General Certificate of Education January 2006 Advanced Subsidiary Examination
MATHEMATICS Unit Statistics 1B
MS/SS1B
STATISTICS Unit Statistics 1B Thursday 12 January 2006
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 * an insert for use in Question 5 (enclosed) 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. All necessary working should be shown; 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. Fill in the boxes at the top of the insert. * *
* * *
*
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, formulae may be quoted, without proof, from the booklet.
*
P80684/Jan06/MS/SS1B 6/6/6/
MS/SS1B
2
Answer all questions.
1 At a certain small restaurant, the waiting time is defined as the time between sitting down at a table and a waiter first arriving at the table. This waiting time is dependent upon the number of other customers already seated in the restaurant. Alex is a customer who visited the restaurant on 10 separate days. The table shows, for each of these days, the number, x, of customers already seated and his waiting time, y minutes. x
9
3
4
10
8
12
7
11
2
6
y
11
6
5
11
9
13
9
12
4
7
(a) Calculate the equation of the least squares regression line of y on x in the form y a bx. (4 marks) (b) Give an interpretation, in context, for each of your values of a and b.
(2 marks)
(c) Use your regression equation to estimate Alex's waiting time when the number of customers already seated in the restaurant is: (i) 5; (ii) 25.
(2 marks)
(d) Comment on the likely reliability of each of your estimates in part (c), given that, for the regression line calculated in part (a), the values of the 10 residuals lie between +1.1 minutes and 71.1 minutes. (3 marks)
P80684/Jan06/MS/SS1B
3
2 Xavier, Yuri and Zara attend a sports centre for their judo club's practice sessions. The probabilities of them arriving late are, independently, 0.3, 0.4 and 0.2 respectively. (a) Calculate the probability that for a particular practice session: (i) all three arrive late;
(1 mark)
(ii) none of the three arrives late;
(2 marks)
(iii) only Zara arrives late.
(2 marks)
(b) Zara's friend, Wei, also attends the club's practice sessions. The probability that Wei arrives late is 0.9 when Zara arrives late, and is 0.25 when Zara does not arrive late. Calculate the probability that for a particular practice session: (i) both Zara and Wei arrive late;
(2 marks)
(ii) either Zara or Wei, but not both, arrives late.
(3 marks)
3 When an alarm is raised at a market town's fire station, the fire engine cannot leave until at least five fire-fighters arrive at the station. The call-out time, X minutes, is the time between an alarm being raised and the fire engine leaving the station. The value of X was recorded on a random sample of 50 occasions. The results are summarised below, where x denotes the sample mean. X X x 286:5
x x2 45:16 (a) Find values for the mean and standard deviation of this sample of 50 call-out times. (2 marks) (b) Hence construct a 99% confidence interval for the mean call-out time.
(4 marks)
(c) The fire and rescue service claims that the station's mean call-out time is less than 5 minutes, whereas a parish councillor suggests that it is more than 6 12 minutes. Comment on each of these claims.
(2 marks)
P80684/Jan06/MS/SS1B
s
Turn over
4
4 The time, x seconds, spent by each of a random sample of 100 customers at an automatic teller machine (ATM) is recorded. The times are summarised in the table. Time (seconds)
Number of customers
20 < x 4 30
2
30 < x 4 40
7
40 < x 4 60
18
60 < x 4 80
27
80 < x 4 100
23
100 < x 4 120
13
120 < x 4 150
7
150 < x 4 180
3
Total
100
(a) Calculate estimates for the mean and standard deviation of the time spent at the ATM by a customer. (4 marks) (b) The mean time spent at the ATM by a random sample of 36 customers is denoted by Y . (i) State why the distribution of Y is approximately normal.
(1 mark)
(ii) Write down estimated values for the mean and standard error of Y .
(2 marks)
(iii) Hence estimate the probability that Y is less than 112 minutes.
(3 marks)
P80684/Jan06/MS/SS1B
5
5 [Figure 1, printed on the insert, is provided for use in this question.] The table shows the times, in seconds, taken by a random sample of 10 boys from a junior swimming club to swim 50 metres freestyle and 50 metres backstroke. Boy
A
B
C
D
E
F
G
H
I
J
Freestyle (x seconds)
30.2
32.8
25.1
31.8
31.2
35.6
32.4
38.0
36.1
34.1
Backstroke (y seconds)
33.5
35.4
37.4
27.2
34.7
38.2
37.7
41.4
42.3
38.4
(a) On Figure 1, complete the scatter diagram for these data.
