Psych 2810 November Exam A - UWO Math...
Transcript of Psych 2810 November Exam A - UWO Math...
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Psych 2810 November Exam Solutions
Section A
Example 1.
a) quantitative, continuous
b) qualitative, unordered
c) qualitative, ordered
d) qualitative, unordered
e) quantitative, discrete
f) qualitative, ordered
g) qualitative, ordered
A1.
a) Quantitative, discrete
b) Qualitative, unordered
c) Quantitative, continuous
d) Qualitative, ordered
e) Qualitative, unordered
Section B
Example 1. 34, 45, 56, 67, 76, 76, 77, 77, 89, 90
⋯ 68.7
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mode = 76,77
median = 76
Example 2.
Temperature Frequency Midpoint
(‐10,0) 100 ‐5
(0,10) 65 5
(10,20) 70 15
(20,30) 65 25
300
5 100 5 65 15 70 25 65
300
8. 33
Example 3. In a right skewed curve, the mean is larger than the median.
See p. 14 of the booklet
B1. ?=200‐50‐80‐30
?=40
Midpoint 5
15
25
35
5 50 15 80 25 40 35 30
200
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17.5
B2. 34, 44, 50, 56, 66, 67, 67, 78, 78, 88, 98
Mode =67,78 (bimodal)
Median=67
Mean=⋯
=66
B3. Before 4th mark
80 median =80 70, 80 ,90
After 4th mark 70, 80 ,80, 90 median = 80
Answer is E
B4. 78.75
245 4 78.75
245 315
70
B5. Answer is (iii) Think of the normal distribution
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Section C
97th percentile means that 97% of kids his age are shorter than him, or
only 3% are taller than him.
75th percentile means that 75% of kids his age weigh less than him and
25% weigh more than him.
Example 1. ⋯ 65.3
∑
...definitional...same answer with
computational
. . . . . . .
2853.436
475.5717
√475.5717 21.8
Example 2. ∑ ∑
...computational formula
5 ∑ 20 14 16 18 20 88
∑ 20 14 16 18 20 1576
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. 6.8
√6.8 2.6
Example 3. 6 ⋯ 70.3
∑
67 70.3 87 70.3 90 70.3 56 70.3 45 70.377 70.3
5
1567.345
313.5
Example 4. 34, 45, 56, 56, 78,98
median = 2 56
1 45
3 78
C1.4,5,6,3,2
inorder2,3,4,5,6
Mean=(4+5+...+2)/5=20/5=4
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Variance=[(4‐4)2+(5‐4)2+(6‐4)2+(3‐4)2+(2‐4)2]/4
Variance=s2=2.5
s=√2.5=1.58
C2.5,10,10,10,12,15,18,20,20
Q1=10
Q2=median=12
Q3=18
C3.
24,46,49,51,64,64,**67,**81,88,89,97,103,120
Ifwelook,wecanseethatthedataisinorder.Thereare13datapoints,sothemedianisthe7thdatawhichis67.
Q1isinbetweenthe3rdand4thdatapoints,andQ3isbetweenthe10thand11thdatapoints.
501 Q
median=67
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933 Q
9624120 Range
435093 IQR
Wecanusetheformulasabovetocalculatethemeanandsamplestandarddeviation:
5.7213
120103974624
x
8.2612
25.8615
12
)5.72120()5.72103()5.7297()5.7246()5.7224( 22222
s
Wenowcheckforoutliers:
5.157)43(5.1935.13 IQRQ
5.14)43(5.1505.11 IQRQ
Sincenodatapointisabove157.5orbelow‐14.5,therearenooutliersinthedataset.
C4.Calculatethemean,median,mode,range,standarddeviationandvarianceofthefollowingmarks:50,65,70,55,80,95,35,65,80
Writethenumbersinascendingorder:n=9
35,50,55,65,65,70,80,80,95
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Mode=65
Median=middle#=5thnumber=65
Mean=35 50 ⋯ 95
966
Range=max‐min=95‐35=60
Makeacharttofindthestandarddeviation:
235‐66 96150‐66 25655‐66 12165‐66 165‐66 170‐66 1680‐66 19680‐66 19695‐66 841
2
125898
17.99
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C5.Theanswerisc).
C6.Theanswerisb).
C7.Theanswerisa).
C8.34,32,26,28,27,40,23,21,34,37,38,41
Inorder...
21,23,26,27,28,32,34,34,37,38,40,41
21 23 ⋯ 41
1231.75
Makeacharttofindthestandarddeviation.
