Statistics and Probability

workbook markscheme
~ categorized past IB Paper 1 and Paper 2 examination questions ~

IB DP Mathematics Standard Level

Topic 5
This workbook contains past Paper 1 and Paper 2 IB examination questions categorized
according to major concepts in this topic.

Contents

Statistics
• 5.1 Measures of Central Tendency
Paper 1 Questions page 1
4 questions; 21 marks
Paper 2 Questions page 2
6 questions; 42 marks

• 5.2 Measures of Spread

Paper 1 Questions page 5
9 questions; 53 marks
Paper 2 Questions page 7
2 questions; 12 marks

• 5.3 Cumulative Frequency

Paper 1 Questions page 8
8 questions 47 marks
Paper 2 Questions page 11
12 questions; 131 marks

Probability
• 5.4 Simple Probability
Paper 1 Questions page 18
16 questions 86 marks
Paper 2 Questions page 23
4 questions; 31 marks

• 5.5 Conditional Probability

Paper 1 Questions page 25
17 questions 118 marks
Paper 2 Questions page 31
8 questions; 90 marks
• 5.6 Discrete Probability Distributions
Paper 1 Questions page 38
7 questions 68 marks
Paper 2 Questions page 42
6 questions; 71 marks

• 5.7 Binomial Distribution

Paper 1 Questions page 46
2 questions 16 marks
Paper 2 Questions page 47
12 questions; 92 marks

• 5.8 Normal Distribution

Paper 1 Questions page 52
3 questions; 21 marks
Paper 2 Questions page 53
27 questions; 247 marks

The use of GDC is not permitted for Paper 1 but is required for Paper 2 questions.
5.1 Measures of Central Tendency Paper 1
1. (a) 18 A1 N1
(b) (i) 10 A2 N2
(ii) 44 A2 N2
[5]

2. (a) Mean =
∑fx
∑f
∑fx = (1) (0) + (2) (4) + (3) (6) + (4) (k) + (5) (8) + (6)
(6) + (7) (6) (A1)

∑f k + 30 (A1)

144 + 4k
Using mean 4.6 = (M1)
k + 30
4.6k + 138 = 144 + 4k (A1)
0.6k = 6
k = 10 (A1) (C5)
(b) Mode = 4 (A1) (C1)
(accept 5, if k < 8 )
[6]
(10 × 1) + (20 × 2) + (30 × 5) + (40 × k ) + (50 × 3)
3. = 34 (M1)(A1)
k + 11
40k + 350
= 34 (A1)
k + 11
⇒k=4 (A1) (C4)
[4]
4. List of frequencies with p in the middle
23 + p
eg 5 + 10, p, 6 + 2 ⇒ 15, 8, or 15 < , or p > 7. (M1)
2
Consideration that p < 10 because 2 is the mode or discretionary for
further processing. (M1)
Possible values of p are 8 and 9 (A2)(A2) (C6)
[6]

1
5.1 Measures of Central Tendency Paper 2
x  2230 
1. (a) =
mean ∑= 
n  45 
 (M1)

x = 49.6 (Accept 50) (A1) (C2)

(b) y= ∑ y (may be implied) (M1)

n+2

∑ y= 2230 + 37 + 30 (A1)

2297
y= (A1)
47
= 48.9 (Accept 49) (A1) (C4)
[6]
2. Jan–Sept ∑ = 630 × 9 = 5670 (M1)(A1)
Oct–Dec ∑ = 810 × 3 = 2430 (M1)(A1)
5670 + 2430
x= (M1)
12
mean = 675 (A1) (C6)
[6]
3. (a) Median = middle number of 75 (M1)
= 38th number
=4 (A1) (C2)
5 + 18 + 48 + 72 + 100 + 42
(b) Mean = (M1)
75
285
=
75
= 3.8 (A1) (C2)
[4]
(72 × 1.79) + (28 × 1.62)
4. Mean = (M1)(M1)(M1)
100
= 1.7424 (= 1.74 to 3 sf) (A1) (C4)
[4]

2
5. (a) (i) 10 (A1)
(ii) 14 + 10 = 24 (A1) 2
(b)
xi fi

 15 1 
 25 5 
 35 7 
 45 9 
 
(A1) 55 10 (A1)
 65 16 
 75 14 
 85 10 
 
 95 8 
80 (AG)
(i) µ = 63 (A1)
(ii) σ = 20.5 (3 sf) (A1) 4
(c) Assymetric diagram/distribution (A1) 1
(d)

60

cumulative
frequency (A1)
40.5

{
(A1) points and
curve
20

50 60 65 70 80
length (cm) (A1) 3
OR Median = 65 (A3) 3
Note: This answer assumes appropriate use of a calculator with correct
arguments.
OR Linear interpolation on the table: (M1)
 48 − 40.5   40.5 − 32 
  × 60 +   × 70 = 65 (2sf) (A1)(A1) 3
 48 − 32   48 − 32 
[10]

3
6. (a) Correct mid interval values 14, 23, 32, 41, 50 (A1)

Substituting into
∑ fw M1
∑f
7(14) + 12(23) + 13(32) + 10(41) + 8(50)
eg w =
50
1600
w = A1
50
w = 32 (kg) AG N0
(b) Total weight of other boxes = 1600 − 50x (A1)
Total number of other boxes = 50 − x (A1)
Setting up their equation M1
1600 − 50 x
eg = 30, 1600 − 50x = 1500 − 30x
50 − x
x=5 A1 N3
(c) Setting up their inequality M1
Correct substitution A1
98 + 276 + 416 + 41 (10 + y) + 400 1600 − 41 y
eg < 33, < 33
50 + y 50 + y

1600 + 41y < 1650 + 33y (A1)

8y < 50 (y < 6.25) A1
6 A1 N1
Note: If candidates don’t use the mid-interval values, but
assume that all the new boxes weigh the minimum
amount for Class D, award marks as follows:
Setting up their inequality M1
Correct substitution A1
1600 − 36.5 y
eg < 33
50 + y

1600 + 36.5y < 1650 + 33y (A1)

3.5y < 50 (y < 14.28...) A1
14 A1 N1
[12]

4
5.2 Measures of Spread Paper 1
1. (a) evidence of using ∑ fi = 100 (M1)
k=4 A1 N2
(b) (i) evidence of median position (M1)
e.g. 50th item, 26 + 10 + 20 = 56
median = 3 A1 N2
(ii) Q1 = and Q3 = 5 (A1)(A1)
interquartile range = 4 (accept 1 to 5 or 5−1, etc.) A1 N3
[7]
2. (a) A = 18, B = 19, C = 23, D = 31, E = 36 A1A1A1A1A1 N5
(b) IQR = 12 A1 N1
[6]
3. b = 3, c = 3 A1A1 N2
 a+b+c+d 
using mean  = 4 M1
 4 
using range (d − a = 6) M1
a = 2, d = 8 A1A1 N2
[6]
4. (a) (i) r = 10 A2 N2
(ii) s = 13 A2 N2

(b) Using
∑ x =10 A1
12
t = 18 A1 N1
[6]
5. (a) 3 A1 N1
(b) 6 A2 N2
(c) Recognizing the link between 6 and the upper quartile (M1)
eg 25% scored greater than 6,
0.25 × 32 (A1)
8 A1 N3
[6]
6. d = 11; c = 11 (A1)(A1)(C1)(C1)
d – a = 8 (or 11 – a = 8) (A1)
a=3 (A1) (C2)
3 + b + 11 + 11  sum 
= 8  or = 8 (A1)
4  4 
b=7 (A1) (C2)
[6]

5
7.
x f Σf
4 2 2
5 5 7
6 4 11
7 3 14
8 4 18
10 2 20
12 1 21

(a) m=6 (A2) (C2)

(b) Q1 = 5 (A2) (C2)
(c) Q3 = 8 (A1)
IQR = 8 – 5 (M1)
= 3 (accept 5 – 8 or [5, 8]) (C2)
[6]
8. Median = middle value => b = 11 (A1)
a + b + c a + 11 + c
Mean = = = 9 => a + 11 + c = 27 (M1)
3 3
=> a + c = 16 (A1)
Range = c – a = 10 (M1)(A1)
Solving equations simultaneously gives a = 3 (A1) (C6)
[6]
300
9. (a) m= (M1)
25
= 12 (A1) (C2)

 625 
(b) s=   (M1)
 25 
=5 (A1) (C2)
[4]

6
5.2 Measures of Spread Paper 2
1. (a) Σfx = 1(2) + 2(4) + ... + 7(4), Σfx = 146 + 5x (seen anywhere) A1

evidence of substituting into mean =

∑ fx
(M1)
∑ f

correct equation A1
146 + 5 x
e.g. = 4.5, 146 + 5x = 4.5(34 + x)
34 + x
x = 14 A1 N2
(b) σ = 1.54 A2 N2
[6]
2. (a) σ = 1.61 A2 N2
(b) median = 4.5 A1 N1
(c) Q1 = 3, Q3 = 5 (may be seen in a box plot) (A1)(A1)
IQR = 2 (accept any notation that suggests the interval 3 to 5) A1 N3
[6]

7
5.3 Cumulative Frequency Paper 1
1. (a) correct end points (A1)(A1)
max = 27, min = 4
range = 23 A1 N3 3
(b) Graph 3 A2 N2 2
[5]
2.

(a) Lines on graph (M1)

100 students score 40 marks or fewer. A1 N2
(b) Identifying 200 and 600 A1
Lines on graph (M1)
a = 55, b = 75 A1A1 N1N1
[6]
3. (a) (i) m = 165 A1 N1
(ii) Lower quartile (1st quarter) = 160 (A1)
Upper quartile (3rd quarter) = 170 (A1)
IQR = 10 A1 N3
(b) Recognize the need to use the 40th percentile, or 48th student (M1)
eg a horizontal line through (0, 48)
a = 163 A1 N2
[6]
4. (a) D B C A1A1A1 N3
(b) B A C A1A1A1 N3
[6]

8
5.

800

700

600
Number
of 500
candidates
400

300

200

100

10 20 30 40 50 60 70 80 90 100

Mark
(a) Lines on graph (M1)
100 students score 40 marks or fewer. A1 2
(b) Identifying 200 and 600 A1
Lines on graph. (M1)
a = 55, b = 75. A1A1 4
[6]
6. (a)
Mark (x) 0 ≤ x < 20 20 ≤ x < 40 40 ≤ x < 60 60 ≤ x < 80 80 ≤ x < 100
Number of
22 50(±1) 66(±1) 42(±1) 20
Students

(A1)(A1)(A1) (C3)
(b) 40th Percentile ⇒ 80th student fails, (mark 42%) (M2)
Pass mark 43% (Accept mark > 42.) (A1) (C3)
[6]
7. (a) line(s) on graph (M1)
median is 183 (A1) (C2)
(b) Lower quartile Q1 = 175 (A1)

Upper quartile Q3 = 189 (A1)

IQR is 14
(Accept 189 – 175, 175 to 189, 189 to 175 and 175 – 189) (M1)(A1) (C4)
[6]

9
8.