(2 marks)
(b) Hence: (i) give two distinct comments on what your scatter diagram reveals;
(2 marks)
(ii) state, without calculation, which of the following 3 values is most likely to be the value of the product moment correlation coefficient for the data in your scatter diagram. 0.912
0.088
0.462
(1 mark)
(c) In the sample of 10 boys, one boy is a junior-champion freestyle swimmer and one boy is a junior-champion backstroke swimmer. Identify the two most likely boys.
(2 marks)
(d) Removing the data for the two boys whom you identified in part (c): (i) calculate the value of the product moment correlation coefficient for the remaining 8 pairs of values of x and y ; (3 marks) (ii) comment, in context, on the value that you obtain.
(1 mark)
P80684/Jan06/MS/SS1B
s
Turn over
6
6 Plastic clothes pegs are made in various colours. The number of red pegs may be modelled by a binomial distribution with parameter p equal to 0.2. The contents of packets of 50 pegs of mixed colours may be considered to be random samples. (a) Determine the probability that a packet contains: (i) less than or equal to 15 red pegs;
(2 marks)
(ii) exactly 10 red pegs;
(2 marks)
(iii) more than 5 but fewer than 15 red pegs.
(3 marks)
(b) Sly, a student, claims to have counted the number of red pegs in each of 100 packets of 50 pegs. From his results the following values are calculated. Mean number of red pegs per packet 10:5 Variance of number of red pegs per packet 20:41 Comment on the validity of Sly's claim.
7
(4 marks)
(a) The weight, X grams, of soup in a carton may be modelled by a normal random variable with mean 406 and standard deviation 4.2. Find the probability that the weight of soup in a carton: (i) is less than 400 grams;
(3 marks)
(ii) is between 402.5 grams and 407.5 grams.
(4 marks)
(b) The weight, Y grams, of chopped tomatoes in a tin is a normal random variable with mean m and standard deviation s. (i) Given that P
Y < 310 0:975, explain why: 310
m 1:96s
(3 marks)
(ii) Given that P
Y < 307:5 0:86, find, to two decimal places, values for m and s. (4 marks)
END OF QUESTIONS
P80684/Jan06/MS/SS1B
Surname
Other Names
Centre Number
Candidate Number
Candidate Signature
General Certificate of Education January 2006 Advanced Subsidiary Examination
MATHEMATICS Unit Statistics 1B
MS/SS1B
STATISTICS Unit Statistics 1B
Insert Thursday 12 January 2006
1.30 pm to 3.00 pm
Insert for use in Question 5. Fill in the boxes at the top of this page. Fasten this insert securely to your answer book.
Turn over for Figure 1
P80684/Jan06/MS/SS1B 6/6/6/
s
Turn over
2
Figure 1 (for use in Question 5) Scatter Diagram for Freestyle and Backstroke Swimming Times c
y 44 ±
I 42 ±
H
40 ± J G
Backstroke (seconds)
38 ±
F
36 ± E 34 ±
32 ±
30 ±
28 ±
26 ±
24 ± c
±
±
±
±
±
±
±
±
±
x
24
26
28
30
32
34
36
38
40
Freestyle (seconds) Copyright Ó 2006 AQA and its licensors. All rights reserved. P80684/Jan06/MS/SS1B
MS/SS1B – AQA GCE Mark Scheme, 2006 January series
Key To Mark Scheme And Abbreviations Used In Marking M
mark is for method
m or dM A B E
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
or ft or F CAO CSO AWFW AWRT ACF AG SC OE A2,1 –x EE NMS PI SCA
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.