221‐31.75 115.5623‐31.75 76.5626‐31.75 33.0627‐31.75 22.5628‐31.75 14.0632‐31.75 0.0634‐31.75 5.0634‐31.75 5.0637‐31.75 27.5638‐31.75 39.06
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40‐31.75 68.0641‐31.75 85.56
478.1611
6.6
Thevarianceis 2 43.5
D. Grand Mean
Example 1.
120 67 100 76 85 79
120 100 85
73.3
Example 2.
57.7
Grand median = can’t be calculated since we don’t know all of the
values (scores) of each student.
Translation and Scale Theorems
Example 3.
2 16.5 9 3
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3 2
∴ 3/2 16.5 24.75
D1.
77.1
D2. 57.8
Grand median = impossible to tell because we don’t have
all of the data to find the middle score.
Scale Theorem: If we multiply every score (x) by a constant
(c),then the new mean is and the new standard deviation is and the new variance is
D3. 1.5 4
(start) 2
15.2
∴
2 1.5
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2 1.5 1. 3
∴ 1. 3 15.2 ≐ 20.3
Section E
Example 1. 75% lie between 120‐2(5) and 120+2(5)
110 lb and 130 lb
89% lie between 120‐3(5) and 120+3(5)
105 lb and 135 lb
Example 2. 30, 70, 50, 3
d= distance that "a" or "b" are from the mean,
6. 6
% 100 1 100 1.
97.75%
Example 3. 45,75 45, 75, 60, 10
1.5
1 120 1.
67
Therefore, 67 scores are between 45 and 75.
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Example 4. a) 200 80[68,92] is 75% of data
0.75 1
0.25
0.25 1.
4
2
80 2 68
2 12
6
or use and solve for ...where d= 12 the distance from the "a"
and "b" from the mean
212
2 12
6
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b) 72, 88
, 72 88
80
86
1.33
200 1.
87
E1. 5 162
75% lie between 162‐2(5), 162+2(5)
=152 and 172
89% lie between 162‐3(5), 162+3(5)
=147 and 177
E2. 20,50
↙↘
∴ 4 20 5020 50
235
∴ 35 4 20
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4 20 35
4 15
3.75 ∴ 100 1
100 1.
≐ 93%
E3. 2 10 30 20
∴ 20 2 10
2 10 5
∴ 100 1
100 1 = 96%
Section F
Example 1. a) 60 9 3
60 3 63 ← 84 ∴ 84% 63
b)
2 60 2 3 66 ← 97.5 ∴97.5% 66
c) 60 3 57 ← 16 ∴ 16% 66
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d)
20 60 2 3 54 ← 2.5 ∴ 2.5% 54
100 2.5 97.5% 54
Example 2.
100 60 25 5
a) 20 60 2 5 70 ← 97.5
∴ 97.5% 2.5% 70 0.025 100 2.5
b) What score is in the 70th percentile?
84
60 5 65 16
60 5 55
Remember, the number you are finding must be in the middle of the
chart!
5 ∆60 50
7065 84
2034
∆ ∆ 2.94 ∴ 60 2.94 62.94
62.9
∴ 70 62.9
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c 48, 72
48 72 60 5
2.4
1
100 11
2. 482.6 ∴ 82
Example 3.
12084 120 1 140
20
0.5
20 15120 50135 ∆140 84
34
Remember, the number you are finding must be in the middle of the
chart!
∆ ∆ 25.5 ∴ 50 25.5 75.5
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∴ 75.5 ∴ 100 75.524.5% 135
number of scores=0.245x150=36 people
c) 120 20 100 ← 16
Remember, the number you are finding must be in the middle of the
chart!
20 8100 16108120 50
∆ 34 ∆ 13.6
∴ 16 13.6 29.6
∴ 29.6% 108 70.4% 108
=0.704x 150=105.6 scores
Example 4 200 98 16 4
120
1
120 200 1
1
0.4 2.5 1.58
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∴
98 1.58 4 ∴ 91.7
98 1.58 4
104.3
or use
1.58=d/4
d=6.32
98 ‐ 6.32, 98 + 6.32
∴ 91.7 104.3
F1 . 40 100 10
a) 2 .
40 1 10 50
x 2s 40 2 10 60 ∴ 97.5%arebelow60
100 97.5 2.5%
0.025x100=2.5 scores are greater or equal to 60
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b) Remember, the number you are finding must be in the middle of
the chart!
10 Δ 40 5075
25 34
5084
Δ 7.4 ∴ 40 7.4 47.4
Therefore, 47 is the score of the 75th percentile
c) Chebyshev’s not MSSD
22,58
40 10(same)
40 10 22 10 18 1.8
∗
100 11
100 11
1. 869.1%
0.691x100=69.1 scores between 22 and 58
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F2. 200 100
a) 120 → 84
120 100 ∴ 20
0.75
b)
20 10120 84 ∆130 13.5
140 97.5
∆
.
