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Mathematics Practice Question Paper
STATISTICS 1, S1
Practice Paper S1-A
TIME 1 hour 30 minutes
INSTRUCTIONS



Write your Name on each sheet of paper used or the front of the booklet used.
Answer all the questions.
You may use a graphical calculator in this paper.
INFORMATION




The number of marks is given in brackets [] at the end of each question or part-question.
You are advised that you may receive no marks unless you show sufficient detail of the working to
indicate that a correct method is being used.
Final answers should be given to a degree of accuracy appropriate to the context.
The total number of marks for this paper is 72.
Section A (36 marks)
1
The box-and-whisker plot in Fig. illustrates the scores, out of 80, of 120 people in a diving
competition.
38
50
60
65
72
Draw a cumulative frequency graph to illustrate these data.
2
[4]
100 people attend a music festival. They are asked which, if any, of the instruments piano, cello,
violin they play.
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Their answers are illustrated in Fig.
Piano
Cello
2
30
9
5
8
24
10
12
Violin
A person is chosen at random from those attending the festival and asked which of the three instruments
he or she plays.
Find the probability that this person plays
(i)
3
the piano,
(ii) exactly one of the other instruments given that he or she plays the piano.
In a year group of three classes the distribution of sexes is given in the table below.
Males
Females
Class 1
10
15
Class 2
11
9
Class 3
9
9
Three students are selected, one from each class, at random.
Find the probability that
(i)
all 3 are male,
(ii)
only one is male.
4
[4]
[2]
[3]
A train company runs a non-stop service from Oxbridge to Camford. The numbers of passengers on
the 07:30 service on 20 weekdays were as follows.
184
193
195
189
173
175
171
178
174
163
184
162
171
154
199
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217
(i)
Mathematics Practice Question Paper
187
169
183
186
Calculate the median and the inter-quartile range.
[3]
(ii) Using the inter-quartile range, show that there is just one outlier. Find the effect of its removal
on the median and the inter-quartile range.
5
[4]
A random sample of cyclists were asked how many days they had used their bicycles in the last
week. The results are given in the following table.
(i)
Number of days (x)
0
1
2
3
4
5
6
7
Frequency (f)
15
10
9
5
7
24
8
2
Illustrate the distribution using a suitable diagram and describe its shape.
[3]
(ii) Calculate the mean and the standard deviation, s, of the data. Give your answers to 4 decimal
places.
6
[3]
(iii) As a reward for taking part in the survey, the cyclists’ names are entered for a draw. There are
3 identical prizes. In how many ways can the 3 winners be chosen ?
[2]
In one turn of the game of Polopoly a player throws three ordinary dice, the score being the largest of
the numbers appearing face up. The score, X, is given by the probability distribution given in the
following table.
r
P(X = r)
(i)
1
2
3
4
5
6
1
216
7
216
19
216
37
216
61
216
91
216
Find E(X) and Var(X) .
[4]
(ii) Find the probability that the player will score a total of exactly 10 in two turns.
[4]
Section B (36 marks)
7
A survey is conducted to find which type of property people live in and whether the property is
owned or rented by its occupier. The results for a particular region of the country are as follows.
Type of Property
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Proportion of
Proportion of properties
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each type
Owned
Rented
Detached / semi-detached
45%
75%
25%
Terraced house
35%
50%
50%
Flat / bedsit
20%
35%
65%
A property is chosen at random.
(i)
Construct a tree diagram to represent the information in the table.
[3]
(ii)
Find the probability that the property is owned.
[3]
(iii) Find the probability that the property is a terraced house or rented.
[4]
(iv)Given that the property is owned, calculate the probability that it is a terraced house.
[3]
Two properties are now chosen at random.
(v)
8
Find the probability that they are
(A)
of the same type,
(B)
of different types.
[5]
Phil likes rifle shooting at an amusement arcade. He reckons that he can hit the target on 3 out of 4
shots on average. Each “go” at the amusement arcade consists of 10 independent shots at a moving
target. A prize is awarded if at least 9 shots hit the target.
(i)
Show that the probability that Phil wins a prize in one “go” is 0.244, correct to 3 significant
figures.
[2]
(ii) Phil has 3 “goes”. Find the probability that he wins
(A)
exactly one prize,
(B)
at least one prize.
[6]
(iii) How many “goes” does Phil need to have so that the probability of winning at least one prize is
more than 90% ?
[4]
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Val is less experienced at rifle shooting. She thinks that she has an even chance of hitting the target
with one shot. Phil thinks that she has a better chance of hitting the target. He conducts a hypothesis
test at the 10% significance level by getting Val to have 10 shots at the target.
(iv) Write down suitable hypotheses for this test in terms of p, the probability that Val hits the
target, giving a reason for your alternative hypothesis.
[3]
(v) Find the least number of times Val should hit the target to suggest that Phil is correct.
[3]
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Practice Paper S1-A
MARK SHEME
Qu
Answer
Mark
Comment
Section A
1
G1
Correctly scaled axes,
with attempted ogive.
G1
Maximum &
minimum points
plotted
G1
Median plotted
G1
4 Quartiles plotted
Curve or line
segments accepted
2
(i)
(ii)
3
(i)
(ii)
P(plays piano) =
P(plays one other instrument | plays piano) =
P(all 3 male) =
P(1 male) =
=
4
(i)
M1
A1
45
or 0.45
100
2
10
2
=
45
9
10 11 9
11



