i) Draw up a probability distribution table for X, in terms of k, and find the value of k. [3]

(ii) Find E(X). [2]

3 It was found that 68% of the passengers on a train used a cell phone during their train journey. Of

those using a cell phone, 70% were under 30 years old, 25% were between 30 and 65 years old and

the rest were over 65 years old. Of those not using a cell phone, 26% were under 30 years old and

64% were over 65 years old.

(i) Draw a tree diagram to represent this information, giving all probabilities as decimals. [2]

(ii) Given that one of the passengers is 45 years old, find the probability of this passenger using a

cell phone during the journey. [3]

4 Delip measured the speeds, x km per hour, of 70 cars on a road where the speed limit is 60 km per hour.

His results are summarised by Σ(x − 60) = 245.

(i) Calculate the mean speed of these 70 cars. [2]

His friend Sachim used values of (x − 50) to calculate the mean.

(ii) Find Σ(x − 50). [2]

(iii) The standard deviation of the speeds is 10.6 km per hour. Calculate Σ(x − 50)2. [2]

© UCLES 2010 9709/63/O/N/10



3

5 The following histogram illustrates the distribution of times, in minutes, that some students spent

taking a shower.

0 2 4 6 8 10 12 14 16 18 20

5

0

10

15

20

25

30

35

40

Frequencydensity

Time inminutes

(i) Copy and complete the following frequency table for the data. [3]

Time (t minutes) 2 < t ≤ 4 4 < t ≤ 6 6 < t ≤ 7 7 < t ≤ 8 8 < t ≤ 10 10 < t ≤ 16

Frequency

(ii) Calculate an estimate of the mean time to take a shower. [2]

(iii) Two of these students are chosen at random. Find the probability that exactly one takes between

7 and 10 minutes to take a shower. [3]

[Questions 6 and 7 are printed on the next page.]

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4

6Windows

Windows

AisleFrontBack

A small aeroplane has 14 seats for passengers. The seats are arranged in 4 rows of 3 seats and a back

row of 2 seats (see diagram). 12 passengers board the aeroplane.

(i) How many possible seating arrangements are there for the 12 passengers? Give your answer

correct to 3 significant figures. [2]

These 12 passengers consist of 2 married couples (Mr and Mrs Lin and Mr and Mrs Brown), 5 students

and 3 business people.

(ii) The 3 business people sit in the front row. The 5 students each sit at a window seat. Mr and Mrs

Lin sit in the same row on the same side of the aisle. Mr and Mrs Brown sit in another row on

the same side of the aisle. How many possible seating arrangements are there? [4]

(iii) If, instead, the 12 passengers are seated randomly, find the probability that Mrs Lin sits directly

behind a student and Mrs Brown sits in the front row. [4]

7 The times spent by people visiting a certain dentist are independent and normally distributed with a

mean of 8.2 minutes. 79% of people who visit this dentist have visits lasting less than 10 minutes.

(i) Find the standard deviation of the times spent by people visiting this dentist. [3]

(ii) Find the probability that the time spent visiting this dentist by a randomly chosen person deviates

from the mean by more than 1 minute. [3]

(iii) Find the probability that, of 6 randomly chosen people, more than 2 have visits lasting longer

than 10 minutes. [3]

(iv) Find the probability that, of 35 randomly chosen people, fewer than 16 have visits lasting less

than 8.2 minutes. [5]






© UCLES 2010 9709/63/O/N/10




GCE Advanced Subsidiary Level and GCE Advanced Level

MARK SCHEME for the October/November 2010 question paper

for the guidance of teachers

9709 MATHEMATICS

9709/61 Paper 6, maximum raw mark 50

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers.

Mark schemes must be read in conjunction with the question papers and the report on the examination.

• CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2010 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.



Page 2 Mark Scheme: Teachers’ version Syllabus Paper

GCE A LEVEL – October/November 2010 9709 61

© UCLES 2010

Mark Scheme Notes Marks are of the following three types:

M Method mark, awarded for a valid method applied to the problem. Method marks are not lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. Correct application of a formula without the formula being quoted obviously earns the M mark and in some cases an M mark can be implied from a correct answer.