∆ 6.75
∴ 84 6.75 90.75
∴ 130 90.75% 130 100 90.75%
=9.25 % above
∴ # 200 0.0925 18.5 19
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c)
20 580 1685 ∆100 50
34
∆34
520
∆ 8.5
16 8.5 24.5 ∴ 85 24.5
∴ 100 24.5 75.5%
∴ 0.755 200 151
d) Chebyshev’s
85,115
100 20
0.75
100 20 85 And solve for k
∴ 100 11
=100 1.
77.8%
∴ 1 0.778 200 156 ople
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F3. Chebyshev’s not MSSD
200 16 ∴ 4 95
11
150 200 1
1
0.75 1
0.25
0.25 1 √4 2
∴
103
Other z value 2 87
∴ 87 103
Section G
Example 1. 10! = 3628800
Example 2. 3! 103
3!
!
! !3! !
!720
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Example 3. 10
631!
! ! !840
Example 4. !
!
!
!27907200
Example 5. Each week Natalie and three friends pool their money and
buy 12 lottery tickets. In how many ways can the tickets be divided
equally amongst Natalie and her friends?
123333
12!3! 3! 3! 3!
369600
Example 6. a) 155
!
! !3003
b) !! !
!! ! 0.0419
Example 7. a) 9
432!
! ! !1260
b) 3! 6321
! !
! ! !360
↳ 3 3
Section G
G1. 8! 40320
G2. 253
!
! !2300
G3. =!
!
!
!60
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G4. Arrange =permutations
!
!
!
!5040
G5. 5M, 6W → 4
a) = 3 women or 4 women
= 63
64
20 15 35
b) = Total – no men
= 114
50
64
330 15 315
G6. a) 525
2598960
b)135
1287
c)41
135
5148
↑ 1 4
G7. 7 scientist, 3 lab tech Project #1 ‐ 5
Project #2 ‐ 3
Project #3 – 2
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a) 10532
!
! ! !2520
5 people to project 1 3 people to project 2
and 2 people to project 3
b)3! 7421
arrange 3 lab tech’s – 1 to each project
project 1 needs 4 more people; project 2 needs 2
more people and project 3 needs 1 more person
OR
31
21
11
74
32
11
G8. 8M, 7W
# ways at least 1 women = Total ways ‐ # ways no women
=155
80
75
= 3003 – 21
= 2982
G9. exactly 2 men = 2 men, 3 women
= 72
83
1176
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G10. 4M, 5H
93
84
G11. 53
you can only choose from the history books
G12. ‐ permutation since you want to assign duties
153
3! 2730 15 3
G13. Invite A, not B or invite B, not A or invite neither
64 6
4 6
5
The answer is E
G14. 7 scientist, 3 lab tech Project #1 ‐ 5
Project #2 ‐ 3
Project #3 – 2
a) 10532
!
! ! !2520
5 people to project 1 3 people to project 2
and 2 people to project 3
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b)3! 7421
arrange 3 lab tech’s – 1 to each project
project 1 needs 4 more people; project 2 needs 2
more people and project 3 needs 1 more person
OR
31
21
11
74
32
11
Section H
Example 1. a) 1/2 b) 2/52=1/26
Example 2.
Pr(A or B)=Pr(A) + Pr(B)‐ Pr(A and B)
0.8=0.6 + 0.5 ‐ Pr(A and B)
Pr(A and B)= 0.3...not mutually exclusive since Pr(A and B) 0
Pr(A)xPr(B)=0.6(0.5)=0.30=Pr(A and B)...yes, independent
The answer is ii)
Example 3. a) 16
862!
! ! !360360
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b)
0.0000777
OR
88
86
22
16862
Example 4. a)
↘ 2
!
! ! !!
! ! !
0.000357
OR
b) Pr 5 Pr 5 Pr 6
Example 5. Pr 1 1 Pr
1
*draw a tree diagram if it helps to see all of the outcomes
S={BBB, BBG, BGB, BGG,GBB, GGB, GBG, GGG}
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Example 6. a) Pr(2H, 2C) !
! !!
! !!
! !!
! !
0.0609
b) Pr 1 1 Pr
1 0.78
Example 7. a) Pr Pr Pr
0.3 0.4 0.12
b) Pr(E or F) Pr Pr Pr
Pr Pr Pr Pr
0.3 0.4 0.3 0.4
0.7 0.12
0.58
c)Pr 0.18...draw a Venn diagram
Example 8.
SM __ __ __ __ __ __ ! !
!0.25
Example 9.
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P __ __ __ __ __ __ __ __ M ! !
!0.022..the 2! is since the
books could switch to opposite ends
Example 10.