or 0.11
25 20 18
100
10 9 9 15 11 9 15 9 9

 

 


25 20 18 25 20 18 25 20 18
39
or 0.39
100
Median = 180.5
Inter-quartile range
[ or
For (30 + 8 + 5 + 2)
= 188 – 171 = 17
= 188.5 – 171 = 17.5 ]
M1
A1
For
2
M1
A1
n
45
Product of 3 terms
2
M1
M1
Product of 3 terms
Digits correct on top
of at least one
A1
3
B1
For median
M1
A1
For sensible attempt at
finding IQR
3
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(ii)
Mathematics Practice Question Paper
Q1 – 1.5  IQR = 171 – 1.5  17 = 145.5
If 217 is removed, median drops to 178
B1
For showing 217 is >
1.5 IQR above Q3
For showing there are
no values < 1.5 IQR
below Q1
For effect on median
IQR becomes 187 – 171 = 16 or 187.5 – 171 = 16.5
B1
For effect on IQR
Q3 + 1.5  IQR = 188 + 1.5  17 = 213.5
E1
E1
Hence only data item outside the interval [145.5, 213.5]
is 217.
or 186.25 – 170.5 = 15.75
5
4
(i)
G1
G1
For linear scales on
Distribution is bimodal
B1
both axes
3
For heights of lines of
vertical line chart
(ii)
Mean =
253
= 3.1625 days (to 4 d.p.)
80
1189  80  3.16252
79
4.922626582 = 2.2187 (to 4 d.p.)
Standard deviation =
=
B1
For comment
For mean
M1
For variance
A1
3
(iii) Number of ways of choosing the 3 winners
For nC3
M1
A1
2
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=
(i)
6
80
C3 = 82160
E(X) = rP( X  r ) =
=
Mathematics Practice Question Paper
1071
216
1
216
(1 x 1 + 2 x 7 + ... + 6 x 91)
A1
= 4.96 (to 3 s.f.)
1
r 2 P( X  r ) =
216
(12 x 1 + 22 x 7 + ... + 62 x 91)
 1071 