A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained.

Accuracy marks cannot be given unless the associated method mark is earned (or implied).

B Mark for a correct result or statement independent of method marks.

• When a part of a question has two or more "method" steps, the M marks are generally independent unless the scheme specifically says otherwise; and similarly when there are several B marks allocated. The notation DM or DB (or dep*) is used to indicate that a particular M or B mark is dependent on an earlier M or B (asterisked) mark in the scheme. When two or more steps are run together by the candidate, the earlier marks are implied and full credit is given.

• The symbol √ implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A or B marks are given for correct work only. A and B marks are not given for fortuitously "correct" answers or results obtained from incorrect working.

• Note: B2 or A2 means that the candidate can earn 2 or 0. B2/1/0 means that the candidate can earn anything from 0 to 2.

The marks indicated in the scheme may not be subdivided. If there is genuine doubt whether a candidate has earned a mark, allow the candidate the benefit of the doubt. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored.

• Wrong or missing units in an answer should not lead to the loss of a mark unless the

scheme specifically indicates otherwise. • For a numerical answer, allow the A or B mark if a value is obtained which is correct to 3 s.f.,

or which would be correct to 3 s.f. if rounded (1 d.p. in the case of an angle). As stated above, an A or B mark is not given if a correct numerical answer arises fortuitously from incorrect working. For Mechanics questions, allow A or B marks for correct answers which arise from taking g equal to 9.8 or 9.81 instead of 10.





© UCLES 2010

The following abbreviations may be used in a mark scheme or used on the scripts:

AEF Any Equivalent Form (of answer is equally acceptable) AG Answer Given on the question paper (so extra checking is needed to ensure that

the detailed working leading to the result is valid) BOD Benefit of Doubt (allowed when the validity of a solution may not be absolutely

clear) CAO Correct Answer Only (emphasising that no "follow through" from a previous error

is allowed) CWO Correct Working Only – often written by a ‘fortuitous' answer ISW Ignore Subsequent Working MR Misread PA Premature Approximation (resulting in basically correct work that is insufficiently

accurate) SOS See Other Solution (the candidate makes a better attempt at the same question) SR Special Ruling (detailing the mark to be given for a specific wrong solution, or a

case where some standard marking practice is to be varied in the light of a particular circumstance)

Penalties

MR –1 A penalty of MR –1 is deducted from A or B marks when the data of a question or

part question are genuinely misread and the object and difficulty of the question remain unaltered. In this case all A and B marks then become "follow through √" marks. MR is not applied when the candidate misreads his own figures – this is regarded as an error in accuracy. An MR–2 penalty may be applied in particular cases if agreed at the coordination meeting.

PA –1 This is deducted from A or B marks in the case of premature approximation. The

PA –1 penalty is usually discussed at the meeting.





© UCLES 2010

1 mean = 18.2

sd = 50/876

= 4.19

B1

M1

A1

[3]

Correct unsimplified expression seen

Correct answer

2 mean = 200 × 2/15 (= 26.67) (80/3)

variance = 200 × 2/15 × 13/15 (= 23.11)(208/9)

P(21 < X < 35) =

11.23

67.265.34

11.23

67.265.21P

−<<

−z

= P(–1.075 < z < 1.629)

= 0.8589 + 0.9483 – 1

= 0.807

B1

M1

M1

M1

A1

[5]

mean and variance correct

standardising, ±, with or without cc, must

have sq rts

continuity corrections 20.5 or 21.5, 34.5

or 35.5

Φ1 + Φ2 – 1

answer rounding to 0.807

3 (i) P(X > 20) = P(z > –6.4/3.7)

= P(z > –1.730)

= 0.9582

Number of students = 335 or 336

M1

A1

A1ft

[3]

Standardising no cc no sq rt

Prob rounding to 0.958

Correct answer ft their prob, must be

integer

(ii) P(very slow) = 0.05

P(0, 1, 2) =

(0.95)8 + 8C1(0.05)1(0.95)7 + 8C2(0.05)2(0.95)6

= 0.6634 + 0.2793 + 0.0515

= 0.994

B1

M1

M1

A1

[4]