2
AQA GCE Mark Scheme, 2006 January series – MS/SS1B
MS/SS1B Q
Solution
Marks
1(a) Gradient, b = 0.886 to 0.887 b = 0.88 to 0.89 Intercept, a = 2.31 to 2.33 a = 2.3 Attempt at Σx Σx2 Σy Σxy or Attempt at Sxx Sxy Attempt at a correct formula for b b = 0.886 to 0.887 a = 2.31 to 2.33
AWFW AWFW
B2 (B1)
AWFW AWRT
(M1)
72, 624, 87, 720 105.6, 93.6
(m1) (A1) (A1)
(b) a: average waiting time of 2.32 minutes (139 seconds) when entering empty restaurant
b: average increase in waiting time of 0.886 minutes (53 seconds) for each customer in restaurant on entry or
y = a + 25b
(i) For x = 5
y = 6.6 to 6.8
(ii) For x = 25
y = 24.3 to 24.6
(ii) Unreliable as extrapolation
AWFW AWFW
4 B1
B1
OE; accept minimum waiting time
2
OE
2
Both; AWFW
M1
A1
(d)(i) Reliable as interpolation and small residuals or Reliable as interpolation but large percentage residuals so inconclusive or Large percentage residuals so unreliable
Comments
B2 (B1)
Accept a & b interchanged only if y = ax + b stated or subsequently used correctly in either (b) or (c)
(c) Use of y = a + 5b
Total
B1 B1
Within range OE OE
(B1) (B1) (B1) B1
Total
3 11
3
Outside range OE
MS/SS1B – AQA GCE Mark Scheme, 2006 January series
MS/SS1B (cont) Q
Solution
Marks
Total
(i) P(X ∩ Y ∩ Z) = 0.3 × 0.4 × 0.2 = 0.024
M1
1
(ii) P(X' ∩ Y' ∩ Z') = 0.7 × 0.6 × 0.8
M1 A1
2(a) P(X) = 0.3
P(Y) = 0.4
P(Z) = 0.2
= 0.336 (iii) P(X' ∩ Y' ∩ Z) = 0.7 × 0.6 × 0.2
(b) P(W | Z) = 0.9
Comments
2
At least 2 correct terms CAO
M1
Correct numerical expression
= 0.084
A1
CAO
= 0.18
M1 A1
P(W | Z') = 0.25
(i) P(Z ∩ W) = 0.2 × 0.9
2
Correct numerical expression CAO
(ii) P((Z ∩ W') ∪ (Z′ ∩ W)) or 1 – [P((Z ∩ W) ∪ (Z′ ∩ W'))] = 0.2 × (1 – 0.9) + (1 – 0.2) × 0.25
M1
0.2 × 0.9 or (b)(i)
M1
(1 – 0.2) × (1 – 0.25) Cannot score an M1 in both methods
= 0.02 + 0.20 = 0.22
A1 Total
4
3 11
1 – (0.18 + 0.60) CAO
AQA GCE Mark Scheme, 2006 January series – MS/SS1B
MS/SS1B (cont) Q
Solution 286.5 = 5.73 3(a) Mean = 50 45.16 Standard deviation = = 49 or 50
Marks
B1
0.95 to 0.961 (b) 99% ⇒ z = 2.57 to 2.58
CI for µ is
Thus
B1
(σ
or s )
2
Use of
!
5.37 to 5.39, 6.07 to 6.09)
A1
(c) CI excludes both values of 5 and 6½ so Neither claim appears valid
(
Dependent 4
B1 ! B1
AWFW on (b); OE Dependent on (b); OE
or
CI excludes 5 so claim not valid and CI excludes 6½ so claim not valid
(B1 ) (B1 ) Total
on (b); OE 2 8
5
)
on z and s 2 > 0 but not on x Accept only 50 or 49 for n
A1
5.73 ± (0.34 to 0.36)
2.5758
Must have ÷ n with n > 1
( 0.95 to 0.961) 50
AWFW AWFW
M1
n
Comments
CAO
B1
x ± z×
5.73 ± 2.5758 ×
Total
on (b); OE
MS/SS1B – AQA GCE Mark Scheme, 2006 January series
MS/SS1B (cont) Q
Solution 4(a)
∑
fx = 8025
∑
fx 2 = 739975
Marks
Comments
Mean ( x ) = 80.2 to 80.3
B2
AWFW
80.25
Standard Deviation (sn, sn-1) = 30.9 to 31.2
B2
AWFW
30.97882 or 31.13489
MPs (x): 25, 35, 50, 70, 90, 110, 135, 165
(B1)
Mean ( x ) =
fx 100 ∑
(b)(i) Large (n > 30) sample or Central Limit Theorem (ii) Mean (Y ) = 80.2 to 80.3