90
80

Cumulative frequency
70
60
50
40
30
20
10
0
0 5 10 15 20 25 30 35 40 45
LQ = 14 M = 20 UQ = 24
Diameter (mm)
(a) (i) Correct lines drawn on graph, (A1)(C1)
median = 20 (A1)(C1)
(ii) Correct lines drawn on graph, (A1)(C1)
UQ = Q3 = 24 (A1)(C1)
(b) IQR = Q3 – Q1(or UQ – LQ) (M1)
= 10 (accept 14 to 24) (A1) (C2)
[6]

10
5.3 Cumulative Frequency Paper 2
1. (a) (i) evidence of appropriate approach (M1)
e.g. 9 + 25 + 35, 34 + 35
p = 69 A1 N2
(ii) evidence of valid approach (M1)
e.g. 109 – their value of p, 120 – (9 + 25 + 35 + 11)
q = 40 A1 N2
(b) evidence of appropriate approach (M1)

e.g. substituting into

fx∑, division by 120
n
mean = 3.16 A1 N2
(c) 1.09 A1 N1
[7]
2. (a) (i) p = 65 A1 N1
(ii) for evidence of using sum is 125 (or 99 − p) (M1)
q = 34 A1 N2
(b) evidence of median position (M1)
125
e.g. 63rd student,
2
median is 17 (sit-ups) A1 N2

(c) evidence of substituting into

∑ f ( x) (M1)
125
15(11) + 16(21) + 17(33) + 18(34 ) + 19(18) + 20(8) 2176
e.g . ,
125 125
mean = 17.4 A1 N2
[7]
3. (a) median m = 32 A1 N1
(b) lower quartile Q1 = 22, upper quartile Q3 = 40 (A1)(A1)
interquartile range = 18 A1 N3
(c)
Time (minutes) Number of students
0 ≤ t < 10 5
10 ≤ t < 20 11
20 ≤ t < 30 20
30 ≤ t < 40 24
40 ≤ t < 50 14
50 ≤ t < 60 6
A1A1 N2
[6]

11
4. (a) (i) 50 (accept 49, “fewer than 50”) A1 N1
(ii) Cumulative frequency (7) = 90 (A1)
90 − 50 (M1)
= 40 A1 N2
(iii) 75th or 75.5th person A1
median = 6.25 (min), 6 min 15 secs A1 N1
(b) Evidence of finding 40% (60%) of 150 M1
Number spending less than k minutes is (150 − 60) = 90 (A1)
k=7 A1 N2
(c) (i)

t (minutes) 0≤t<2 2≤t<4 4≤t<6 6≤t<8 8 ≤ t < 10 10 ≤ t < 12

Frequency 10 23 37 38 27 15

A1A1A1 N3
(ii) Evidence of using all correct mid-interval values (1, 3, 5, 7, 9, 11) A1
 1×10 + 3 × 23 + 5 × 37 + 7 × 38 + 9 × 27 + 11×15 
mean =  
 150 
= 6.25 (min), 6 min 15 secs A1 N1
[14]
5. (a)
Houses

100
91±1

90

80 75

70

60
350 000

50

40

30
25

20
135±5

240±5

10

100 200 300 400 500 Selling price ($1000)

(A1)(A2)(A1) 4

12
 (b) Q1 = 135 ± 5 Q3 = 240 ± 5 (M1)(A1)
Interquartile range = 105 ± 10. (Accept 135 – 240 or 240 – 135.) (A1) 3
Note: Award (M1) for the correct lines on the graph.
(c) a = 94 – 87 = 7, b = 100 – 94 = 6 (A1)(A1) 2
12(50 ) + 46(150 ) + 29(250 ) + 7(350 ) + 6(450 )
(d) mean = (M1)
100
= 199 or $199000 (A1)
OR
mean = 199 or $199000 (G2) 2
(e) (i) $350000 => 91.5
Number of De luxe houses ∼100 – 91.5 (M1)
= 9 or 8 (A1)
65 5 6  5  15
(ii) P (both > 400000) =   = or   = (M1)(A1) 4
9  8  12 8  7  28
[15]
6. (a) (i) median fare = $24 (±0.5) (A1)
(ii) fare ≤ $35 => number of cabs is 154 (or 153) (A1) 2
(b) 40% of cabs = 80 cabs (A1)
fares up to $22 (A1)
distance = $22 ÷ $0.55 (M1)
a = 40 km (A1) 4
(c) Distance 90 km => fare = 90 × $0.55
= $49.50 (A1)
Fare $49.50 => number of cabs = 200 –186 (M1)
= 14 (A1)
14
Thus percentage is = 7% (A1) 4
200
[10]
7. (a) s = 7.41(3 sf) (G3) 3
(b)
Weight (W) W ≤ 85 W ≤ 90 W ≤ 95 W ≤ 100 W ≤ 105 W ≤ 110 W ≤ 115
Number of
5 15 30 56 69 76 80
packets

(A1) 1
(c) (i) From the graph, the median is approximately 96.8.
Answer: 97 (nearest gram). (A2)
(ii) From the graph, the upper or third quartile is approximately 101.2.
Answer: 101 (nearest gram). (A2) 4
(d) Sum = 0, since the sum of the deviations from the mean is zero. (A2)
OR
 ∑Wi 
∑ ∑
(Wi − W ) = Wi −  80
 80 
 =0 (M1)(A1) 2
 

13
 (e) Let A be the event: W > 100, and B the event: 85 < W ≤ 110
P( A ∩ B)
P(AB) = (M1)
P( B )
20
P(A ∩ B) = (A1)
80
71
P(B) = (A1)
80
P(AB) = 0.282 (A1)
OR
71 packets with weight 85 < W ≤ 110. (M1)
Of these, 20 packets have weight W > 100. (M1)
20
Required probability = (A1)
71
= 0.282 (A1) 4
[14]
8. (a) (Using mid-intervals)
65(7) + 75(25) + ... + 135(5)
v= (M1)
7 + 25 + ... + 5
29450
= = 98.2 km h–1 (A1)
300
OR
v = 98.2 (G2) 2
(b) (i) a = 165, b = 275 (A1)
(ii)

320

CUMULATIVE FREQUENCY
280
255 Cars
Number of cars

240

200

160

120

80
105 kmh –1

40

20 40 60 80 100 120 140 v (kmh–1)

(A4) 5

14
 (c) (i) Vertical line on graph at 105 km h–1 (M1)
300 − 200
× 100% = 33.3(±1.3%) (A1)
300
OR
33.3(±1.3%) (A2)
(ii) 15% of 300 = 45 300 – 45 = 255
Horizontal line on graph at 255 cars (M1)
Speed = 114(± 2 km h–1) (A1)
OR
Speed = 114(± 2 km h–1) (A2) 4
[11]

9. (a) x = $59 (G2)

OR
10 × 24 + 30 × 16 + ... + 110 × 10 + 130 × 4
x= (M1)
24 + 16 + ... + 10 + 4
7860
=
134
= $59 (A1) 2
(b)
Money ($) <20 <40 <60 <80 <100 <120 <140
Customers 24 40 62 102 120 130 134 (A1)

140

120
number
of 100
customers
80

60

40

20

20 40 60 80 100 120 140

money ($) (A4) 5

15
 (c) (i) t = 2d2/3 + 3
Mean d = 59 (M1)
Mean t ≈ 2 × (59)2/3 + 3 (M1)
≈ 33.3 min. (3 sf) (accept 33.2) (A1)
(ii) t > 37 ⇒ 2d2/3 + 3 > 37 (M1)
2d2/3 > 34
d2/3 > 17 (A1)
d > (17)3/2
d > 70.1
From the graph, when d = 70.1, n = 82 (A1)
number of shoppers = 134 – 82 (A1)
= 52 (A1) 8
[15]
10. (a)
x 15 45 75 105 135 165 195 225
f 5 15 33 21 11 7 5 3 (M1)
_
x = 97.2 (exactly) (A1) 2
(b)
x 30 60 90 120 150 180 210 240
Σf 5 20 53 74 85 92 97 100 (A1) 1
(c)

100
Σf 90
80
70
60
customers

50
40
30
20
10 Q3
a n
20 40 60 80 100 120 140 160 180 200 220 240

(A4) 4
(d) Median = 87 ± 2 (A1)
Lower quartile = 65 ± 2 (A1)
Upper quartile = 123 ± 2 (A1) 3
[10]

16
11. (a)
Mark ≤ 10 ≤ 20 ≤ 30 ≤ 40 ≤ 50 ≤ 60 ≤ 70 ≤ 80 ≤ 90 ≤ 100
No. of Candidates 15 65 165 335 595 815 905 950 980 1000

(A3) 3
(b)

1000

900

800

700

600
No. of candidates

500

400

300

200

100

10 20 30 40 50 60 70 80 90 100
Marks
(c) (i) Median = 46 (M1)(A1)
(ii) Scores < 35: 240 candidates (M1)(A2)
(iii) Top 15% ⇒ Mark ≥ 63 (M1)(A1)(A1) 8
[16]
12. (a) 76 (mice) (A1) (N1)
(b) 11.2 (seconds) (A1) (N1)
(c) (i) p = 76 − (16 + 22) = 38 (allow ft from (ii) (a)) (A1) (N1)
q = 132 − 76 = 56 (A1) (N1)
7.5 × 16 + .....14.5 × 23  3363 
(ii) x= =  (M1)
16 + ...23  300 
= 11.2 (accept 11.21) (A1) (N2)
[6]

17
5.4 Simple Probability Paper 1
1. (a) (i) n = 0.1 A1 N1
(ii) m = 0.2, p = 0.3, q = 0.4 A1A1A1 N3 4
(b) appropriate approach
e.g. P(B′) =1– P(B), m + q, 1–(n + p) (M1)
P(B′) = 0.6 A1 N2 2
[6]
2. (a) P(A ∩ B) = P(A) × P(B) (= 0.6x) A1 N1
(b) (i) evidence of using P(A ∪ B) = P(A) + P(B) – P(A)P(B) (M1)
correct substitution A1
e.g. 0.80 = 0.6 + x – 0.6x, 0.2 = 0.4x
x = 0.5 A1 N2
(ii) P(A ∩ B) = 0.3 A1 N1
(c) valid reason, with reference to P(A ∩ B) R1 N1
e.g. P(A ∩ B) ≠ 0
[6]

3. (a)
19
(= 0.158) A1 N1
120
(b) 35 − (8 + 5 + 7)(= 15) (M1)
15  3 1 
Probability = = = = 0.125  A1 N2
120  24 8 
(c) Number studying = 76 (A1)
Number not studying = 120 − number studying = 44 (M1)
44  11 
Probability = = = 0.367  A1 N3
120  30 
[6]
4. (a)
3/9 A

A
4/10
6/9
B

4/9 A
6/10
B

5/9
B

A1A1A1 N3
 4 6  6 4
(b)  × + ×  M1M1
 10 9   10 9 
48  8 
=  , 0.533  A1 N1
90  15 
[6]