 Var(X) = 5593


216
 216 
2
= 1.31 (to 3 s.f.)
= P(4, 6) + P(5, 5) + P(6, 4)
37
216

91
216

61
216

61
216

91
216

A1
4
For  2 pairs soi
For a product of 2
correct probabilities
For sum of 3 correct
products
M1
M1
P(score exactly 10 in 2 turns)
=
For r 2 P( X  r )
M1
= 5593
216
(ii)
For rP( X  r )
M1
M1
37
216
A1
= 0.224 (to 3 s.f.)
4
Total = 36
Qu
Answer
Section B
7 (i)
Mark
0.75
Owned
Detached
or semi-d
0.45
0.35
0.20
(ii)
Comment
0.25
Rented
0.50
Owned
0.50
Rented
0.35
Owned
0.65
Rented
Terraced
B1
For overall structure
B1
For 1st set branches
B1
For 2nd set branches
Flat or
bedsit
P(property is owned)
= 0.45  0.75 + 0.35  0.50 + 0.20  0.35
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3
M1
M1
For one product
For sum of 3 prods
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= 0.5825
A1
(iii) P(property terraced or rented)
= P(terraced) + P(rented) – P(terraced and rented)
= 0.35 + (1 – 0.5825)  0.35  0.50
= 0.5925
or
0.45  0.25 + 0.35 + 0.20  0.65 = 0.5925
(iv) P(property terraced | owned)
P(property terraced and owned)
=
P(property owned)
0.35  0.5
=
= 0.30 (2 s.f.)
0.5825
(v)
P(each is the same type of property)
= 0.452 + 0.352 + 0.202
3
M1
M1 A1
A1
or
M1 A1
M1
A1
4
M1
M1
For “addition law”
for terms
or
For 2 products
For sum
For numerator
For quotient
A1
3
M1
M1
A1
For “p2”
For sum of 3 squares
M1
For “1 – their 0.365”
= 0.365
P(each is a different type of property)
= 1 – 0.365
A1
= 0.635
[ or 2  0.45  0.35 + 2  0.45  0.20 + 2  0.35  0.20
= 0.635 ]
8
(i)
5
[ Let X ~ B(10, 0.75) ]
P(Phil wins a prize) =
P(X  9) = 1 – P(X  8) = 1 – 0.7560
[ or
= 10  0.759  0.25 + 0.7510
= 0.1877… + 0.0563… ]
M1
A1
P(X  9) = 0.244 (to 3 s.f.)
(ii)(A) [ Let Y ~ B(3, 0.244) ]
P(Y = 1) = 3  0.244  0.7562
2
P(Y  1) = 1 – P(Y = 0) = 1 – 0.7563
Maths Practice Question Paper
For “0.244  0.7562”
For “3  p  q2”
M1
M1
A1
= 0.418 (3 s.f.)
(ii)(B)
For use of tables
3
M1
M1
For “0.7563”
For “1 – p3”
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= 0.568 (3 s.f.)
A1
3
(iii)
[ Let n represent the number of goes, then ]
Require 1 – 0.756n > 0.9  0.756n < 0.10
By trial:
or by logs:

1 – 0.7568 = 0.893 < 0.90
1 – 0.7569 = 0.919 > 0.90
(v)
For “1 – 0.7563”
M1
For inequality
M1
For attempt at solving
inequality
n log(0.756) < log(0.10)
n>
log(0.10)
= 8.23
log(0.756)
hence Phil needs to have 9 goes.
(iv)
M1
A1
4
H0: p = 0.5
B1
For null hypothesis
H1: p > 0.5
B1
For alternative hypothesis
since we want to see if Val is more likely to hit the
target than not.
E1
Using binomial tables for n = 10:
M1
For one comparison
P(X  7) = 1 – P(X  6) = 1 – 0.8281
= 0.1719 > 0.10
M1
For 2nd comparison
For reason
3
P(X  8) = 1 – P(X  7) = 1 – 0.9453
= 0.0547 < 0.10
A1
So Val should hit the target at least 8 times.
3
Total = 36
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Practice Paper S1-B
TIME 1 hour 30 minutes
INSTRUCTIONS



Write your Name on each sheet of paper used or the front of the booklet used.
Answer all the questions.
You may use a graphical calculator in this paper.
INFORMATION




