General Certificate of Education June 2007 Advanced Subsidiary Examination
MATHEMATICS Unit Statistics 1B
MS/SS1B
STATISTICS Unit Statistics 1B Thursday 14 June 2007
1.30 pm to 3.00 pm
For this paper you must have: * an 8-page answer book * the blue AQA booklet of formulae and statistical tables. You may use a graphics calculator.
Time allowed: 1 hour 30 minutes Instructions Use blue or black ink or ball-point pen. Pencil should only be used for drawing. Write the information required on the front of your answer book. The Examining Body for this paper is AQA. The Paper Reference is MS/SS1B. Answer all questions. Show all necessary working; otherwise marks for method may be lost. The final answer to questions requiring the use of tables or calculators should normally be given to three significant figures. * *
* * *
Information The maximum mark for this paper is 75. The marks for questions are shown in brackets. Unit Statistics 1B has a written paper only.
* * *
Advice Unless stated otherwise, you may quote formulae, without proof, from the booklet.
*
P94346/Jun07/MS/SS1B 6/6/6/6/
MS/SS1B
2
Answer all questions.
1 The table shows the length, in centimetres, and maximum diameter, in centimetres, of each of 10 honeydew melons selected at random from those on display at a market stall. Length
24
25
19
28
27
21
35
23
32
26
Maximum diameter
18
14
16
11
13
14
12
16
15
14
(a) Calculate the value of the product moment correlation coefficient.
(3 marks)
(b) Interpret your value in the context of this question.
(2 marks)
2 The British and Irish Lions 2005 rugby squad contained 50 players. The nationalities and playing positions of these players are shown in the table. Nationality
Playing position
English
Welsh
Scottish
Irish
Forward
14
5
2
6
Back
8
7
2
6
(a) A player was selected at random from the squad for a radio interview. Calculate the probability that the player was: (i) a Welsh back; (ii) English; (iii) not English;
(1 mark) (2 marks) (1 mark)
(iv) Irish, given that the player was a back;
(2 marks)
(v) a forward, given that the player was not Scottish.
(2 marks)
(b) Four players were selected at random from the squad to visit a school. Calculate the probability that all four players were English. (3 marks)
P94346/Jun07/MS/SS1B
3
3
(a) A sample of 50 washed baking potatoes was selected at random from a large batch. The weights of the 50 potatoes were found to have a mean of 234 grams and a standard deviation of 25.1 grams. Construct a 95% confidence interval for the mean weight of potatoes in the batch. (4 marks) (b) The batch of potatoes is purchased by a market stallholder. He sells them to his customers by allowing them to choose any 5 potatoes for £1. Give a reason why such chosen potatoes are unlikely to represent a random sample from the batch. (1 mark)
4 A library allows each member to have up to 15 books on loan at any one time. The table shows the numbers of books currently on loan to a random sample of 95 members of the library. Number of books on loan
0
1
2
3
4
5–9
10–14
15
Number of members
4
13
24
17
15
11
5
6
(a) For these data: (i) state values for the mode and range;
(2 marks)
(ii) determine values for the median and interquartile range;
(4 marks)
(iii) calculate estimates of the mean and standard deviation.
(4 marks)
(b) Making reference to your answers to part (a), give a reason for preferring: (i) the median and interquartile range to the mean and standard deviation for summarising the given data; (1 mark) (ii) the mean and standard deviation to the mode and range for summarising the given data. (1 mark)
P94346/Jun07/MS/SS1B
s
Turn over
4
5 Bob, a gardener, measures the time taken, y minutes, for 60 grams of weedkiller pellets to dissolve in 10 litres of water at different set temperatures, x °C. His results are shown in the table. x
16
20
24
28
32
36
40
44
48
52
56
y
4.7
4.3
3.8
3.5
3.0
2.7
2.4
2.0
1.8
1.6
1.1
(a) State why the explanatory variable is temperature.
(1 mark)
(b) Calculate the equation of the least squares regression line y ¼ a þ bx.
(4 marks)
(c)
(2 marks)
(i) Interpret, in the context of this question, your value for b.