18
5. (a) Independent ⇒ P(A ∩ B) = P(A) × P(B) (= 0.3 × 0.8) (M1)
= 0.24 A1 N2
(b) P(A ∪ B) = P(A) + P(B) − P(A ∩ B) (= 0.3 + 0.8 − 0.24) M1
= 0.86 A1 N1
(c) No, with valid reason A2 N2
eg P(A ∩ B) ≠ 0 or P(A ∪ B) ≠ P(A) + P(B) or correct
numerical equivalent
[6]
6. (a) For attempting to use the formula (P(E ∩ F) = P(E)P(F)) (M1)
Correct substitution or rearranging the formula A1
1
1 2 P (E ∩ F )
eg = P(F), P(F) = , P(F) = 3
3 3 P (E ) 2
3
1
P(F) = A1 N2
2
(b) For attempting to use the formula (P(E ∪ F) = P(E) + P(F)
− (P(E ∩ F)) (M1)
2 1 1
P(E ∪ F) = + − A1
3 2 3

=
5
(= 0.833) A1 N2
6
[6]
7. Total number of possible outcomes = 36 (may be seen anywhere) (A1)
(a) P ( E ) =P (1,1) + P (2, 2) + P (3, 3) + P (4, 4) + P (5, 5) + P (6, 6)

6
= (A1) (C2)
36
(b) P ( F ) = P (6, 4) + P (5, 5) + P (4, 6)

3
= (A1) (C1)
36
(c) P ( E ∪ F )= P ( E ) + P ( F ) − P ( E ∩ F )

1
P (E ∩ F ) = (A1)
36
6 3 1  8 2 
P(E ∪ F ) = + − = = , 0.222  (M1)(A1) (C3)
36 36 36  36 9 
[6]

19
  13   12   11   10 
8. Correct probabilities   ,   ,   ,   (A1)(A1)(A1)(A1)
 24   23   22   21 

 13 12 11 10 
Multiplying  × × ×  (M1)
 24 23 22 21 
17160  65 
P(4 girls) = = = 0.0673  (A1) (C6)
255024  966 
[6]
9. For using P(A ∪ B) = P(A) + P(B) − P(A ∩ B) (M1)
Let P(A) = x then P(B) = 3x
P(A ∩ B) = P(A) × 3P(A) (= 3x2) (A1)
0.68 = x + 3x − 3x2 (A1)
3x2 − 4x + 0.68 = 0
x = 0.2 ( x = 1.133, not possible) (A2)
P(B) = 3x = 0.6 (A1) (C6)
[6]

7 6 7 
10. P(RR) = × =  (M1)(A1)
12 11  22 
5 4 5
P(YY) = × =  (M1)(A1)
12 11  33 
P (same colour) = P(RR) + P(YY) (M1)
31
= (= 0.470, 3 sf) (A1) (C6)
66
[6]
11. (a) P (A ∪ B) = P (A) + P (B) – P (A ∩ B) ⇒ P (A ∩ B) = P (A) +
P (B) – P (A ∪ B) (M1)
3 4 6
= + – (M1)
11 11 11
1
= (0.0909) (A1) (C3)
11
(b) For independent events, P (A ∩ B) = P (A) × P (B) (M1)
3 4
= × (A1)
11 11
12
= (0.0992) (A1) (C3)
121
[6]
12. P(different colours) = 1 – [P(GG) + P(RR) + P(WW)] (M1)
 10 9 10 9 6 5 
=1–  × + × + ×  (A1)
 6 25 26 25 26 25 
 210 
=1–   (A1)
 650 
44
= (= 0.677, to 3 sf) (A1) (C4)
65

20
 OR
P(different colours) = P(GR) + P(RG) + P(GW) + P(WG) + P(RW) + P(WR) (A1)
 10 6   10 10 
= 4 ×  + 2 ×  (A1)(A1)
 26 25   26 25 
44
= (= 0.677, to 3 sf) (A1) (C4)
65
[4]
13. (a) U

B
A

(A1) (C1)
(b) n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
65 = 30 + 50 – n(A ∩ B)
⇒ n(A ∩ B) = 15 (may be on the diagram) (M1)
n(B ∩ A′) = 50 – 15 = 35 (A1) (C2)
n( B ∩ A′) 35
(c) P(B ∩ A′) = = = 0.35 (A1) (C1)
n(U ) 100
[4]
14. (a)
6, 6 1
6
1– 36
6
6
1–
6 –5
6 5
not 6 6, not 6
36
not 6, 6 5
6
1– 36
6
–5
6 not 6

–5
6
not 6 not 6, not 6 25
36
(M2) (C2)
1 1 1 5 5 1  5 5
(b) P(one or more sixes) = × + × + × or 1 − ×  (M1)
6 6 6 6 6 6  6 6
11
= (A1) (C2)
36
[4]

21
15. (a)

A B

(A1) (C1)
(b) (i) n(A ∩ B) = 2 (A1) (C1)
2  1
(ii) P(A ∩ B) =  or  (allow ft from (b)(i)) (A1) (C1)
36  18 
(c) n(A ∩ B) ≠ 0 (or equivalent) (R1) (C1)
[4]
16. (a) p(A ∩ B) = 0.6 + 0.8 – 1 (M1)
= 0.4 (A1) (C2)
(b) p( A ∪ B) = p( (A ∩ B)) = 1 – 0.4 (M1)
= 0.6 (A1) (C2)
[4]

22
5.4 Simple Probability Paper 2
1. (a) evidence of valid approach involving A and B (M1)
e.g. P(A ∩ pass) + P(B ∩ pass), tree diagram
correct expression (A1)
e.g. P(pass) = 0.6 × 0.8 + 0.4 × 0.9
P(pass) = 0.84 A1 N2 3
(b) evidence of recognizing complement (seen anywhere) (M1)
e.g. P(B) = x, P(A) = 1 – x, 1 – P(B), 100 – x, x + y =1
evidence of valid approach (M1)
e.g. 0.8(1 – x) + 0.9x, 0.8x + 0.9y
correct expression A1
e.g. 0.87 = 0.8(1 – x) + 0.9x, 0.8 × 0.3 + 0.9 × 0.7 = 0.87, 0.8x + 0.9y = 0.87
70 % from B A1 N2 4
[7]
2. METHOD 1
for independence P(A ∩ B) = P(A) × P(B) (R1)
expression for P(A ∩ B), indicating P(B) = 2P(A) (A1)
e.g. P(A) × 2P(A), x × 2x
substituting into P(A ∪ B) = P(A) + P(B) – P(A ∩ B) (M1)
correct substitution A1
e.g. 0.52 = x + 2x – 2x2, 0.52 = P(A) + 2P(A) – 2P(A)P(A)
correct solutions to the equation (A2)
e.g. 0.2, 1.3 (accept the single answer 0.2)
P(B) = 0.4 A1 N6
METHOD 2
for independence P(A ∩ B) = P(A) × P(B) (R1)
1
expression for P(A ∩ B), indicating P(A) = P(B) (A1)
2
1 1
e.g. P(B) × P(B), x × x
2 2
substituting into P(A ∪ B) = P(A) + P(B) – P(A ∩ B) (M1)
correct substitution A1
e.g. 0.52 = 0.5x + x – 0.5x2, 0.52 = 0.5P(B) + P(B) – 0.5P(B)P(B)
correct solutions to the equation (A2)
e.g. 0.4, 2.6 (accept the single answer 0.4)
P(B) = 0.4 (accept x = 0.4 if x set up as P(B)) A1 N6
[7]

23
3. (a) evidence of binomial distribution (seen anywhere) (M1)
 1
e.g. X ~ B  3, 
 4

mean =
3
(= 0.75) A1 N2
4

 3  1   3 
2

(b) P(X = 2) =       (A1)

 2  4   4 

 9 
P(X = 2) = 0.141 =  A1 N2
 64 
(c) evidence of appropriate approach M1
e.g. complement, 1 − P(X = 0), adding probabilities
 27 
P(X = 0) = (0.75)3  = 0.422 ,  (A1)
 64 

 37 
P(X ≥ 1) = 0.578 =  A1 N2
 64 
[7]
4. (a)

(A1)(A1)(A1)
1 4  4 
(b) (i) P(R ∩ S) = × = = 0.267  (A1) (N1)
3 5  15 
1 4 2 1
(ii) P(S) = × + × (A1)(A1)
3 5 3 4
13
= (= 0.433) (A1) (N3)
30
4
(iii) P(R S) = 15 (A1)(A1)
13
30
8
= (= 0.615) (A1) (N3)
13
[10]

24
5.5 Conditional Probability Paper1
1. (a) (i) s=1 A1 N1
(ii) evidence of appropriate approach (M1)
e.g. 21–16, 12 + 8 – q =15
q=5 A1 N2
(iii) p = 7, r = 3 A1A1 N2 5
5
(b) (i) P(art|music) = A2 N2
8
(ii) METHOD 1
12  3 
P(art ) = =  A1
16  4 
evidence of correct reasoning R1
3 5
e.g. ≠
4 8
the events are not independent AG N0
METHOD 2
96  3 
P(art) × P(music) = =  A1
256  8 
evidence of correct reasoning R1
12 8 5
e.g. × ≠
16 16 16
the events are not independent AG N0 4
3
(c) P(first takes only music) = = (seen anywhere) A1
16
7
P(second takes only art)= (seen anywhere) A1
15
evidence of valid approach (M1)
3 7
e.g. ×
16 15
21  7 
P(music and art)= =  A1 N2 4
240  80 
[13]
2. (a) (i) p = 0.2 A1 N1
(ii) q = 0.4 A1 N1
(iii) r = 0.1 A1 N1
2
(b) P(A│B′) = A2 N2
3
(c) valid reason R1
2
e.g. ≠ 0.5, 0.35 ≠ 0.3
3
thus, A and B are not independent AG N0
[6]

25
 4
3. (a) p= A1 N1
5
(b) multiplying along the branches (M1)
1 1 12
e.g. × ,
5 4 40
adding products of probabilities of two mutually exclusive paths (M1)
1 1 4 3 1 12
e.g. × + × , +
5 4 5 8 20 40
14  7 
P(B) = =  A1 N2
40  20 
(c) appropriate approach which must include A′ (may be seen on diagram) (M1)
P( A′ ∩ B )  P( A ∩ B ) 
 do not accept 
P( B ) 
e.g.
P( B ) 
4 3
×
P(A′│B) = 5 8 (A1)
7
20
12  6 
P(A′│B) = =  A1 N2
14  7 
[7]
1
4. (a) P(A) = A1 N1
11
2
(b) P(B│A) = A2 N2
10
(c) recognising that P(A ∩ B) = P(A) × P(B│A) (M1)
correct values (A1)
1 2
e.g. P(A ∩ B) = ×
11 10
2
P(A ∩ B) = A1 N3
110
[6]
3
5. (a) A1 N1
4
(b) P(A ∪ B) = P(A) + P(B) − P(A ∩ B) (M1)
P(A ∩ B) = P(A) + P(B) − P(A ∪ B)
2 3 7
= + − A1
5 4 8
11
= (0.275) A1 N2
40
 11 
P( A ∩ B )  40 
(c) P(A  B) = = A1
P( B )  3 
 