The number of marks is given in brackets [] at the end of each question or part-question.
You are advised that you may receive no marks unless you show sufficient detail of the working to
indicate that a correct method is being used.
Final answers should be given to a degree of accuracy appropriate to the context.
The total number of marks for this paper is 72.
Section A (36 marks)
1
In a certain region of the country the percentages of blood donors in each of four different blood
groups are as follows.
Group
Percentage
(i)
O
A
B
AB
45%
43%
9%
3%
Two of the region’s donors are chosen at random. Find the probability that
(A)
at least one has blood group O,
[2]
(B)
both have the same blood group.
[3]
(ii) Show that at least 8 donors would have to be chosen, at random, to be 99% sure of finding at
least one donor of blood type O.
[3]
2
A company which hires out equipment by the day has three mowers. The number, X, of mowers
which are hired on any one day has the following probability distribution.
1
P(X = r) = k  
 2
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r
for r = 0, 1, 2 and 3.
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Mathematics Practice Question Paper
8
.
15
(i)
Show that k =
[2]
(ii)
Sketch the probability distribution of X .
[2]
(iii) Calculate the expectation and variance of X.
3
[4]
Next September I intend to buy a new car. Its registration plate will be of the form
HW 55 MSD
where HW is the local area code for the Isle of Wight, 55 represents the second half of the year
2005, and the last three letters are chosen at random.
Both parts of the question refer to the last three letters of the registration plate. You may assume that
each of the 26 letters in the alphabet, of which 5 are vowels and 21 are consonants, is available for
each of the random choices.
(i)
Find the probability that the random letters on the plate are MSD, appearing in that order.
[2]
(ii) Find the probability that the letters are M, S, D in any order.
4
[2]
New-born babies are tested for a mild illness which affects 1 in 500 babies. The result of a test is
either positive or negative. A positive result suggests that the baby has the illness. However, the test
is not perfect.