0.05 or 0.95 seen

Binomial term with 8Cr pr (1 – p) 8 – r seen

any p

Correct expression for P(0, 1, 2), p close

to 0.05

Answer rounding to 0.994

4 (i) 3 = 2x / 10

x = 15

height = freq / class width

= x / 20 = 0.75 cm

M1

A1

M1

A1

[4]

Attempt at using freq density = freq / cw

Correct answer

Attempt at using fd = freq / cw with

different cw from above

Correct answer

(ii) mean wt =

(5.5 × 30 + 15.5 × 60 + 23 × 45 + 28 × 75

+ 40.5 × 60 + 60.5 × 15) / 285

= 26.6 grams

M1

M1

A1

[3]

Using freqs or frequency ratios and mid-

points, attempt not ucb, not cw (can do it

without x)

Correct unsimplified answer can have fr

ratios

Correct answer





© UCLES 2010

5 (i)

A B C D

Rick 1/3 2/9 2/9 2/9

Brenda 1/4 1/4 1/4 1/4

Ali 2/35 2/35 2/7 3/5

P(Rick B, Brenda B, Ali not B)

+ P(Rick B, Brenda not B, Ali B)

+ P(Rick not B, Brenda B, Ali B)

= 11/210 + 2/210 + 1/90 = 23/315

P(Rick B, Brenda B, Ali B) = 1/315

Prob(at least 2 at entrance B)

= 24/315 (8/105) (0.0762)

M1

M1

M1

A1

[4]

Obtaining probs of each person for each

entrance (can be implied or awarded in

part (i) or part (ii))

Considering options 2 meet 1 doesn’t,

must have at least two 3-factor terms

Adding option all three meet, must be

added to a prob

Correct answer

(ii) P(entrance A) = 1/210 (0.00476)

P(entrance B) = 1/315 (0.00317)

P(entrance C) = 1/63 (0.0159)

P(entrance D) = 1/30 (0.0333)

P(same entrance) = 2/35 (0.0571)

M1

M1

A1

A1

[4]

Obtaining a three-factor prob for any

entrance

Adding four three-factor probabilities for

the 4 entrances

Two or more correct entrance

probabilities

Correct answer

6 (i) 6P4 = 6!/2!

= 360

B1

[1]

Correct answer

(ii) 4!/2! = 12 B1

[1]

Correct answer

(iii) 4! × 6C4 = 360 or 6P4 B1

[1]

Correct final answer

(iv) e.g. 2R 1B 1G, 1R 2B 1G, 1R 1B 2G

= !2

!4 +

!2

!4+

!2

!4 = 36, mult by 6C3

total = 720

M1

M1

A1

[3]

4!/2! seen

Mult by 6C3

Correct answer

(v) 2R 2B = 4!/2!2! = 6

Mult by 6C2, total = 90

Answer = 360 + 720 + 90 = 1170

M1

A1

A1ft

[3]

Considering 2 colours e.g. RRBB or

RBBR or...

mult by 6C2

Ft their (iii) + (iv) + (v)





© UCLES 2010

7 (i) If y = P(odd number) then P(even number) = 2y

3y + 6y = 1 so y = 1/9 oe. OR prob = 1/3

M1

A1

[2]

2P(Odd) shown = P(Even) and summed

to 1

correct answer accept either

(ii) Score of 8 means throwing a 6

6 is even so P(8) = 2/9 (AG)

B1

B1

[2]

legit justification of use of 2/9

(iii) Var(X) = (48 + 36 + 98 + 128 + 100)/9 – (58/9)2

= 4.02 accept 4.025 (326/81)

M1

A1

[2]

Correct method no dividings, 6.44

squared subt numerically

Correct answer

(iv) P(score 6,10) + P(score 10,6) + P(score 8,8)

= 1/81 + 1/81 + 4/81

= 6/81 (2/27) (0.0741)

M1

A1

[2]

Summing two different 2-factor

probabilities

Correct answer

(v) P(score 6, 10) = 1/81

P(1st score 6 given total 16)

= (1/81) ÷ (6/81)