 4 
11
= (0.367) A1 N1
30
[6]
26
6. (a) (i) evidence of substituting into n(A ∪ B) = n(A) + n(B) – n(A ∩ B) (M1)
e.g. 75 + 55 – 100, Venn diagram
30 A1 N2
(ii) 45 A1 N1
(b) (i) METHOD 1
evidence of using complement, Venn diagram (M1)
e.g. 1 – p, 100 – 30
70  7 
=  A1 N2
100  10 
METHOD 2
attempt to find P(only one sport), Venn diagram (M1)
25 45
e.g. +
100 100
70  7 
=  A1 N2
100  10 

45  9 
(ii) =  A2 N2
70  14 
(c) valid reason in words or symbols (R1)
e. g. P(A ∩ B) = 0 if mutually exclusive, P(A ∩ B) if not mutually exclusive
correct statement in words or symbols A1 N2
e.g. P(A ∩ B) = 0.3, P(A ∪ B) ≠ P(A) + P(B), P(A) + P(B) > 1, some
students play both sports, sets intersect
(d) valid reason for independence (R1)
e.g. P(A ∩ B) = P(A) × P(B), P(B│A) = P(B)
correct substitution A1A1 N3
e.g. 30 75 55 30 75
≠ × , ≠ [12]
100 100 100 55 100
7. (a) (i) correct calculation (A1)
9 5 2 4 + 2 + 3+ 3
e.g. + − ,
20 20 20 20
12  3 
P(male or tennis) = =  A1 N2
20  5 
(ii) correct calculation (A1)
6 11 3 + 3
e.g. ÷ ,
20 20 11
6
P(not football | female) = A1 N2
11
11 10
(b) P(first not football) = , P(second not football) = A1
20 19
11 10
P(neither football) = × A1
20 19
110  11 
P(neither football) = =  A1 N1
380  38 
[7]

27
8. (a) (i) number of ways of getting X = 6 is 5 (A1)
5
P(X = 6) = A1 N2
36
(ii) number of ways of getting X > 6 is 21 (A1)
21  7 
P(X > 6) = =  A1 N2
36  12 

6  3
(iii) P(X =7|X > 5) = =  A2 N2
26  13 
(b) evidence of substituting into E(X) formula (M1)
10
finding P(X < 6) = (seen anywhere) (A2)
36
evidence of using E(W) = 0 (M1)
correct substitution A2
 5   21   10 
e.g. 3  + 1  − k   = 0 , 15 + 21 – 10k = 0
 36   36   36 
36
k= (= 3.6) A1 N4
10
[13]

9. (a)
46
(= 0.474) A1A1 N2
97

(b)
13
(= 0.255) A1A1 N2
51

(c)
59
(= 0.608) A2 N2
97
[6]
20
10. (a) P(PC) = A1
20 + 40

1
= A1 N1
3
30
(b) P(PC′) = A1
30 + 60

1
= A1 N1
3
(c) Investigating conditions, or some relevant calculations (M1)
P is independent of C, with valid reason A1 N2
eg P(PC) = P(PC′), P(PC) = P(P),
20 50 60
= × (ie P(P ∩ C) = P(P) × P(C))
150 150 150
[6]

28
11. (a) P(A ∪ B) = P(A) + P(B) – P(A ∩ B) (M1)
1 3 7
P(A ∩ B) = + −
2 4 8
3
= (A1) (C2)
8
 3
 
P( A ∩ B )  8
(b) P(AB) = = (M1)
P( B )  3
 
 4
1
= (A1) (C2)
2
(c) Yes, the events are independent (A1) (C1)
P(A ∩ B) = P(A)P(B) (R1) (C1)
[6]

120  1 
12. (a) = = 0.333  (A1)(A1) (C2)
360  3 
90 + 120  210 7 
(b) = = = 0.583  (A2) (C2)
360  360 12 

90  3   1 
(c) =
 = 0.429   Accept
4
 (A1)(A1) (C2)
210  7  
7
12 
[6]
13. (a) Independent (I) (C2)
(b) Mutually exclusive (M) (C2)
(c) Neither (N) (C2)
[6]
22
14. (a) P= (= 0.957 (3 sf)) (A2) (C2)
23
(b)
R
21 3
24 23

R
G
22
25

etc
3
25
G

(M1)
OR
P = P (RRG) + P (RGR) + P (GRR) (M1)
22 21 3 22 3 21 3 22 21
× × + × × + × × (M1)(A1)
25 24 23 25 24 23 25 24 23
693
= (= 0.301 (3 sf)) (A1) (C4)
2300
[6]

29
15. (a)

0.6

0.4 0.4

0.6 0.5

0.5
(A1) (C1)
(b) P(B) = 0.4(0.6) + 0.6 (0.5) = 0.24 + 0.30 (M1)
= 0.54 (A1) (C2)
P( B ∩ C ) 0.24 4
(c) P(CB) = = = (= 0.444, 3 sf) (A1) (C1)
P( B ) 0.54 9
[4]
16. (a)
Males Females Totals
Unemployed 20 40 60
Employed 90 50 140
Totals 110 90 200

40 1
(b) (i) P(unemployed female) = = (A1)
200 5
90 9
(ii) P(male I employed person) = = (A1)
140 14
[4]
17. (a)
Boy Girl Total
TV 13 25 38
Sport 33 29 62
Total 46 54 100

38
P(TV) = (A1) (C2)
100
13
(b) P(TV | Boy) = (= 0.283 to 3 sf) (A2) (C2)
46
[4]

30
5.5 Conditional Probability Paper 2
1. (a) appropriate approach (M1)
e.g. tree diagram or a table
P(win) = P(H ∩ W) + P(A ∩ W)) (M1)
= (0.65)(0.83) + (0.35)(0.26) A1
= 0.6305 (or 0.631) A1 N2
(b) evidence of using complement (M1)
e.g. 1 – p, 0.3695
choosing a formula for conditional probability (M1)
P(W ′ ∩ H )
e.g. P(H│W′) =
P(W ′)
correct substitution
(0.65)(0.17)  0.1105 
e.g. =  A1
0.3695  0.3695 
P(home) = 0.299 A1 N3
[8]
2. Sample space ={(1, 1), (1, 2) ... (6, 5), (6, 6)}
(This may be indicated in other ways, for example, a grid or a tree diagram, partly or fully completed)
1
2
3.
..
1
1
2
2 3.
..

3
4
5

6
6 + 5 + 4 + 3 + 2 +1
(a) P (S < 8) = (M1)
36
7
= (A1)
12
OR
7
P (S < 8) = (A2)
12
1+1+ 6 +1+1+1
(b) P (at least one 3) = (M1)
36
11
= (A1)
36
OR
11
P (at least one 3) = (A2)
36

31
 P(at least one 3 ∩ S < 8)
(c) P (at least one 3S < 8) = (M1)
P(S < 8)
7
=
36 (A1)
7
12
1
= (A1)
3
[7]
3. (a) P(F ∪ S) = 1 − 0.14 (= 0.86) (A1)
Choosing an appropriate formula (M1)
eg P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Correct substitution
eg P(F ∩ S) = 0.93 − 0.86 A1
P(F ∩ S) = 0.07 AG N0
(b) Using conditional probability (M1)
 P (F ∩ S ) 
eg P(F | S)  = 
 P ( S ) 

0.07
P(F | S) = (A1)
0.62
= 0.113 A1 N3
(c) F and S are not independent A1 N1
EITHER
If independent P(F | S) = P(F), 0.113 ≠ 0.31 R1R1 N2
OR
If independent P(F ∩ S) = P(F) P(S), 0.07 ≠ 0.31 × 0.62 (= 0.1922) R1R1 N2
(d) Let P(F) = x
P(S) = 2P(F) (= 2x) (A1)
For independence P(F ∩ S) = P(F)P(S) (= 2x2) (R1)
Attempt to set up a quadratic equation (M1)
eg P(F ∪ S) = P(F)P(S) − P(F)P(S), 0.86 = x + 2x − 2x2
2x2 − 3x + 0.86 = 0 A2
x = 0.386, x = 1.11 (A1)
P(F) = 0.386 (A1) N5
[16]

32
4. (a)
3
R
5

M
1
3
2
5 G
2
R
10
2
3
N

8
10 G
A1A1A1 N3
1 2 2
(b) (i) P(M and G) = × (= = 0.133) A1 N1
3 5 15
1 2 2 8
(ii) P(G) = × + × (A1)(A1)
3 5 3 10
10  2 
=  = = 0.667  A1 N3
15  3 
2
P( M ∩ G ) 15
(iii) P(M  G) = = (A1)(A1)
P(G ) 2
3
1
= or 0.2 A1 N3
5
2 1
(c) P(R) = 1 − = (A1)
3 3
Evidence of using a correct formula M1
1 2  1 3 2 2 1 2 2 8
E(win) = 2 × + 5 ×  or 2 × × + 2 × × + 5 × × + 5 × ×  A1
3 3  3 5 3 10 3 5 3 10 

 12 60 
= $4  accept ,  A1 N2
 3 15 
[14]

33
 80  8 
5. (a) (i) P (=
A) = = 0.381 (A1) (N1)
210  21 
35  1 
(ii) =
P (year 2 art) = = 0.167  (A1) (N1)
210  6 
(iii) No (the events are not independent, or, they are dependent) (A1) (N1)
EITHER
P ( A ∩ B )= P ( A) × P ( B ) (to be independent) (M1)

100  10 
P ( B=
) = = 0.476  (A1)
210  21 
1 8 10
≠ × (A1)
6 21 21
OR
P (A)=P (A B ) (to be independent) (M1)

35
P (A B ) = (A1)
100
8 35
≠ (A1)
21 100
OR
P (B )=P (B A) (to be independent) (M1)

100  10  35
P ( B=
) = = 0.476  , P (B A) = (A1)
210  21  80

35 100
≠ (A1) 6
80 210
Note: Award the first (M1) only for a mathematical interpretation of
independence.
(b) n (history) = 85 (A1)

50  10 
=
P (year 1 history) =  = 0.588  (A1)(N2)
85  17 
2
 110 100   100 110   110 100 
(c)  × + ×  =2× ×  (M1)(A1)(A1)
 210 209   210 209   210 209 

200
= =( 0.501)
399
(A1) (N2) 4
[12]

34
6. (a)
1
4 L

7
W
8
3
4 L'
3
5 L
1
8
W'
2
5 L' (A1)(A1)(A1)(A1) 4
 7 1 3
Note: Award (A1) for the given probabilities  , ,  in the correct
8 4 5
positions, and (A1) for each bold value.
7 1 1 3
(b) Probability that Dumisani will be late is × + × (A1)(A1)
8 4 8 5
47
= (0.294)
160
(A1) (N2) 3
P(W ∩ L)
(c) P(WL) =
P( L)
7 1
P(W ∩ L) = × (A1)
8 4
47
P(L) = (A1)
160
7
P(WL) = 32 (M1)
47
160
35
= (= 0.745) (A1) (N3) 4
47
[11]