(ii) Explain why no sensible practical interpretation can be given for your value of a. (2 marks) (d)
(i) Estimate the time taken to dissolve 60 grams of weedkiller pellets in 10 litres of water at 30 °C. (2 marks) (ii) Show why the equation cannot be used to make a valid estimate of the time taken to dissolve 60 grams of weedkiller pellets in 10 litres of water at 75 °C. (2 marks)
6 Each weekday, Monday to Friday, Trina catches a train from her local station. She claims that the probability that the train arrives on time at the station is 0.4, and that the train’s arrival time is independent from day to day. (a) Assuming her claims to be true, determine the probability that the train arrives on time at the station: (i) on at most 3 days during a 2-week period (10 days);
(2 marks)
(ii) on more than 10 days but fewer than 20 days during an 8-week period. (3 marks) (b)
(i) Assuming Trina’s claims to be true, determine the mean and standard deviation for the number of times during a week (5 days) that the train arrives on time at the station. (3 marks) (ii) Each week, for a period of 13 weeks, Trina’s travelling colleague, Suzie, records the number of times that the train arrives on time at the station. Suzie’s results are 2
2
4
1
2
3
3
2
2
0
Calculate the mean and standard deviation of these values. (iii) Hence comment on the likely validity of Trina’s claims.
P94346/Jun07/MS/SS1B
3
2
0 (3 marks) (2 marks)
5
7
(a) Electra is employed by E & G Ltd to install electricity meters in new houses on an estate. Her time, X minutes, to install a meter may be assumed to be normally distributed with a mean of 48 and a standard deviation of 20. Determine: (i) PðX < 60Þ;
(2 marks)
(ii) Pð30 < X < 60Þ;
(3 marks)
(iii) the time, k minutes, such that PðX < kÞ ¼ 0:9 .
(4 marks)
(b) Gazali is employed by E & G Ltd to install gas meters in the same new houses. His time, Y minutes, to install a meter has a mean of 37 and a standard deviation of 25. (i) Explain why Y is unlikely to be normally distributed.
(2 marks)
(ii) State why Y , the mean of a random sample of 35 gas meter installations, is likely to be approximately normally distributed. (1 mark) (iii) Determine PðY > 40Þ.
END OF QUESTIONS
P94346/Jun07/MS/SS1B
(4 marks)
Version 1.0 0607
abc General Certificate of Education
Mathematics 6360 Statistics 6380 MS/SS1B Statistics 1B
Mark Scheme 2007 examination - June series
MS/SS1B - AQA GCE Mark Scheme 2007 June series
Mark schemes are prepared by the Principal Examiner and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation meeting attended by all examiners and is the scheme which was used by them in this examination. The standardisation meeting ensures that the mark scheme covers the candidates’ responses to questions and that every examiner understands and applies it in the same correct way. As preparation for the standardisation meeting each examiner analyses a number of candidates’ scripts: alternative answers not already covered by the mark scheme are discussed at the meeting and legislated for. If, after this meeting, examiners encounter unusual answers which have not been discussed at the meeting they are required to refer these to the Principal Examiner. It must be stressed that a mark scheme is a working document, in many cases further developed and expanded on the basis of candidates’ reactions to a particular paper. Assumptions about future mark schemes on the basis of one year’s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper.
Further copies of this Mark Scheme are available to download from the AQA Website: www.aqa.org.uk Copyright © 2007 AQA and its licensors. All rights reserved. COPYRIGHT AQA retains the copyright on all its publications. However, registered centres for AQA are permitted to copy material from this booklet for their own internal use, with the following important exception: AQA cannot give permission to centres to photocopy any material that is acknowledged to a third party even for internal use within the centre. Set and published by the Assessment and Qualifications Alliance.
The Assessment and Qualifications Alliance (AQA) is a company limited by guarantee registered in England and Wales (company number 3644723) and a registered charity (registered charity number 1073334). Registered address: AQA, Devas Street, Manchester M15 6EX Dr Michael Cresswell Director General
MS/SS1B - AQA GCE Mark Scheme 2007 June series
Key to mark scheme and abbreviations used in marking M m or dM A B E or ft or F CAO CSO AWFW AWRT ACF AG SC OE A2,1 –x EE NMS PI SCA
mark is for method mark is dependent on one or more M marks and is for method mark is dependent on M or m marks and is for accuracy mark is independent of M or m marks and is for method and accuracy mark is for explanation follow through from previous incorrect result correct answer only correct solution only anything which falls within anything which rounds to any correct form answer given special case or equivalent 2 or 1 (or 0) accuracy marks deduct x marks for each error no method shown possibly implied substantially correct approach
MC MR RA FW ISW FIW BOD WR FB NOS G c sf dp
mis-copy mis-read required accuracy further work ignore subsequent work from incorrect work given benefit of doubt work replaced by candidate formulae book not on scheme graph candidate significant figure(s) decimal place(s)
No Method Shown Where the question specifically requires a particular method to be used, we must usually see evidence of use of this method for any marks to be awarded. However, there are situations in some units where part marks would be appropriate, particularly when similar techniques are involved. Your Principal Examiner will alert you to these and details will be provided on the mark scheme. Where the answer can be reasonably obtained without showing working and it is very unlikely that the correct answer can be obtained by using an incorrect method, we must award full marks. However, the obvious penalty to candidates showing no working is that incorrect answers, however close, earn no marks. Where a question asks the candidate to state or write down a result, no method need be shown for full marks. Where the permitted calculator has functions which reasonably allow the solution of the question directly, the correct answer without working earns full marks, unless it is given to less than the degree of accuracy accepted in the mark scheme, when it gains no marks. Otherwise we require evidence of a correct method for any marks to be awarded.