= 1/6

B1

M1

A1

[3]

1/81 seen in numerator

Dividing by their (iv)

Correct answer







9709 MATHEMATICS









© UCLES 2010

















© UCLES 2010








Penalties









© UCLES 2010

1 4p + 5p2 + 1.5p + 2.5p + 1.5p = 1 10p2 + 19p – 2 = 0 p = 0.1 or –2 p = 0.1

M1 A1 A1

[3]

Summing 5 probs to = 1 can be implied For 0.1 seen with or without –2 Choosing 0.1 must be by rejecting –2

2 (i) Σ(x – 50) = 824 – 16 × 50 = 24 B1 Correct answer

2

22

5.616

)50(

16

)50(=

−Σ−

−Σ xx

M1

Consistent substituting in the correct coded variance formula OR valid method for Σx2 then expanding Σ(x – 50)2, 3 terms at least 2 correct

Σ(x – 50)2 = 712 A1 [3]

Correct answer

(ii) new mean = 896/17 (= 52.7) B1 Correct answer

new var = 22

17

)5072(24

17

22712

−+−

+

M1

Using the correct coded variance formula with n = 17 and new coded mean2 OR their (Σx2 + 722)/17 – their new mean2

new sd = 7.94 A1

[3]

Rounding to correct answer, accept 7.95 or 7.98 or 7.91





© UCLES 2010

3 P(E and 12) = 36

4

5

2× =

180

8(2/45)

M1

2/5 or 3/5 mult by dice-related probability seen anywhere

A1

36

4

5

2× seen oe

P(12) = 180

8

36

1

5

3+× =

180

11(0.0611)

M1 A1ft

Summing two 2-factor probs involving 2/5 and 3/5 3/5 × 1/36 + their P(E and 12), ft their P(E 12)

)12(P

)12(P)12(P

andEE =

M1dep

Subst in condit prob formula, must have a fraction

= 11

8(0.727)

A1 [6]

Correct answer

OR list Even: 2 and (4,3) or (3,4) or (2,6) or (6,2) 4 and ditto Gives 8 options Odd: 1 and (6,6) or 3 and (6,6) or 5 and (6,6) Gives 3 options Prob(E│12) = 8/11

M1 A1 M1 A1 M1 A1

List attempt evens 8 options List attempt odds 3 options (Their even)/(their total) Correct answer

4 (i)

sugar flour 194 1 5 9 195 3 8 1 196 2 4 7 197 7 9 4 3 198 4 199 8 200 7 4 1 201 key 1 196 2 means 1.961 kg for sugar and 1.962 kg for flour

B1 B1 B1 B1ft

[4]

Correct stem must be integers. (stem and leaves can be in reverse order) Correct leaves flour must be single and ordered Correct leaves sugar must be single and ordered Correct key needs all this, ft if single leaves and 1.96 etc in stem

(ii) med = 1.989 kg IQ range = 2.011 – 1.977 = 0.034 kg

B1 M1 A1

[3]

correct median subt their LQ from their UQ, UQ > med, LQ < med Correct answer





© UCLES 2010

5 (i) Zotoc: 176.26.21

320367=

−

=z

Ganmor: 267.25.7

350367=

−

=z

P(Zotoc) = 0.985 P(Ganmor) = 0.988

M1 A1 A1

[3]

Standardising either car’s fuel, no cc, no sq, no √ Correct answer Correct answer

(ii) z = 0.23

6.21

32023.0

−

=

x

x = 324.968 d = 4.97

B1 M1 M1ind A1

[4]

± 0.23 seen Standardising either car, no cc, no sq rt, no sq 320 + d – 320 i.e. just d on num Correct answer, –4.97 gets A0

6 (i) constant/given prob, independent trials, fixed/given no. of trials, only two outcomes

B1 B1

[2]

One option correct Three options correct

(ii) P(8, 9, 0, 1) = 9C8(0.3)8(0.7) + (0.3)9 + (0.7)9 + 9C1(0.3)(0.7)8 = 0.196

M1 A1 A1

[3]

One term seen involving (0.3)x(0.7)9 – x(9Cx) Correct unsimplified expression Correct answer