Standard error (Y ) =
At least 4 correct
(M1)
4
Use of
B1
1
OE on (a)
B1
30.9 to 31.2 36 M1
= 5.1 to 5.25 (iii)
Total
⎛ 90 − ( 80.2 to 80.3) ⎞ P (Y < 90 ) = P ⎜⎜ Z < ⎟ ( 5.1 to 5.25) ⎟⎠ ⎝
2
s 2 > 0 in (a) ÷
Standardising 90 Using values from (b)(ii) with
M1 M1
s 2 36 > 0 or
= P(Z < 1.84 to 1.93) = 0.967 to 0.974
A1 Total
6
36 or 6
3 10
AWFW
s 2 100 > 0
AQA GCE Mark Scheme, 2006 January series – MS/SS1B
MS/SS1B (cont) Q
Solution
Marks
5(a) Scatter Diagram or or
B2 (B1) (B1)
(b)(i) Positive/linear correlation/relationship except for two unusual values/results
Total
2
B1
(ii) 0.462 (c) C and D
Comments
4 labelled points plotted 3 labelled points plotted 4 unlabelled points plotted OE
B1
2
OE
B1
1
CAO; accept 3rd/final/last value
B1
CAO
C is likely freestyle champion D is likely backstroke champion
B1
Style identified
C is likely freestyle champion D is likely backstroke champion
(B1) (B1)
or
r = 0.912 to 0.913
(d)(i)
2
B3
AWFW
r = 0.91 to 0.92 or 0.46 to 0.47
B2
AWFW
r = 0.9
B1
AWRT
or or
Attempt at or Attempt at
Σx Σx2 Σy Σy2 Σxy
270.4, 9188.46 301.6, 11437.84 10246.53
Sxx Syy Sxy
(M1)
48.94, 67.52, 52.45
Attempt at a correct formula for r
(m1)
r = 0.912 to 0.913
A1
3
AWFW
(ii) Boys are faster/slower at both strokes or Boys are equally good at both strokes
B1
1
OE;do not accept freestyle times are proportional to backstroke times
Total
7
11
MS/SS1B – AQA GCE Mark Scheme, 2006 January series
Question 5(a)
Swimming Times 44.0 I
42.0
H
40.0
Backstroke ( y seconds)
38.0
J G
C
36.0
F
B E
34.0
A
32.0 30.0 28.0 D 26.0 24.0 24.0
26.0
28.0
30.0
32.0
34.0
Freestyle ( x seconds)
(a) Scatter Diagram 4 labelled points plotted 3 labelled points plotted 4 unlabelled points plotted
B2 (B1) (B1)
Graph = 2
8
36.0
38.0
40.0
AQA GCE Mark Scheme, 2006 January series – MS/SS1B
MS/SS1B (cont) Q
Solution
6(a)(i) B(50, 0.2) P(R ≤ 15) = 0.969 to 0.97
Marks
Total
M1 A1
2
(ii) P(R = 10) = P(R ≤ 10) – P(R ≤ 9)
Comments
Use of in (a) AWFW
0.9692
Stated or implied
or
M1
⎛ 50 ⎞ 10 40 P(R = 10) = ⎜ ⎟ ( 0.2 ) ( 0.8 ) ⎝ 10 ⎠
Stated or implied
= 0.5836 – 0.4437 = 0.139 to 0.141
A1
(iii) P(5 < R < 15) = P(R ≤ 14 or 15) = 0.9393 or 0.9692
M1
Accept values to 3 dp
M1
Accept values to 3 dp
A1
AWFW
minus P(R ≤ 5 or 4) = 0.0480 or 0.0185 = 0.89 to 0.893
2
AWFW
0.1399
0.8913
or
B(50, 0.2) expressions stated for at least 3 of 5 ≤ R ≤ 15 Answer (b) Mean, µ = np = 50 × 0.2 = 10
(M1) (A2)
Or implied by a correct answer 3
B1
Either; CAO
B1
CAO
or
Estimate of p, pˆ = 0.21 Variance, σ 2 = np (1 − p ) = 10 × 0.8 = 8 Mean or Estimate of p is similar to that expected but Variance (standard deviation) is different from that expected
B1
Reason to doubt validity of Sly’s claim
B1
10.5 and 10 or 0.21 and 0.2 Either point 20.41 and 8 or 4.5 and 2.8
Total
4 11
9
Must be based on both 10 or 0.2 and 8 or on both 10 or 0.2 and 2.8 correctly
MS/SS1B – AQA GCE Mark Scheme, 2006 January series