3
MS/SS1B - AQA GCE Mark Scheme 2007 June series
MS/SS1B Q Solution 1(a) r = –0.526 to –0.525 or r = –0.53 to –0.52 or r = –0.6 to –0.4
Marks B3
Total
Comments AWFW
(B2)
AWFW; ignore sign
(B1)
AWFW; ignore sign
OR Attempt at ∑ x , ∑ x2 ,
∑y, ∑y
2
and
∑ xy
260, 6970, 143, 2083 and 3671
or Attempt at S xx , S yy and S xy
(M1)
Attempt at a correct formula for r
(m1)
r = –0.526 to –0.525
(A1)
210, 38.1 and –47
3
(b) Weak/some/moderate negative correlation (relationship/association)
AWFW
B1
OE; must qualify strength and indicate negative B0 for strong/poor/reasonable/average B0 if r > 0 or r < –1 B0 if contradictory statements
B1
Context
between length and (maximum) diameter Ignore subsequent comments (as below) only if B1 B1 already scored OR Some evidence that large lengths are associated with small diameters
(B1) (B1)
OE; must qualify strength and indicate negative
(B1) (B1)
OE; must qualify strength and indicate negative
OR Longer melons tend to have smaller diameters / be thinner Total
4
2 5
MS/SS1B - AQA GCE Mark Scheme 2007 June series
MS/SS1B (cont) Q Solution 2 Ratios: Penalise first occurrence only of a correct answer (a)(i) P(Welsh back) =
(ii) P(English) =
7 or 0.14 50
14 + 8 = 50
Marks
Total
B1
1
B1
CAO; OE
Correct expression; PI
B1
22 11 or or 0.44 50 25
Comments
2
CAO; OE
(iii) P(not English) = 1 – (ii) = 28 14 or or 0.56 50 25
(iv) P(Irish | back) = P ( Irish ∩ back ) = P ( back )
B1
6 = ∑ ( back )
6 or 0.26 to 0.261 23
1
Used; may be implied by values or answer
M1
A1
on (ii) if used; 0 < p < 1
2
CAO/AWFW (6/50 ⇒ 0)
(v) P(forward | not Scottish) =
P ( forward ∩ not Scottish ) = P ( not Scottish ) 14 + 5 + 6 27 − 2 = = 50 − 4 50 − 4 25 or 0.54 to 0.544 46
Used; OE May be implied by values or answer
M1
A1
2
CAO/AWFW (25/50 ⇒ 0)
(b) P(4 × English) = ⎛ 22 ⎞ ⎛ 21 ⎞ ⎛ 20 ⎞ ⎛ 19 ⎞ ⎜ ⎟×⎜ ⎟×⎜ ⎟×⎜ ⎟ = ⎝ 50 ⎠ ⎝ 49 ⎠ ⎝ 48 ⎠ ⎝ 47 ⎠
M1 M1
Reducing non-tabulated value 4 times Reducing 50 and multiplying 4 terms (ignore multipliers)
175560 209 or 5527200 6580 A1
or 0.0317 to 0.032 Total
5
3 11
CAO/AWFW
MS/SS1B - AQA GCE Mark Scheme 2007 June series
MS/SS1B (cont) Q Solution 3(a) 95% ⇒ z = 1.96 or 95% ⇒ t = 2.0 to 2.01 (Knowledge of the t–distribution is not required in this unit)
CI for µ is x ± ( z or t ) ×
Note that 25.1×
Marks B1
( sn−1 or sn ) n
Total
CAO
(B1)
AWFW
M1
Used; must have
234 ± (1.96or 2.009 ) ×
n with n > 1
50 = 25.61224 49 Max of B1 M1 A0 A1
( 25.1 or 25.3to 25.4 )
(
(2.009)
25.1×
50 = 25.35483 49
Thus
Comments
50 or 49
)
A1
on z or t only
Hence 234 ± (6.95 to 7.30) ie or
234 ± 7 (227, 241)
(b) Customers are likely to choose large / similar sized potatoes
A1
4
AWRT
B1
1
OE; accept any sensible alternative
Total
5
6
MS/SS1B - AQA GCE Mark Scheme 2007 June series
MS/SS1B (cont) Q 4(a)(i) Mode = 2
Solution
Range = 15 (ii) CF: x:
Marks B1
Total
B1
2
Comments
CAO CAO