35
7. (a)
Grows
0.9

Red
0.4

0.1 Does not grow

Grows
0.8

0.6
Yellow

0.2 Does not grow

(A3) (N3) 3
(b) (i) 0.4 × 0.9 (A1)
= 0.36(A1) (N2)
(ii) 0.36 + 0.6 × 0.8 (= 0.36 + 0.48) (A1)
= 0.84(A1) (N1)
P (red ∩ grows)
(iii) (may be implied) (M1)
P (grows)

0.36
= (A1)
0.84

 3
= 0.429  
7
(A1) (N2) 7
[10]
8. (a)

U(88)
E(32) H(28)

a b c

39

n (E ∪ H) = a + b + c = 88 – 39 = 49 (M1)
n (E ∪ H) = 32 + 28 – b = 49
60 – 49 = b = 11 (A1)
a = 32 – 11 = 21 (A1)
c = 28 – 11 = 17 (A1) 4

36
 11 1
(b) (i) P(E ∩ H) = = (A1)
88 8
21
P(H '∩ E ) 88
(ii) P(H′E) = = (M1)
P (E ) 32
88
21
= (= 0.656) (A1)
32
OR
21
Required probability = (A1)(A1) 3
32
56 × 55 × 54
(c) (i) P(none in economics) = (M1)(A1)
88 × 87 × 86
= 0.253 (A1)
(ii) P(at least one) = 1 – 0.253 (M1)
= 0.747 (A1)
OR
 32 56 55   32 31 56  32 31 30
3  × ×  + 3 × ×  + × × (M1)
 88 87 86   88 87 86  88 87 86
= 0.747 (A1) 5
[12]

37
5.6 Discrete Probability Distributions Paper 1
7
1. (a) (i) A1 N1
24
(ii) evidence of multiplying along the branches (M1)
2 5 1 7
e.g. × , ×
3 8 3 8
adding probabilities of two mutually exclusive paths (M1)
1 7   2 3 1 1  2 5
e.g.  ×  +  × ,  ×  +  × 
3 8  3 8 3 8  3 8
13
P(F) = A1 N2
24
1 1
(b) (i) × (A1)
3 8
1
A1
24
(ii) recognizing this is P(E│F) (M1)
7 13
e.g. ÷
24 24
168  7 
=  A2 N3
312  13 
(c)
X (cost in euros) 0 3 6
1 4 4
P (X)
9 9 9

A2A1 N3
(d) correct substitution into E(X) formula (M1)
1 4 4 12 24
e.g. 0 × + 3× + 6 × , +
9 9 9 9 9
E(X) = 4 (euros) A1 N2
[14]
2. (a)
3, 9 4, 9 5, 9
3, 10 4, 10 5, 10
3, 10 4, 10 5, 10
A2 N2
(b) 12, 13, 14, 15 (accept 12, 13, 13, 13, 14, 14, 14, 15, 15) A2 N2
1 3 3 2
(c) P(12) = , P(13) = , P(14) = , P(15) = A2 N2
9 9 9 9
(d) correct substitution into formula for E(X) A1
1 3 3 2
e.g. E(S) = 12 × + 13 × + 14 × + 15 ×
9 9 9 9
123
E(S) = A2 N2
9

38
 (e) METHOD 1
correct expression for expected gain E(A) for 1 game (A1)
4 5
e.g. × 50 − × 30
9 9
50
E(A) =
9
amount at end = expected gain for 1 game × 36 (M1)
= 200 (dollars) A1 N2
METHOD 2
attempt to find expected number of wins and losses (M1)
4 5
e.g. × 36, × 36
5 9
attempt to find expected gain E(G) (M1)
e.g. 16 × 50 – 30 × 20
E(G) = 200 (dollars) A1 N2
[12]
3
3. (a) (i) P(B) = A1 N1
4
1
(ii) P(R) = A1 N1
4
3
(b) p= A1 N1
4
1 3
s= , t= A1 N1
4 4
(c) (i) P(X = 3)
1 3
= P (getting 1 and 2) = × A1
4 4
3
= AG N0
16
1 1 3  3
(ii) P(X = 2) = × +  or 1 −  (A1)
4 4 4  16 

13
= A1 N2
16
(d) (i)

X 2 3
P(X = x) 13 3
16 16

A2 N2

39
 (ii) evidence of using E(X) = ∑xP(X = x) (M1)
 13   3 
E(X) = 2  + 3  (A1)
 16   16 
35  3
= = 2  A1 N2
16  16 
(e) win $10 ⇒ scores 3 one time, 2 other time (M1)
13 3
P(3) × P(2) = × (seen anywhere) A1
16 16
evidence of recognizing there are different ways of winning $10 (M1)
 13 3 
e.g. P(3) × P(2) + P(2) × P(3), 2 × ,
 16 16 
36 3 36 3
+ + +
256 256 256 256
78  39 
P(win $10) = =  A1 N3
256  128 
[16]

4  2
4. (a) P(X = 2) = =  A1 N1 1
14  7 

1
(b) P(X = 1) = (A1)
14
k2
P(X = k) = (A1)
14
setting the sum of probabilities = 1 M1
1 4 k2
e.g. + + = 1, 5 + k2 = 14
14 14 14

 k2 9
k2 = 9  accept =  A1
 14 14 

k=3 AG N0 4

(c) correct substitution into E( X ) = ∑ xP( X = x) A1

1 4 9

e.g. 1   + 2  + 3 
 14   14   14 
36  18 
E( X ) = =  A1 N1 2
14  7 
[7]

40
5. (a) For summing to 1 (M1)
1 2 1
e.g. + + + x = 1
5 5 10
3
x= A1 N2
10
(b) For evidence of using E(X) = ∑x f (x) (M1)
Correct calculation A1
1 2 1 3
e.g. × 1 + 2 × + 3 × + 4 ×
5 5 10 10
25
E(X) = (= 2.5) A1 N2
10
1 1
(c) × (M1)
10 10
1
A1 N2
100
[7]
6. (a) For summing to 1 (M1)
eg 0.1 + a + 0.3 + b = 1
a + b = 0.6 A1 N2

(b) evidence of correctly using E( X ) = ∑ x f ( x) (M1)

eg 0 × 0.1 + 1 × a + 2 × 0.3 + 3 × b, 0.1 + a + 0.6 + 3b = 1.5

Correct equation 0 + a + 0.6 + 3b = 1.5 (a + 3b = 0.9) (A1)
Solving simultaneously gives
a = 0.45 b = 0.15 A1A1 N3
[6]

7. (a) For using ∑ p =1 (0.4 + p + 0.2 + 0.07 + 0.02 = 1) (M1)

p = 0.31 A1 N2

(b) For using E(X) = ∑ xP( X = x ) (M1)

E(X) = 1(0.4) + 2(0.31) + 3(0.2) + 4(0.07) + 5(0.02) A1

=2 A2 N2
[6]

41
5.6 Discrete Probability Distributions Paper 2
1. (a) evidence of using ∑pi = 1 (M1)
correct substitution A1
e.g. 10k2 + 3k + 0.6 = 1, 10k2 + 3k − 0.4 = 0
k = 0.1 A2 N2
(b) evidence of using E(X) = ∑pixi (M1)
correct substitution (A1)
e.g. − 1 × 0.2 + 2 × 0.4 + 3 × 0.3
E(X) = 1.5 A1 N2
[7]
2. (a) Adding probabilities (M1)
Evidence of knowing that sum = 1 for probability distribution R1
eg Sum greater than 1, sum = 1.3, sum does not equal 1 N2
(b) Equating sum to 1 (3k + 0.7 = 1) M1
k = 0.1 A1 N1
0 +1
(c) (i) P( X = 0 ) = (M1)
20
1
= A1 N2
20
(ii) Evidence of using P(X > 0) = 1 − P(X = 0)
 4 5 10 
 or + +  (M1)
 20 20 20 
19
= A1 N2
20
[8]
3. (a) three correct pairs A1A1A1 N3 3
e.g. (2, 4), (3, 3), (4, 2), R2G4, R3G3, R4G2
1 2 2
(b) p= ,q= ,r= A1A1A1 N3 3
16 16 16
(c) let X be the number of times the sum of the dice is 5
evidence of valid approach (M1)
e.g. X ~ B(n, p), tree diagram, 5 sets of outcomes produce a win
one correct parameter (A1)
e.g. n = 4, p = 0.25, q = 0.75
Fred wins prize is P(X ≥ 3) (A1)
appropriate approach to find probability M1
e.g. complement, summing probabilities, using a CDF function
correct substitution (A1)
243 12 1
e.g. 1– 0.949…, 1 − , 0.046875 + 0.00390625, +
256 256 256
 13 
probability of winning = 0.0508   A1 N3 6
 256 
[12]

42
4. (a) evidence of using mid-interval values (5, 15, 25, 35, 50, 67.5, 87.5) (M1)
σ = 19.8 (cm) A2 N3
(b) (i) Q1 = 15, Q3 = 40 (A1)(A1)
IQR = 25 (accept any notation that suggests the interval 15 to 40) A1 N3
(ii) METHOD 1
60 % have a length less than k (A1)
0.6 × 200 = 120 (A1)
k 30 (cm) A1 N2
METHOD 2
0.4 × 200 = 80 (A1)
200 – 80 = 120 (A1)
k = 30 (cm) A1 N2
(c) l < 20 cm ⇒ 70 fish (M1)
70
P(small) = (= 0.35) A1 N2
200
(d)
Cost $X 4 10 12
P(X = x) 0.35 0.565 0.085
A1A1 N2
(e) correct substitution (of their p values) into formula for E(X) (A1)
e.g. 4 × 0.35 + 10 × 0.565 + 12 × 0.085
E(X) = 8.07 (accept $8.07) A1 N2
[15]
2
5. (a) Using E(X) = ∑ x P(X = x)
0
(M1)

3 6 1
Substituting correctly E(X) = 0 × + 1× + 2 × A1
10 10 10
= 0.8 A1 N2
(b) (i)

A1A1A1 N3

43
 2 1 2
(ii) P(Y = 0) = × = A1
5 5 30
 4 2 2 4
P(Y = 1) = P(RG) + P(GR)  = × + ×  M1
 6 5 6 5
16
= A1
30
4 3 12
P(Y = 2) = × = (A1)
6 5 30
For forming a distribution M1

y 0 1 2
2 16 12
P(Y = y)
30 30 30

N4
2  1
(c) P(Bag A) = =  (A1)
6  3
4  2
P(Bag B) = =  (A1)
6  3
For summing P(A ∩ RR) and P(B ∩ RR) (M1)
1 1 2 12
Substituting correctly P(RR) = × + × A1
3 10 3 30
= 0.3 A1 N3
P( A ∩ RR )
(d) For recognising that P(1 or 6 │ RR) = P(A│RR) = (M1)
P( RR )
1 27
= ÷ A1
30 90
= 0.111 A1 N2
[19]
6. (a) (i) Attempt to set up sample space, (M1)
Any correct representation with 16 pairs A2 N3
eg 1,1 2,1 3,1 4,1
1,2 2,2 3,2 4,2
1,3 2,3 3,3 4,3
1,4 2,4 3,4 4,4