(iii) mean = 90 × 0.3 = 27 var = 18.9

P(X > 35) = 1 – Φ

−

9.18

275.35

= 1 – Φ(1.955) = 0.0253

P(X < 27) =

−Φ

9.18

275.26 = 1 – Φ(0.115)

= 0.4542 Total prob = 0.480 accept 0.48

B1

M1 M1 M1 A1

[5]

Expressions for 27 and 18.9 (4.347) seen

Standardising one expression, must have sq rt in denom, cc not necessary Continuity correction applied at least once (1 – Φ1) + (1 – Φ2) accept (0.0329 + 0.5) if no cc Rounding to correct answer





© UCLES 2010

7 (i) 4M 2W or 5M 1W chosen in 10C4 × 9C2 + 10C5 × 9C1 = 9828

M1 A1 A1

[3]

At least 1 of 10C4 × 9C2 and 10C5 × 9C1 seen Correct unsimplified expression Correct answer

(ii) 9C3 × 8C1 + 9C4 = 798

Prob = 798/9828 = 0.0812

M1 A1

[2]

One of 9C3 × 8C1 and 9C4 × (8C0) seen Correct answer

(iii) Albert + not T... 9C3 × 8C2 + 9C4 × 8C1 = 3360 Tracey + not A... 9C4 × 8C1 + 9C5 = 1134 Number of ways = 4494

M1 A1 A1

[3]

One of 9C3 × 8C2 or 9C4 × 8C1 or 9C5 × (8C0) seen Unsimplified 3360 or 1134 seen Correct final answer

(iv) 6! – 4! × 5 × 2 or 6! – 5! × 2 (= 480) OR 4! × 5 × 4 or 4! × 5P2 (= 480) prob = 480/6! = 2/3 (0.667) OR using probabilities…as above OR Women together 5!/4! (= 5) Women not together = 15 – 5 = 10 total ways MMMMWW = 6!/4!2! = 15 prob = 2/3

B1 M1 A1

[3] B1 M1 A1

6! – 4! × 5 × 2 or 6! – 5! × 2 or 4! × 5 × 4 or 4! × 5P2 dividing by 6! correct answer 5 or 10 seen Dividing by 15 Correct answer







9709 MATHEMATICS









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© UCLES 2010








Penalties









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1 Normal mean 60 kg, variance 90 kg2

B1 B1

[2]

Any sensible values (mean 40–80 kg, variance 16–225 kg2), could give s.d. 4–15 kg

2 (i)

x 1 2 3 4 5 Prob k 2k 3k 4k 5k

15k = 1 k = 1/15 (0.0667)

M1 M1 A1

[3]

1, 2, 3, 4, 5 seen, together with some probabilities involving k but not x summing probs involving k to 1 correct answer

(ii) E(X) = k + 4k + 9k + 16k + 25k = 55k = 11/3 (3.67)

M1 A1ft

[2]

using Σpx no dividing correct answer, ft on 55k, 0 < k < 1

3 (i) Y 0.7 Ph 0.25 M 0.68 0.05 O Y 0.32 0.26 NPh 0.10 M 0.64 O

M1 A1

[2]

Y = young, M = middle-aged, O = old Correct shape with Ph, NPh first All probabilities and correct

(ii) P(Ph│M) = 1.032.025.068.0

25.068.0

×+×

×

B1 M1

For correct numerator using cond prob formula with numerator < denominator For attempt at P(35 – 60 years old), involving the sum of two 2-factor probs, seen anywhere

= 0.842 (170/202) A1 [3]

Correct answer

4 (i) =x 60 + 245/70 = 63.5

M1 A1

[2]

245/70 seen Correct answer

(ii) Σ(x – 50) = Σx – Σ50 = 245 + 70 × 60 – 70 × 50 = 945

M1 A1

[2]

Any valid method, involving 70 Correct answer

(iii) coded mean = 945/70 = 13.5

2

22

6.1070

945

70

)50(=

−

−Σ x

Σ(x – 50)2 = 20623 (20600)

M1 A1

[2]