4 17 41 58 73 84 89 95 0 1 2 3 4 9 14 15
Median (48th) = 3
B2
CAO; B0 if shown method is incorrect
Interquartile Range (72nd – 24th) = 4 – 2 = 2
B2
CAO Allow B1 for identification of 4 and 2 B0 if shown method is incorrect
If neither correct but CF attempted and matched correctly with ≥ 5 x–values (iii) Mean ( x ) = 4.2
(M1) (A1)
Allow for median = 2 + 4 CAO
B2
x 17
∑ fx ∑ fx
Standard Deviation ( sn , sn−1 ) = 3.88 to 3.91
AWFW
B2
If neither correct but mid-points of 7 and 12 seen ∑ fx and use of mean ( x ) = 95
(B1)
(b)(i) Unknown values (16) have no effect on median and IQR or median and IQR are exact values but x and s are estimates (ii) Use all available data or Enable further analyses
(M1)
4
B1
1
B1
1
Total
12
7
Allow for 4.1 ≤ x ≤ 4.3
2
= 399 = 3111
(3.887 or 3.907)
MS/SS1B - AQA GCE Mark Scheme 2007 June series
MS/SS1B (cont) Q Solution 5(a) Time taken depends upon temperature
Marks B1
Total 1
Comments OE; not x set values
(b) b (gradient) = –0.0873 to –0.087 b (gradient) = –0.09 to –0.08
B2 (B1)
AWFW AWFW; −8.73−02 ⇒ B0
a (intercept) = 5.94 to 5.96 a (intercept) = 5.6 to 6.1
B2 (B1)
AWFW AWFW
Attempt at
∑x, ∑x , ∑y 2
and
∑ xy
or Attempt at S xx and S xy Attempt at correct formula for b b = –0.0873 to –0.087 a = 5.94 to 5.96
!!) ( −0.08727 !!) ( 5.9509
396, 16016, 30.9 and 958.8 (M1) 1760 and –153.6 (m1) (A1) (A1)
4
AWFW AWFW
Accept a and b interchanged only if then identified correctly later in question (c)(i) Each 1 ºC rise in temperature results in an (average) decrease of 0.087 m (5 s) in time taken for pellets to dissolve
B1 B1
(ii) a is y–value at x = 0 at which water is solid/ice/frozen so pellets cannot dissolve
B1 B1
(d)(i) When x = 30 y = 3.3 to 3.4 y = 2.9 to 3.7
2
2
B2 (B1)
If B0, use of their equation with x = 30 (ii) When x = 75 y < 0 or negative which is impossible
(M1)
B1 ↑Dep↑ B1 Total
8
Quantified rise in x (results in) Decrease in y OE Indication that it is y at x = 0 Mention of solid or ice or frozen AWFW AWFW
2
OE 2 13
OE; not extrapolation
!!) ( 3.3327
MS/SS1B - AQA GCE Mark Scheme 2007 June series
MS/SS1B (cont) Q Solution 6(a) Use of binomial in (a) or (b)(i)
Marks M1
Total
(i) P(T10 ≤ 3) = 0.38 to 0.383
B1
2
(ii) P(10 < T40 < 20) =
0.8702 or 0.9256
M1
Allow 3 dp accuracy
minus
0.0352 or 0.0156
M1
Allow 3 dp accuracy
0.83 to 0.84
A1
AWFW
=
Comments
PI AWFW
(0.3823)
(0.835)
OR B(40, 0.40) expressions stated for at least 3 terms within 10 ≤ T40 ≤ 20
(M1)
Answer = 0.83 to 0.84
(A2)
(b)(i) n = 5
Or implied by a correct answer 3
AWFW
p = 0.4
Mean, µ = np = 2
B1
CAO
Variance, σ 2 = np(1 – p) = 1.2
M1
Use of np(1 – p) even if SD
Standard deviation = 1.2 or = 1.09 to 1.1
A1
(ii) Mean ( x ) = 2
use of mean ( x ) =
∑x 13
(iii) Means are same and SDs are similar/same B1 Means are same but SDs are different so ↑Dep↑ Trina’s claims appear valid / invalid B1 Total