1
(ii) Probability of two 4s is (= 0.0625) A1 N1
16

44
(b)

x 0 1 2
P(X = x) 9 6 1
16 16 16

A1A1A1 N3
(c) Evidence of selecting appropriate formula for E(X) (M1)
2
eg E(X) = ∑ x P( X = x ) ,
0
E(X) = np

Correct substitution
9 6 1 1
eg E(X) = 0 × +1× +2× , E(X) = 2 ×
16 16 16 4
8  1
E(X) = =  A1 N2
16  2 
[10]

45
5.7 Binomial Distribution Paper 1
3
1
1. (a) (i) Attempt to find P(3H) =   (M1)
3
1
= A1 N2
27
(ii) Attempt to find P(2H, 1T) (M1)
2
1 2
= 3  A1
 3 3
2
= A1 N2
9
1 
(b) (i) Evidence of using np  ×12  (M1)
 3 
expected number of heads = 4 A1 N2
(ii) 4 heads, so 8 tails (A1)
E(winnings) = 4 × 10 – 8 × 6 (= 40 – 48) (M1)
= –$ 8 A1 N1
[10]
2.
(a) Evidence of binomial formula (M1)
5
 5  1 
P(X = 3) =     A1
 3  2 

=
5
(= 0.313) A1 N2
16
(b) METHOD 1
P(at least one head) = 1 − P(X = 0) (M1)
5
1
= 1−   A1
2

=
31
(= 0.969) A1 N2
32
METHOD 2
P(at least one head) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
+ P(X = 5) (M1)
= 0.15625 + 0.3125 + 0.3125 + 0.15625 + 0.03125 A1
= 0.969 A1 N2
[6]

46
5.7 Binomial Distribution Paper 2
1. (a) X ~ B(100,0.02)
E(X) = 100 × 0.02 = 2 A1 1
100 
(b) P(X = 3) =   (0.02)3(0.98)97 (M1)
 3 
= 0.182 A1 2
(c) METHOD 1
P(X > 1) = 1 – P(X ≤ 1) = 1 – (P(X = 0) + P(X = 1) M1
= 1 – ((0.98)100 + 100(0.02)(0.98)99) (M1)
= 0.597 A1 2
METHOD 2
P(X > 1) = 1 – P (X ≤ 1) (M1)
= 1 – 0.40327 (A1)
= 0.597 A1 2
P(X ≥ 1) A0
= 1 – P(X<2) = 1 – 0.67668 M1(ft)
= 0.323 A1(ft) 2
[6]
2. (a) correct substitution into formula for E(X) (A1)
e.g. 0.05× 240
E(X) =12 A1 N2 2
(b) evidence of recognizing binomial probability (may be seen in part (a)) (M1)
 240 
e.g.   (0.05)15 (0.95)225, X ~ B(240,0.05)
 15 
P(X =15) = 0.0733 A1 N2 2
(c) P(X ≤ 9) = 0.236 (A1)
evidence of valid approach (M1)
e.g. using complement, summing probabilities
P(X ≥10) = 0.764 A1 N3 3
[7]
3. (a) evidence of recognizing binomial probability (may be seen in (b) or (c)) (M1)
7
e.g. probability =   (0.9)4(0.1)3, X ~ B(7, 0.9), complementary
 4
probabilities
probability = 0.0230 A1 N2
(b) correct expression A1A1 N2
7
e.g.   p4(1 – p)3, 35p4(1 – p)3
 4
(c) evidence of attempting to solve their equation (M1)
7
e.g.   p4(1 – p)3 = 0.15, sketch
 4
p = 0.356, 0.770 A1A1 N3
[7]

47
4. (a) 36 outcomes (seen anywhere, even in denominator) (A1)
valid approach of listing ways to get sum of 5, showing at least two pairs (M1)
e.g. (1, 4)(2, 3), (1, 4)(4, 1), (1, 4)(4, 1), (2, 3)(3, 2) , lattice diagram
4  1
P(prize) = =  A1 N3
36  9 
(b) recognizing binomial probability (M1)
 8  1   8 
3 5
 1
e.g. B  8,  , binomial pdf,     
 9  3  9   9 
P(3 prizes) = 0.0426 A1 N2
[5]
5. (a) (i) valid approach (M1)
1
e.g. np, 5 ×
5
E(X) = 1 A1 N2
(ii) evidence of appropriate approach involving binomial (M1)
 1
e.g. X ~ B  5, 
 5
recognizing that Mark needs to answer 3 or more questions correctly (A1)
e.g. P(X ≥ 3)
valid approach M1
e.g. 1 – P(X ≤ 2), P(X = 3) + P(X =4) + P(X = 5)
P(pass) = 0.0579 A1 N3
(b) (i) evidence of summing probabilities to 1 (M1)
e.g. 0.67 + 0.05 + (a + 2b) + ... + 0.04 = 1
some simplification that clearly leads to required answer
e.g. 0.76 + 4a + 2b = 1 A1
4a + 2b = 0.24 AG N0
(ii) correct substitution into the formula for expected value (A1)
e.g. 0(0.67) + 1(0.05) + ... + 5(0.04)
some simplification (A1)
e.g. 0.05 + 2a + 4b + ... + 5(0.04) = 1
correct equation A1
e.g. 13a + 5b = 0.75
evidence of solving (M1)
a = 0.05, b = 0.02 A1A1 N4
(c) attempt to find probability Bill passes (M1)
e.g. P(Y ≥ 3)
correct value 0.19 A1
Bill (is more likely to pass) A1 N0
[17]

48
6. (a) E(X) = 2 A1 N1
(b) evidence of appropriate approach involving binomial (M1)
10 
e.g.   (0.2)3, (0.2)3(0.8)7, X ~ B(10, 0.2)
3
P(X = 3) = 0.201 A1 N2
(c) METHOD 1
P(X ≤ 3) = 0.10737 + 0.26844 + 0.30199 + 0.20133 (= 0.87912...) (A1)
evidence of using the complement (seen anywhere) (M1)
e.g. 1 – any probability, P(X > 3) = 1 – P(X ≤ 3)
P(X > 3) = 0.121 A1 N2
METHOD 2
recognizing that P(X > 3) = P(X ≥ 4) (M1)
e.g. summing probabilities from X = 4 to X = 10
correct expression or values (A1)
10
10 
e.g. ∑   (0.2)10–r(0.8)r
r =4  r 
0.08808 + 0.02642 + 0.005505 + 0.000786 + 0.0000737 + 0.000004 + 0.0000001
P(X > 3) = 0.121 A1 N2
[6]
7. (a) evidence of binomial distribution (may be seen in parts (b) or (c)) (M1)
e.g. np, 100 × 0.04
mean = 4 A1 N2
100 
(b) P(X = 6) =   (0.04 )6 (0.96 )94 (A1)
 6 
= 0.105 A1 N2
(c) for evidence of appropriate approach (M1)
e.g. complement, 1 − P(X = 0)
P(X = 0) = (0.96)100 = 0.01687... (A1)
P(X ≥ 1) = 0.983 A1 N2
[7]
8. (a) evidence of using binomial probability (M1)
7
e.g. P(X = 2) =   (0.18)2(0.82)5
 2
P(X = 2) = 0.252 A1 N2
(b) METHOD 1
evidence of using the complement M1
e.g. 1 – (P(X ≤ 1))
P(X ≤ 1) = 0.632 (A1)
P(X ≥ 2) = 0.368 A1 N2
METHOD 2
evidence of attempting to sum probabilities M1
e.g. P(2 heads) + P(3 heads) + ... + P(7 heads), 0.252 + 0.0923 + …
correct values for each probability (A1)
e.g. 0.252 + 0.0923 + 0.0203 + 0.00267 + 0.0002 + 0.0000061
P(X ≥ 2) = 0.368 A1 N2
[5]

49
9. (a) X ~ B(100, 0.02)
E(X) = 100 × 0.02 = 2 A1 N1
100 
(b) P(X = 3) =   (0.02)3 (0.98)97 (M1)
 3 
= 0.182 A1 N2
(c) METHOD 1
P(X > 1) = 1 – P(X ≤ 1) = 1 – (P(X = 0) + P(X = 1)) M1
= 1 – ((0.98)100 + 100(0.02)(0.98)99) (M1)
= 0.597 A1 N2
METHOD 2
P(X > 1) = 1 – P(X ≤ 1) (M1)
= 1 – 0.40327 (A1)
= 0.597 A1 N2
P(X ≥ 1) A0
= 1 – P(X ≤ 2) = 1 – 0.67668 M1(FT)
= 0.323 A1(FT) N0
[6]
10. (a)
Second die
in pair
First die
in pair 1 four
6

four
1
6 5
6 not
four

1 four
5 6
6 not
four

5
6 not
four
A1A1A1 N3
1 5 5 1  5 5 
(b) P(E) = × + × = +  (A2)
6 6 6 6  36 36 

10  5 
=  = or 0.278  A1 N3
36  18 
(c) Evidence of recognizing the binomial distribution (M1)
 5 5 13
eg X ~ B 5 ,  or p = , q =
 18  18 18

 5   5   13 
3 2

P(X = 3) =       (or other evidence of correct setup) (A1)

 3   18   18 
= 0.112 A1 N3

50
 (d) METHOD 1
Evidence of using the complement M1
eg P(X ≥ 3) = 1 − P(X ≤ 2)
Correct value 1 − 0.865 (A1)
= 0.135 A1 N2
METHOD 2
Evidence of adding correct probabilities M1
eg P(X ≥ 3) = P(X = 3) + P(X = 4) + P(X = 5)
Correct values 0.1118 + 0.02150 + 0.001654 (A1)
= 0.135 A1 N2
[12]
35 7 5 1
11. p(Red) = = p(Black) = =
40 8 40 8

 8  1   7 
1 7

(a) (i) p(one black) =      (M1)(A1)

 1  8   8 
= 0.393 to 3 sf (A1) 3
(ii) p(at least one black) = 1 – p(none) (M1)
 8  1   7 
0 8

= 1 –      (A1)
 0  8   8 
= 1 – 0.344
= 0.656 (A1) 3
400
(b) 400 draws: expected number of blacks = (M1)
8
= 50 (A1) 2
[8]

8− 4
 8  1   1 
4

12. (a) p (4 heads) =      (M1)

 4  2   2 
8
8× 7× 6×5  1 
= × 
1× 2 × 3 × 4  2 
70
= ≅ 0.273 (3 sf) (A1) 2
256
8−3
 8  1   1 
3 8
8× 7× 6  1 
(b) p (3 heads) =      = × 
 3  2   2  1× 2 × 3  2 
56
= ≅ 0.219 (3 sf) (A1) 1
256
(c) p (5 heads) = p (3 heads) (by symmetry) (M1)
p (3 or 4 or 5 heads) = p (4) + 2p (3) (M1)
70 + 2 × 56 182
= =
256 256
≈ 0.711 (3 sf) (A1) 3
[6]