Using variance formula with coded mean Correct answer





© UCLES 2010

5 (i)

2 to 4 4 to 6 6 to 7 7 to 8 8 to 10 10 to 16 20 44 34 30 30 36

M1 A1 A1

[3]

Using fd to evaluate freqs Any four correct All correct

(ii) mid-points 3, 5, 6.5, 7.5, 9, 13 E(X) = (3 × 20 + 5 × 44 + 6.5 × 34 + 7.5 × 30 +

9 × 30 + 13 × 36) / 194 = 1464/194 = 7.55

M1 A1ft

[2]

5 or 6 correct mid-points Correct answer, ft on 6 correct mid-points and the frequencies in their table

(iii) p = 60/194 (0.309) P(1) = 2 × (60/194)(134/193) = 8040/18721 (0.429)

B1ft M1 A1

[3]

60/194 seen, ft on (their 30 + their 30) / their total multiplying a probability by 2 Correct answer

6 (i) 14P12 = 4.36 × 1010

M1 A1

[2]

14P12 seen oe Correct answer

(ii) business people 3! = 6 students 5! = 120 married couples 3P2 × 2 × 2 = 24 total ways = 17280

B1 B1 B1 B1

[4]

3! oe seen, not in denominator 5! oe seen, not in denominator 24 oe seen, not in denominator correct final answer

(iii) Mrs Brown 3 Mrs Lin 10 Student 5 Prob = 3 × 10 × 5 × 11P9 / (i) = 0.0687 OR1 3/14 × 10/13 × 5/12 = 150/2184 (0.0687) OR2 1 − 3/14 = 11/14 1 − 11/14 × 5/13 = 127/182 8/14(4/13 × 12/12 + 9/13 × 7/12) + 3/14(3/13 × 12/12 + 10/13 × 7/12) = 1206/2184 1 − (1524 + 1716 − 1206)/2184 = 150/2184

B1 B1 M1 A1

[4] B1 B1 M1 A1 B1 B1 M1 A1

any 2 of 3, 10, 5 oe seen, not in denominator 11P9 seen multiplied dividing by their (i) correct answer any 2 of numerators 3, 10, 5 oe seen denominators 14, 13, 12 of 3 fractions multiplying 3 separate fractions correct answer 1 − 3/14 seen 1 − 11/14 × 5/13 seen attempt to find P(Mrs Lin not behind a student and Mrs Brown not in front row), involving 8/14 × prob + 3/14 × prob correct answer





© UCLES 2010

7 (i) z = 0.807

0.807 = σ

2.810 −

s = 2.23

B1 M1 A1

[3]

0.807 seen standardising, must have σ, no sq rt, no cc and a z-value correct answer

(ii) P(> 1 min from mean) = P(mod z >23.2

1)

= P ( )4484.0>z = (1 – 0.6729) × 2 = 0.654

M1 M1 A1

[3]

standardising, their sd, no cc and adding two areas using 1 – Φ(z) correct answer

(iii) P(> 2 longer) = 1 – P(0, 1, 2 longer) = 1 – {(0.79)6 + 6C1(0.21)(0.79)5 +

6C2(0.21)2(0.79)4} = 0.112

M1 A1 A1

[3]

binomial term 6Cxpx(1 − p)6 − x

correct unsimplified answer correct answer

(iv) µ = 35 × 0.5 = 17.5 σ

2 = 35 × 0.5 × 0.5 = 8.75

P(X < 16) =

−Φ

75.8

5.175.15

B1

M1 M1

17.5 and 8.75 or 75.8 seen

standardising, with or without cc, must have sd in denom continuity correction 15.5 or 16.5 only, seen

= 1 – Φ(0.676) = 1 – 0.7505 = 0.2495 (0.249 or 0.250) OR 35C00.500.535 + 35C10.510.534 + 35C20.520.533 +... = 8582372584/235 = 0.250

M1 A1

[5] M1 A1 M1 A1 A1

using 1 – Φ(z) correct answer binomial term 35Cx0.5x0.535 − x at least 2 correct terms (x Þ 0) seen summing 16 or 17 terms correct expression correct answer



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Maths Oct Nov 2010