MS/SS1B (cont) Q
Solution
Marks
Comments
Weight, X ~ N(406, 4.22)
7 (a)
400 − 406 ⎞ ⎛ P(X < 400) = P ⎜ Z < ⎟ 4.2 ⎝ ⎠
(i)
= P(Z < –1.428 to –1.43) = 1 – P(Z < 1.428 to 1.43) = 0.076 to 0.077
P(Z < 0.36) – P(Z < –0.83) = 0.64058 – (1 – 0.79673) = 0.433 to 0.44
or
σ
m1
Φ(–z) = 1 – Φ(z) 3
0.07636
Difference OE
B2,1
AWRT; ignoring signs 4
AWFW
0.43731
Accept explanation in words Standardising 310 using µ and σ
M1 Accept in words
Equating
= 1.96 ⇒ result
310 = µ + 1.96σ
AWRT
M1
M1
310 − µ ⎞ ⎛ P(Y < 310) = P ⎜ Z < σ ⎟⎠ ⎝ or x = µ + / ± zσ 310 − µ
Standardising (399.5, 400 or 400.5) with 406 and ( 4.2 , 4.2 or 4.22) and/or (406 − x)
A1
(b)(i) 0.975 ⇒ z = 1.96
Thus
M1
A1
(ii) P(402.5 < X < 407.5) = P(X < 407.5) – P(X < 402.5) =
m1
AG
⇒ result
Substitution
NB: Working backwards from given equation ⇒ at most M1 M0 mo (ii)
Total
3
0.86 ⇒ z = 1.08
B1
AWRT
1.0803
M1
Attempt at solving 2 equations each of form x – µ = zσ
σ = 2.84 to 2.842
A1
AWFW
2.841
µ = 304.4 to 304.5
A1
AWFW
304.43
310 – µ = 1.96σ 307.5 – µ = 1.08σ 2.5 = 0.88σ
Total TOTAL
10
4 14 75
AQA January 2006 Examinations Scaled Mark Component Grade Boundaries (GCE Specifications) Comp. Code MM2B MPC1 MPC2 MPC3 MPC4 MS/SS1A/W MS/SS1A/C MS1B MS2A/W MS2A/C MS2B
Component Title GCE MATHEMATICS UNIT M2B GCE MATHEMATICS UNIT PC1 GCE MATHEMATICS UNIT PC2 GCE MATHEMATICS UNIT PC3 GCE MATHEMATICS UNIT PC4 GCE MATHEMATICS UNIT S1A - WRITTEN GCE MATHEMATICS UNIT S1A - COURSEWORK GCE MATHEMATICS UNIT S1B GCE MATHEMATICS UNIT S2A - WRITTEN GCE MATHEMATICS UNIT S2A - COURSEWORK GCE MATHEMATICS UNIT S2B
Maximum Scaled Mark 75 75 75 75 75 75 25 75 75 25 75
MAD2 MAM2/W MAM2/C MAM3 MAM4/W MAM4/C MAP2 MAP3 MAP4 MAP5 MAP6 MAS1/W MAS1/C MAS2/W MAS2/C MAS3 MAS4/W
GCE MATHEMATICS A UNIT D2 GCE MATHEMATICS A UNIT M2 - WRITTEN GCE MATHEMATICS A UNIT M2 - COURSEWORK GCE MATHEMATICS A UNIT M3 GCE MATHEMATICS A UNIT M4 - WRITTEN GCE MATHEMATICS A UNIT M4 - COURSEWORK GCE MATHEMATICS A UNIT P2 GCE MATHEMATICS A UNIT P3 GCE MATHEMATICS A UNIT P4 GCE MATHEMATICS A UNIT P5 GCE MATHEMATICS A UNIT P6 GCE MATHEMATICS A UNIT S1 - WRITTEN GCE MATHEMATICS A UNIT S1 - COURSEWORK GCE MATHEMATICS A UNIT S2 - WRITTEN GCE MATHEMATICS A UNIT S2 - COURSEWORK GCE MATHEMATICS A UNIT S3 GCE MATHEMATICS A UNIT S4 - WRITTEN
60 70 30 60 70 30 60 60 60 60 60 70 30 70 30 60 70
Page 8 of 12
A 58 61 58 61 62 60 20 58 60 20 59 48 56 24 45
47 48 46 46 50 55 24 55 24
klm
Scaled Mark Grade Boundaries B C D 50 42 34 53 45 38 51 44 37 54 47 40 54 46 38 53 45 38 18 15 13 50 43 36 53 45 38 18 15 13 51 43 35 42 36 30 49 42 35 21 18 15 39 34 29 no candidates entered for this component no candidates entered for this component 41 35 29 42 36 30 40 34 29 40 34 28 44 38 32 48 41 35 21 18 15 48 41 34 21 18 15 no candidates entered for this component no candidates entered for this component
E 27 31 30 33 31 31 10 29 30 10 27 24 28 12 24
24 24 24 22 26 29 12 27 12