51
5.8 Normal Distribution Paper 1
1. (a)

A1A1 N2
Note: Award A1 for vertical line to right of mean, A1
for shading to right of their vertical line.
(b) evidence of recognizing symmetry (M1)
e.g. 105 is one standard deviation above the mean so d is one standard
deviation below the mean, shading the corresponding part,
105 – 100 = 100 – d
d = 95 A1 N2
(c) evidence of using complement (M1)
e.g. 1 – 0.32, 1 – p
P(d < X < 105) = 0.68 A1 N2
[6]
2. METHOD 1
(a) σ = 10 (A1)
1.12 × 10 = 11.2 A1
11.2 + 100 (M1)
x = 111.2 A1 N2
(b) 100 – 11.2 (M1)
= 88.8 A1 N2
[6]
METHOD 2
(a) σ = 10 (A1)
Evidence of using standardisation formula (M1)
x − 100
= 1.12 A1
10
x = 111.2 A1 N2
100 − x
(b) = 1.12 A1
10
x = 88.8 A1 N2
[6]
3. (a) Evidence of using the complement e.g. 1 – 0.06 (M1)
p = 0.94 A1 N2
(b) For evidence of using symmetry (M1)
Distance from the mean is 7 (A1)
e.g. diagram, D = mean – 7
D = 10 A1 N2
(c) P(17 < H < 24) = 0.5 – 0.06 (M1)
= 0.44 A1
E(trees) = 200 × 0.44 (M1)
= 88 A1 N2
[9]

52
5.8 Normal Distribution Paper 2
1. (a) 0.0668 A2 N2
(b) Using the standardized value 1.645 (A1)
k = 26.1 kg A1 N2
(c)

A1A1 N2
[6]
2. (a) P(H < 153) = 0.705 ⇒ z = 0.538(836...) (A1)
153 − µ
Standardizing (A1)
5
153 − µ
Setting up their equation 0.5388... = M1
5
µ =150.30 
= 150 (to 3sf) A1 N3
153 − µ
(b) Z= = 1.138... (accept 1.14 from µ = 150.3, or 1.2
5
from µ = 150) (A1)
P(Z > 1.138) = 0.128 (accept 0.127 from z = 1.14, or 0.115
from z = 1.2) A1 N2
[6]
180 − 160
3. (a) z= =1 (A1)
20
φ (1) = 0.8413 (A1)
P(height > 180) = 1 − 0.8413
= 0.159 A1 N3
(b) z = −1.1800 (A1)
d −160
Setting up equation −1.18 = (M1)
20
d = 136 A1 N3
[6]

53
4. X ~ N(µ, σ2), P(X < 3) = 0.2, P(X > 8) = 0.1
P(X < 8) = 0.9 (M1)
Attempt to set up equations (M1)
3−µ 8−µ
= − 0.8416 , =1.282 A1A1
σ σ
3 − µ = −0.8416σ
8 − µ = 1.282σ
5 = 2.1236σ
σ = 2.35, µ = 4.99 A1A1 N4
[6]

5. X ~ N(µ, σ2), P(X > 90) = 0.15 and P(X < 40) = 0.12 (M1)
Finding standardized values 1.036, –1.175 A1A1
90 − µ 40 − µ
Setting up the equations 1.036 = , – 1.175 = (M1)
σ σ
µ = 66.6, σ = 22.6 A1A1
[6]
6. (a) σ=3 (A1)
evidence of attempt to find P(X ≤ 24.5) (M1)
24.5 − 20
e.g. z =1.5,
3
P(X ≤ 24.5) = 0.933 A1 N3 3
(b) (i)

A1A1 N2
(ii) z = 1.03(64338) (A1)
attempt to set up an equation (M1)
k − 20 k − 20
e.g. = 1.0364, = 0.85
3 3
k = 23.1 A1 N3 5
[8]

54
7. (a) symmetry of normal curve (M1)
e.g. P(X < 25) = 0.5
P(X > 27) = 0.2 A1 N2 2
(b) METHOD 1
finding standardized value (A1)
27 − 25
e.g.
σ
evidence of complement (M1)
e.g. 1– p, P(X < 27), 0.8
finding z-score (A1)
e.g. z = 0.84…
attempt to set up equation involving the standardized value M1
27 − 25 X −µ
e.g. 0.84 = ,0.84 =
σ σ
σ = 2.38 A1 N3 5
METHOD 2
set up using normal CDF function and probability (M1)
e.g. P(25 < X < 27) = 0.3, P(X < 27) = 0.8
correct equation A2
e.g. P(25 < X < 27) = 0.3, P(X > 27) = 0.2
attempt to solve the equation using GDC (M1)
e.g. solver, graph, trial and error (more than two trials must be shown)
σ = 2.38 A1 N3 5
[7]
8. (a) evidence of attempt to find P(X ≤ 475) (M1)
e.g. P(Z ≤ 1.25)
P(X ≤ 475) = 0.894 A1 N2
(b) evidence of using the complement (M1)
e.g. 0.73, 1 – p
z = 0.6128 (A1)
setting up equation (M1)
a − 450
e.g. = 0.6128
20
a = 462 A1 N3
[6]

55
9. (a) evidence of appropriate approach (M1)
e.g. 1 – 0.85, diagram showing values in a normal curve
P(w ≥ 82) = 0.15 A1 N2
(b) (i) z = –1.64 A1 N1
(ii) evidence of appropriate approach (M1)
x − µ 68 − 76.6
e.g. –1.64 = ,
σ σ
correct substitution A1
68 − 76.6
e.g. –1.64 =
σ
σ = 5.23 A1 N1
(c) (i) 68.8 ≤ weight ≤ 84.4 A1A1A1 N3
(ii) evidence of appropriate approach (M1)
e.g. P(–1.5 ≤ z ≤1.5), P(68.76 < y < 84.44)
P(qualify) = 0.866 A1 N2
(d) recognizing conditional probability (M1)
P( A ∩ B)
e.g. P(A│B) =
P( B)
P(woman and qualify) = 0.25 × 0.7 (A1)
0.25 × 0.7
P(woman│qualify) = A1
0.866
P(woman│qualify) = 0.202 A1 N3
[15]

10. A ~ N(46, 102) B ~ N(µ, 122)

(a) P(A > 60) = 0.0808 A2 N2
(b) correct approach (A1)
 60 − µ 
e.g. P  Z <  = 0.85, sketch
 12 
60 − µ
= 1.036... (A1)
12
µ = 47.6 A1 N2
(c) (i) route A A1 N1
(ii) METHOD 1
P(A < 60) = 1 – 0.0808 = 0.9192 A1
valid reason R1
e.g. probability of A getting there on time is greater than
probability of B
0.9192 > 0.85 N2
METHOD 2
P(B > 60) = 1 – 0.85 = 0.15 A1
valid reason R1
e.g. probability of A getting there late is less than probability of B
0.0808 < 0.15
N2

56
 (d) (i) let X be the number of days when the van arrives before 07:00
P(X = 5) = (0.85)5 (A1)
= 0.444 A1 N2
(ii) METHOD 1
evidence of adding correct probabilities (M1)
e.g. P(X ≥ 3) = P(X = 3) + P(X = 4) + P(X = 5)
correct values 0.1382 + 0.3915 + 0.4437 (A1)
P(X ≥ 3) = 0.973 A1 N3
METHOD 2
evidence of using the complement (M1)
e.g. P(X ≥ 3) = 1 – P(X ≤ 2), 1 – p
correct values 1 – 0.02661 (A1)
P(X ≥ 3) = 0.973 A1 N3
[13]

11. X ~ N (7, 0.52)

(a) (i) z=2 (M1)
P(X < 8) = P(Z < 2) = 0.977 A1 N2
(ii) evidence of appropriate approach (M1)
e.g. symmetry, z = −2
P(6 < X < 8) = 0.954 (tables 0.955) A1 N2
(b) (i)

d
A1A1 N2
(ii) z = − 1.645 (A1)
d −7
= −1.645 (M1)
0.5
d = 6.18 A1 N3
(c) Y ~ N(µ, 0.52)
P(Y < 5) = 0.2 (M1)
z = − 0.84162... A1
5−µ
= − 0.8416 (M1)
0.5
µ = 5.42 A1 N3
[13]

57
12. (a)

12.92% 10.38%
B

T t
A1A1 N2
(b) METHOD 1
P(X < r) = 0.1292 (A1)
r = 6.56 A1 N2
1 − 0.1038 (= 0.8962) (may be seen later) A1
P(X < t) = 0.8962 (A1)
t = 7.16 A1 N2
METHOD 2
finding z-values −1.130..., 1.260... A1A1
evidence of setting up one standardized equation (M1)
r − 6.84
e.g. = − 1.13 , t =1.260 × 0.25 + 6.84
0.25
r = 6.56, t = 7.16 A1A1 N2N2
[7]
13. (a) evidence of approach (M1)
23.7 − 21
e.g. finding 0.84…, using
σ
correct working (A1)
23.7 − 21
e.g. 0.84... = , graph
σ
σ = 3.21 A1 N2
(b) (i) evidence of attempting to find P(X < 25.4) (M1)
e.g. using z = 1.37
P(X < 25.4) = 0.915 A1 N2
(ii) evidence of recognizing symmetry (M1)
e.g. b = 21 – 4.4, using z = –1.37
b = 16.6 A1 N2
[7]

58
14. Note: Candidates may be using tables in this question, which leads to a
variety of values. Accept reasonable answers that are consistent
with working shown.
W ~ N(2.5, 0.32)
(a) (i) z = − 1.67 (accept 1.67) (A1)
P(W < 2) = 0.0478 (accept answers between 0.0475 and
0.0485) A1 N2
(ii) z=1 (A1)
P(W > 2.8) = 0.159 A1 N2
(iii)

2.5 kg

A1A1 N2
(iv) Evidence of appropriate calculation M1
eg 1 − (0.047790 + 0.15866), 0.8413 − 0.0478
P = 0.7936 AG N0
(b) (i) X ~ B(10, 0.7935...)
Evidence of calculation M1
eg P(X = 10) = (0.7935...)10
P(X = 10) = 0.0990 (3 sf) A1 N1
(ii) METHOD 1
Recognizing X ~ B(10, 0.7935...) (may be seen in (i)) (M1)
P(X ≤ 6) = 0.1325... (or P(X = 1) + ... + P(X = 6)) (A1)
evidence of using the complement (M1)
eg P(X ≥ 7) = 1 − P(X ≤ 6), P(X ≥ 7) = 1 − P(X < 7)
P(X ≥ 7) = 0.867 A1 N3
METHOD 2
Recognizing X ~ B(10, 0.7935...) (may be seen in (i)) (M1)
For adding terms from P(X = 7) to P(X = 10) (M1)
P(X ≥ 7) = 0.209235 + 0.301604 + 0.257629 + 0.099030 (A1)
= 0.867 A1 N3
[13]