9
∑x
∑x
B2
(M1)
CAO AWFW CAO
B1
Standard Deviation ( sn , sn−1 ) = 1.1 to 1.16 If neither correct but
3
AWFW
2
= 26 = 68
(1.1094 or 1.1547)
3 Must have scored full marks in (b)(i) and (b)(ii) 2 13
MS/SS1B - AQA GCE Mark Scheme 2007 June series
MS/SS1B (cont) Q Solution 7(a) Time, X ~ N(48, 202)
Marks
60 − 48 ⎞ ⎛ (i) P(X < 60) = P ⎜ Z < ⎟ = 20 ⎠ ⎝
Total
Standardising (59.5, 60 or 60.5) with 48 and ( 20 ,20 or 202 ) and/or (48 − x)
M1
P(Z < 0.6) = 0.725 to 0.73
A1
(ii) P(30 < X < 60) = P(X < 60) – P(X < 30) = (i) – P(X < 30) = (i) – P(Z < –0.9) =
Comments
2
AWFW
(0.72575)
M1
Difference or equivalent Standardising other than 60 and 30 ⇒ max of M1 m1 A0
(i) – {1 – P(Z < +0.9)} = 0.72575 – {1 – 0.81594} =
m1
Area change
0.54 to 0.542
A1
AWFW
(0.54169)
B1
AWFW
(1.2816)
k − 48 20
M1
Standardising k with 48 and 20
= 1.2816
m1
Equating z-term to z-value; not using 0.9, 0.1, |1 – z| or Φ(0.9) = 0.81594
(iii) 0.9 ⇒ z = 1.28 to 1.282
z=
k = 73.6 to 74
A1
3
4
AWFW
(b) Time, Y ~ N(37, 252) (i) Use of µ – (2 or 3) × σ = 37 – (50 or 75)
< 0 ⇒ likely negative times (ii) Central Limit Theorem or n large / > 30 (iii) Variance of Y =
Or equivalent justification
M1
252 35
B1
2
B1
1
for (likely) negative times
B1
OE; stated or used
⎛ 40 − 37 ⎞ P (Y > 40 ) = P ⎜ Z > ⎟ = 25 35 ⎠ ⎝
M1
Standardising 40 with 37 and 25 and/or (37 − 40)
P(Z > 0.71) = 1 – P(Z < 0.71) =
m1
Area change
0.238 to 0.24 Total
A1
4 16
TOTAL
75
10
AWFW
35
(1 – 0.76115)
klm
Scaled mark component grade boundaries - June 2007 exams GCE Component Code Component Title
Maximum Scaled Mark
A
GCE MATHEMATICS UNIT PC2 GCE MATHEMATICS UNIT PC3 GCE MATHEMATICS UNIT PC4 GCE MATHEMATICS UNIT S03 GCE MATHEMATICS UNIT S04 GCE MATHS/STATISTICS UNIT 1A - COURSEWORK GCE MATHS/STATISTICS UNIT 1A - WRITTEN GCE MATHEMATICS UNIT S1B GCE MATHEMATICS UNIT S2A - COURSEWORK GCE MATHEMATICS UNIT S2A - WRITTEN GCE MATHEMATICS UNIT S2B
75 75 75 75 75 25 75 75 25 75 75
62 62 60 58 60 20 61 59 20 60 62
55 55 53 50 52 18 54 51 18 51 54
48 48 46 43 44 15 46 43 15 44 46
41 41 39 36 37 13 39 36 13 36 38
35 35 33 29 30 10 31 29 10 29 31
MED1 MED2 MED3 MED4 MED5 MED6
GCE MEDIA STUDIES UNIT 1 GCE MEDIA STUDIES UNIT 2 GCE MEDIA STUDIES UNIT 3 GCE MEDIA STUDIES UNIT 4 GCE MEDIA STUDIES UNIT 5 GCE MEDIA STUDIES UNIT 6
60 60 100 60 60 60
41 39 72 43 50 44
36 34 64 38 41 38
31 29 56 34 32 32
26 25 48 30 23 27
21 21 40 26 15 22
HEB1 HEB2
GCE MODERN HEBREW UNIT 1 GCE MODERN HEBREW UNIT 2
100 100
70 70
62 63
54 57
47 51
40 45
MUS1 MUS2 MUS3 MUS4 MUS5 MUS6
GCE MUSIC UNIT 1 GCE MUSIC UNIT 2 GCE MUSIC UNIT 3 GCE MUSIC UNIT 4 GCE MUSIC UNIT 5 GCE MUSIC UNIT 6
100 60 60 120 80 40
66 46 49 76 67 36
55 40 44 68 63 33
45 34 39 60 59 30
35 28 34 52 55 27
25 23 29 45 51 25
MPC2 MPC3 MPC4 MS03 MS04 MS/SS1A/C MS/SS1A/W MS1B MS2A/C MS2A/W MS2B
Scaled Mark Grade Boundaries B C D
E