For babies with the illness, the probability of a positive result is 0.99
For babies without the illness, the probability of a negative result is 0.95
A new-born baby is chosen at random and tested for the illness.
(i)
Copy and complete this probability tree diagram to illustrate the situation.
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(ii) Find the probability that the result is positive.
[3]
(iii) Given that the result of a test is positive, show that the conditional probability that the baby has
the illness is 0.038 (correct to 3 decimal places).
[3]
5
The magazine Nearly Eighteen has a web site, on which it recently ran a pop trivia quiz with ten
questions. The results of the first 1000 entries were analysed. Summary statistics for the numbers of
questions answered correctly, x, and associated frequencies, f, are as follows:
n = 1000
(i)
(ii)
xf = 6100
x2f = 42260
Show that the mean score is 6.1.
[1]
Calculate the mean square deviation and the variance, s2, of the data. Comment on the relative
size of the two answers.
[3]
Each question in the quiz was in fact of the multiple choice variety, with four possible answers.
Three points are awarded for a question answered correctly and one point is deducted for a question
which is not answered correctly.
(iii) Show that, if x questions are answered correctly, the number of points, y, is given by
y = 4x  10 .
[1]
(iv)
Hence find the mean and variance of the points scored.
[3]
Section B (36 marks)
6
A motoring magazine carried out a survey of the value of 60 petrol-driven cars that were five years
old. In the survey, the value of each car was expressed as a percentage of its value when new. The
results of the survey are summarised in the following table.
Percentage of original
value (x)
15  x < 20
Number of cars
20  x < 25
12
25  x < 30
18
30  x < 35
13
35  x < 40
6
40  x < 45
5
45  x < 55
2
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4
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(i)
Mathematics Practice Question Paper
Draw a histogram on graph paper to illustrate the data.
[4]
(ii) Carefully describe the shape of the distribution.
[2]
(iii) Calculate an estimate of the median of the data.
[2]
(iv) Use your calculator to find estimates of the mean and standard deviation of the data, giving your
answers correct to 2 decimal places.
[4]
(v) Hence identify any outliers, explaining your method.
[4]
A similar survey of 60 five-year old diesel-driven cars produced a mean of 34.2% and a standard
deviation of 11.7%.
(vi) Use these statistics to compare the values of petrol and diesel cars, five years after they were
purchased as new. [2]
7
A motoring organisation reports that the proportion of drivers who would fail a basic sight test is 1 in
6.
For parts (i) to (iii) you may assume that the report is correct.
(i)
Write down the value of p, the probability that a driver chosen at random would pass the sight
test.
[1]
(ii) A random sample of 30 drivers is taken.
(A)
How many would be expected to pass the test ?
(B)
Find the probability that exactly this number pass the test.
[4]
(iii) A random sample of n drivers is taken.
(A)
Find the probability that all pass the sight test when n = 13.
(B)
Find the smallest value of n such that the probability of all drivers passing is less than 5%
.
[4]
A journalist wishes to test the accuracy of the motoring organisation’s report by checking the sight of a
random sample of 15 drivers.
(iv) Write down suitable hypotheses for this test in terms of p.
[2]
(v) Find the critical region for the test at the 10% significance level and illustrate it on a number
line.
[5]
(vi) Find the minimum sample size for which the upper tail of the critical region would not be
empty.
[2]
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Mathematics Practice Question Paper
Practice Paper S1-B
MARK SHEME
Qu
Answer
Mark
Comment
Section A
1
(i)
(A)
P(at least one has blood group O)
= 1 – 0.552 = 0.698 (to 3 s.f.)
M1
A1
P(both have same blood group)
M1
M1
A1
For calculation
2
(B)
= 0.452 + 0.432 + 0.092 + 0.032
= 0.396 (to 3.s.f.)
(ii)
For at least 2 squares
For sum of 4 squares
3
For n = 7, P(at least one has blood group O)
= 1 – 0.557 = 0.985 (to 3 s.f.) < 99%
M1
Attempt at least one
evaluation & comparison
For n = 8, P(at least one has blood group O)
A1
Both evaluations correct
E1
Comparisons
= 1 – 0.558 = 0.992 (to 3 s.f.) > 99%
3
2
r
0
1
2
3
P(X = r)
k
1k
2
1k
4
1k
8
15
8
k = 1  k =
8
15
(i)
k 1 12  14  18  = 1 
M1
For forming equation
For solution with fractions
A1
2
(ii)
G1
For lines in proportion
G1
(dependent)
For scaled axes
2
(iii) E(X)
= r P(X = r)
M1
For E(X)
A1
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= 0x
8
15
4
15
+1x
+2x
Mathematics Practice Question Paper
2
15
+3x
1
15
=
Var(X) = E(X2) – [E(X)]2
= 0x
=
3
(i)
194
225
8
15
+1x
4
15
11
15
M1
A1
+4x
2
15
+9x

1
15
 1511 
2
4
= 0.862 (to 3 s.f.)
P(letters on plate are MSD, in that order)
3
1
 1 
=   
= 0.000057 (to 2 s.f.)
17576
 26 
(ii)
For E(X2)
P(letters on plate are M, S, D in any order)
 1 
For  
 26 
M1
A1
3
2
For “3!  their part (i)”
M1
3
6
3
 1 
= 3!    
= 0.00034 (to 2 s.f.)