59
15.
Notes: Accept any suitable notation, as long as the candidate’s
intentions are clear.
The following symbols will be used in the
markscheme.
Girls’ height G ~ N(155, 102), boys’ height B ∼ N(160, 122)
Height H, Female F, Male M.
(a) P(G > 170) = 1 − P(G < 170) (A1)
 170 − 155 
P(G > 170) = P  Z <  (A1)
 10 
P(G > 170) = 1 − Φ (1.5) = 1 − 0.9332
= 0.0668 A1 N3
(b) z = −1.2816 (A1)
Correct calculation (eg x = 155 + −1.282 × 10) (A1)
x = 142 A1 N3
(c) Calculating one variable (A1)
eg P(B < r) = 0.95, z = 1.6449
r = 160 + 1.645(12) = 179.74
= 180 A1 N2
Any valid calculation for the second variable, including use of
symmetry (A1)
eg P(B < q) = 0.05, z = −1.6449
q = 160 − 1.645(12) = 140.26
= 140 A1 N2
(d) P(M ∩ (B > 170)) = 0.4 × 0.2020, P(F ∩ (G > 170)) =
0.6 × 0.0668 (A1)(A1)
P(H > 170) = 0.0808 + 0.04008 A1
= 0.12088 = 0.121 (3 sf) A1 N2
P( F ∩ (H >170 ))
(e) P(F H > 170) = (M1)
P(H >170 )

0.60 × 0.0668  0.0401 0.04008 

= = or  A1
0.121  0.121 0.1208 
= 0.332 A1 N1
[17]

60
16. (i) P ( X > 3 200) =P ( Z > 0.4) (M1)
=
1 − 0.6554 =
34.5 % (=
0.345) (A1) (N2)
(ii) P (2 300 < X < 3 300) = P (−1.4 < Z < 0.6) (M1)
= 0.4192 + 0.2257
= 0.645 (A1)
P= =
(both) (0.645) 2
0.416 (A1) (N2)
(iii) = P ( Z < 0.65)
0.7422 (A1)

d − 3 000
= 0.65 (A1)
500
=d $=
3 325 ( $ 3 330 to 3 s.f.) (Accept $3325.07) (A1) (N3)
[8]
185 −170
17. (a) z= = 0.75 (M1)(A1)
20
P(Z < 0.75) = 0.773 (A1) (N3)
(b) z = −0.47 (may be implied) (A1)
d −170
−0.47 = (M1)
20
d = 161 (A1) (N3)
[6]
18. (a) (i) a = −1 (A1)
b = 0.5 (A1)
(ii) (a) 0.841 (A2)
(b) 0.6915 − 0.1587 (or 0.8413 − 0.3085) (M1)
= 0.533 (3 sf) (A1)
(N2) 6
(b) (i) Sketch of normal curve (A1)(A1)

(ii) c = 0.647 (A2) 4

[10]

61
19. X ~ N (80, 82)
(a) P(X < 72) = P(Z < –1) (M1)
= 1 – 0.8413
= 0.159 (A1)
OR
P(X < 72) = 0.159 (G2) 2
(b) (i) P(72 < X < 90) = P( –1 < Z < 1.25) (M1)
= 0.3413 + 0.3944 (A1)
= 0.736 (A1)
OR
P(72 < X < 90) = 0.736 (G3)
(ii)

72 80 90
(A1)(A1) 5
(c) 4% fail in less than x months
⇒ x = 80 – 8 × Φ–1(0.96) (M1)
= 80 – 8 × 1.751 (A1)
= 66.0 months (A1)
OR
x = 66.0 months (G3) 3
[10]

 350 − 310 
20. (a) P(M ≥ 350) = 1 – P (M < 350) = 1 – P  Z <  (M1)
 30 
= 1 – P(Z < 1.333) = 1 – 0.9088
= 0.0912 (accept 0.0910 to 0.0920) (A1)
OR
P(M ≥ 350) = 0.0912 (G2)
(b)

0.025 0.025

–1.96 0 1.96 Z
P(Z < 1.96) = 1 – 0.025 = 0.975 (A1)
1.96 (30) = 58.8 (M1)
310 – 58.8 < M < 310 + 58.8 ⇒ a = 251, b = 369 (A1)
OR
251 < M < 369 (G3)
[5]

62
21. Note: Where accuracy is not specified, accept answers with greater than 3 sf accuracy,
provided they are correct as far as 3 sf
197 – 187.5
(a) z= = 1.00 (M1)
9.5
P (Z > 1) = 1 – Φ(1) = 1 – 0.8413 = 0.1587
= 0.159 (3 sf) (A1)
= 15.9% (A1)
OR
P (H > 197) = 0.159 (G2)
= 15.9% (A1) 3
(b) Finding the 99th percentile
Φ(a) = 0.99 => a = 2.327 (accept 2.33) (A1)
=> 99% of heights under 187.5 + 2.327(9.5) = 209.6065 (M1)
= 210 (3 sf) (A1)
OR
99% of heights under 209.6 = 210 cm (3 sf) (G3)
Height of standard doorway = 210 + 17 = 227 cm (A1) 4
[7]
22. (a) (These answers may be obtained from a calculator or by finding z in each case and the corresponding
area.)
M ~ N (750, 625)
(i) P (M < 740 g) = 0.345 (G2)
OR
z = –0.4 P(z < –0.4) = 0.345 (A1)(A1)
(ii) P (M > 780 g) = 0.115 (G2)
OR
z = 1.2 P(z > 1.2) = l – 0.885 = 0.115 (A1)(A1)

(iii) P(740 < M < 780) = 0.540 (G1)

OR
1 – (0.345 + 0.115) = 0.540 (A1) 5
(b) Independent events
Therefore, P (both < 740) = 0.3452 (M1)
= 0.119 (A1) 2
(c) 70% have mass < 763 g (G1)
Therefore, 70% have mass of at least 750 – 13
x = 737 g (A1) 2
[9]

63
23. (a) Let X be the random variable for the IQ.
X ~ N(100, 225)
P(90 < X < 125) = P(–0.67 < Z < 1.67) (M1)
= 0.701
70.1 percent of the population (accept 70 percent). (A1)
OR
P(90 < X < 125) = 70.1% (G2) 2
(b) P( X ≥ 125) = 0.0475 (or 0.0478) (M1)
P(both persons having IQ ≥ 125) = (0.0475)2 (or (0.0478)2) (M1)
= 0.00226 (or 0.00228) (A1) 3
(c) Null hypothesis (H0): mean IQ of people with disorder is 100 (M1)
Alternative hypothesis (H1): mean IQ of people with disorder
is less than 100 (M1)
  
  
  95.2 − 100  
P( X < 95.2) = P  Z < = P(Z < –1.6) = 1 – 0.9452
 15 
  
  25  
= 0.0548 (A1)
The probability that the sample mean is 95.2 and the null hypothesis
true is 0.0548 > 0.05. Hence the evidence is not sufficient. (R1) 4
[9]
25 − 25.7
24. (a) Z= = –1.4 (M1)
0.50
P(Z < –1.4) = 1 – P(Z < 1.4)
= 1 – 0.9192
= 0.0808 (A1)
OR
P(W < 25) = 0.0808 (G2) 2
(b) P(Z < –a) = 0.025 ⇒ P(Z < a) = 0.975
⇒ a = 1.960 (A1)
25 − µ
= –1.96 ⇒ µ = 25 + 1.96 (0.50) (M1)
0.50
= 25 + 0.98 = 25.98 (A1)
= 26.0 (3 sf) (AG)
OR
25.0 − 26.0
= –2.00 (M1)
0.50
P(Z < –2.00) = 1 – P(Z < 2.00)
= 1 – 0.9772 = 0.0228 (A1)
≈ 0.025 (A1)
OR
µ = 25.98 (G2)
⇒ mean = 26.0 (3 sf) (A1)(AG) 3
(c) Clearly, by symmetry µ = 25.5 (A1)
25.0 − 25.5
Z= = –1.96 ⇒ 0.5 = 1.96σ (M1)
σ
⇒ σ = 0.255 kg (A1) 3

64
 cement saving
(d) On average, = 0.5 kg (A1)
bag
cost saving
= 0.5(0.80) = $0.40 (M1)
bag
5000
To save $5000 takes = 12500 bags (A1) 3
0.40
[11]
25. (a) Let X be the lifespan in hours
X ~ N(57, 4.42)

a 0 b
(i) a = –0.455 (3 sf) (A1)
b = 0.682 (3 sf) (A1)
(ii) (a) P (X > 55) = P(Z > –0.455)
= 0.675 (A1)
 2 3 
(b) P (55 ≤ X ≤ 60) = P  ≤Z ≤ 
 4.4 4.4 
≈ P(0.455 ≤ Z ≤ 0.682)
≈ 0.6754 + 0.752 – 1 (A1)
= 0.428 (3sf) (A1)
OR
P (55 ≤ X ≤ 60) = 0.428 (3 sf) (G2) 5
(b) 90% have died ⇒ shaded area = 0.9 (M1)

0 (A1)
Hence t = 57 + (4.4 × 1.282) (M1)
= 57 + 5.64 (A1)
= 62.6 hours (A1)
OR t = 62.6 hours (G3) 5
[10]

65
  50 − µ 
26. (a)P(speed > 50) = 0.3 = 1 – Φ   (A1)
 10 
50 − µ
Hence, = Φ–1(0.7) (M1)
10
µ = 50 – 10Φ–1(0.7) (M1)
= 44.75599 …… = 44.8 km/h (3 sf) (accept 44.7) (AG) 3
(b) H1: “the mean speed has been reduced by the campaign”. (A1) 1
(c) One-tailed; because H1 involves only “<”. (A2) 2
(d) For a one-tailed test at 5% level, critical region is
Z < µm – 1.64σm (accept –1.65σm) (M1)
σ 10
Now, µm = µ = 44.75...; σm = = = 2 (allow ft) (A1)
n 25
So test statistic is 44.75... –1.64 ×2 = 41.47 (A1)
Now 41.3 < 41.47 so reject H0, yes. (A1) 4
[10]
27. (a) Area A = 0.1 (A1) 1
(b) Since p (X ≥ 12) = p (X ≤ 8), (M1)
then 8 and 12 are symmetrically disposed around the (M1)(R1)
mean.
8 + 12
Thus mean = (M1)
2
= 10 (A1) 5
 12 − 10 
(c) Φ  = 0.9 (A1)(M1)(A1)
 σ 
2
⇒ = 1.282 (or 1.28) (A1)
σ
2  2 
σ =  or  (A1)
1.282  1.28 
= 1.56 (3 sf) (AG) 5
 11 − 10 
(d) p (X ≤ 11) = p  Z ≤  (or 1.56) (M1)(A1)
 1.561 
= p (Z ≤ 0.6407) (or 0.641 or 0.64) (A1)
= Φ(0.6407) (M1)
= 0.739 (3 sf) (A1) 5
[16]

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