17576
8788
 26 
4
(i)
A1
2
Tree diagram:
0.002
0.998
0.99
Positive
result
0.01
Negative
result
0.05
Positive
result
Has the
illness
Does not
have the
illness
0.95
B1
B1
Negative
result
2
For 3 correct
probabilities
For 2 further correct
probabilities
(ii)
P(result is positive) = 0.002  0.99 + 0.998  0.05
= 0.05188 = 0.052 (to 2 s.f.)
M1
M1
A1
For one product
For sum of 2 products
3
(iii)
P(baby has the illness | positive test result)
=
P(baby has the illness and test result positive)
P(test result positive)
M1
M1 for numerator
M1
M1 for quotient with
their part (ii)
A1
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Mathematics Practice Question Paper
3
0.002  0.99
=
= 0.038 (3 d.p.)
0.05188
5
(i)
(ii)
Mean =
6100
= 6.1
1000
B1
For mean
1
42260  1000  6.12 5050
msd =
=
= 5.05
1000
1000
M1
For Sxx
42260  1000  6.12 5050
variance =
=
= 5.055
999
999
A1
For both values
They are almost the same due to large n.
Number of points = 3x – (10 – x) = 4x – 10
E1
For reference to n
3
For “3x – (10 – x)”
B1
(iii)
1
y = 4 x  10 = 4  6.1 – 10 = 14.4
(iv)
variance(y) = 16  variance(x)
M1
A1
For y
B1
= 16  5.055… = 80.9 (to 3 s.f.)
3
For “16  their
variance(x)”
Total = 36
Qu
Answer
Section B
6
Mark
Comment
G1
For linear scaled axes
G1
For frequency density
or equivalent or key
G1
For heights of first 6
bars (joined); all
correctly positioned
G1
For size of 7th bar
(i)
4
(ii)
Shape of distribution: It has positive skew.
B1
B1
Find median by simple interpolation
M1
For skew
For positive
2
(iii)
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For identifying interval
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= 25 +
(iv)
14.5
 5 = 29
18
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or 25 +
14
 5 = 28.9
18
Mid-interval points:
17.5, 22.5, 27.5, 32.5, 37.5, 42.5, 50
1795
Mean =
= 29.92 (to 2 d.p.)
60
57112.5  60  29.922
= 7.60 (to 2 d.p.)
59
Mean  2 s.d. = 29.92  2 x 7.60 = 14.71 (to 2 d.p.)
containing median
A1
2
B1
For mid-interval points
B1
For mean
M1
A1
For variance
ft their mean
s.d. =
(v)
(vi)
4
M1
For attempt at x  2sd
Mean + 2 s.d. = 29.92 + 2 x 7.60 = 45.13 (to 2 d.p.)
A1
For both values
Hence the 2 percentages in range 45 to 55 could be
outliers, since they may lie more than 2 s.d. from the
mean.
M1
A1
For “2 outliers”
or equivalent
On average cars with diesel engines held their value
better than cars with petrol engines.
E1
Greater variation in the percentage values for cars with
diesel engines compared to cars with petrol engines.
7
For precise value
4
or equivalent
E1
or equivalent
2
(i)
p = 56
B1
(ii)(A)
Expected number to pass = 30  56 = 25
B1
For expected number
M1
For (5/6)r  (1/6)30-r
For probability
1
2 For 30Cr  …
(ii)(B)
(iii)(A)
30
P(exactly 25 pass) =
C25 
 56    16 
25
5
= 0.192 (to 3 s.f.) = 0.19 (to 2 s.f.)
P(all pass sight test)
M1
A1
2
M1
For use of tables or
working out
= 1 – 0.9065 = 0.0935 [using tables]
or
(iii)(B)
=
 56 
13
= 0.0935 (to 3 s.f.)
A1
2
Searching for appropriate n:
for n = 16: P(all pass)
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For attempt at search
M1
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= 1 – 0.9459 [tables] or
for n = 17: P(all pass)
= 1 – 0.9549 [tables] or
 165 
16
 165 
17
= 0.0541
A1
2
= 0.0451
hence smallest sample size is where n = 17
(iv)
(v)
H0: p = 56 ;
B1,1
H1: p  56
For hypotheses
2
Using binomial tables for n = 15:
M1
For at least one
comparison
P(X  9) = 0.0274 < 0.05,
but P(X  10) = 0.0898 > 0.05 A1
So lower tail of crit. reg. is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
M1
P(X  15) = 1 – P(X < 14) = 1 – 0.9351
A1
= 0.0649 > 0.05, So upper tail of crit. reg. is empty.
Critical region
For comparison
Acceptance region
G1
For diagram
5
0
(vi)
9 10
15
From part (iv), for n = 16: P(all pass) =
0.0541 > 5%,
and for n = 17: P(all pass) = 0.0451 <
5%.
M1
For comparisons
A1
2
Hence minimum sample size for which upper tail is not
empty is 17.
Total = 36
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