Pr 2 1 Pr 2
1 ! !
!1 0.25 0.75
F1 F2 __ __ __ __ __ __
Example 11. J = Jessy wins N = Natalie wins
Pr 0.50 0.30 0.15
Example 12. D = has disease
Pr Pr 0.99 0.0099
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H1. a) Pr(EF) Pr Pr 0.2 0.6 0.12
b) Pr(E or F) Pr Pr Pr
0.2 0.6 0.12
0.8 0.12 0.68
Pr ∴ 0.08
H2. Pr 3 , 2
0.0085834
H3. Jessica and Diandra must be on
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∴ 1 10 2 8
Pr , 0.067
22
1
H4. 0.0385
H5. 11 people – choosing 4
a) 0.045 2
b) 0.45
c)
∴ 3
∴ 9
93114
0.25
H6. Pr 2 Pr 2 Pr 3 * draw a tree
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0.5
H7. 8 letters ADG, ___, ___, ___,____,____ ! !
!0.107
H8.
!! ! !
!! ! !!
! ! !
≐ 0.164
H9. 5R, 3M, 2H
a) Pr(1R) = 0.417
b) Pr(at least 1R) = 1‐Pr(no R)
1 1 0.083 0.92
H10. Pr(1 of each) = 0.25
H11. 3P, 5B, 3Bl → 3
Pr(1 of each) = 0.273
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H12. 8R, 4B, 3P, 5O → 5
a) Pr(at least 2P) = 2P or 3P or 4P or 5P
or 1‐Pr(0 purple) – Pr(1 purple)
= 1‐ 1 0.399 0.464 0.14
b) Pr(2 red, 1 blue, 1 purple, 1 orange)
0.108
H13. 16 university 4 not
20 – 8 men, 12 women
a) 12 secretarial, 2 maintenance and 6 management
Pr(all women in secretarial)
=
0.00000794
All 12 women into secretarial and 8 men into 2 maintenance
and 6 management
OR
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Pr(all women in secretarial)
=
b)Pr(at least 3) = Pr(3 management by univ.)+Pr(4 management
by univ.) + Pr(5 management by univ.)
=
Choose 3 for management from 16 university (need 12
secretarial, 2 maintenance,3 more management) and
(need 12 secretarial, 2 maintenance and 2 more management)
H14. Pr(at least 1 man)=1 – Pr (no man)
1 ‐ 1 0.0186 0.98
H15. Pr (3 men or 4 men)
0.326 0.0932
0.419
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H16. 5 Florida, 4 Europe, 3 Australia
a) Pr (none to Florida) = 0.027
b) Pr (all different) 0.273
*H17.
4 dogs, 6 cats, 3 bunnies
a)
!
! !
!
! !!
! !
b) at least 3 cats= 3 cats or 4 cats
=
!
! !
!
! !
!
! !!
! !
c) 42
62
32
4!2! 2!
6!2! 4!
3!2! 1!
=270
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Section I
Example 1.
Pr (E/F)
.
.0.5
Example 2.
Pr(D/+)
. .
. . . .0.0089
Example 3.
Pr(A/B′ Pr
Pr(A/B′ Pr 0.4
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Example 4. ... is 0.70. The probability that a student passes the second
midterm, give they failed the first midterm is 0.5. The probability that a
student fails the second midterm given they passed the first is 0.20.
Find the probability that a randomly selected student passes both of
the first and second midterms.
Pr Pr 0.70 0.80 0.56
Example 5. Seventy percent of customers who buy meat‐lovers get the
combo, while seven percent of all customers buy vegetarian and do not
opt for the combo.
0.10 0.07
0.7
a) 0.60 0.6 0.7 0.3 0.10 0.3
∴ 0.5
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b) Find the probability that a randomly selected pepperoni pizza eater
did not opt for the combo?
a) Pr ∩ Pr 0.6 0.3 0.18 b) Pr(C′/P) 0.5
Example 6.
a) 205555
!
! ! ! !
b)
!! ! !!
! ! ! !
.0.0000645
OR
55
155
105
55
205555
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Example 7.
a) Pr(H)=0.05(0.5) + .75(0.7)+ 0.2(0.2)=0.59
ˆ 80 x 0.59 = 47.2 or 47 times
b) Pr / )=
=
. .
.
.
.0.549
Section I
I1. Pr(E/F)=
.
.0.75
I2. Pr(F/E)=
.
.0.5
I3. If Pr(A or B) = 0.8,Pr(A) = 0.2 and A,B are independent
(Pr(AB) = Pr(A) Pr(B) ) events on the same sample space, find
Pr(A/B) and Pr(B).
Pr(A or B) =Pr(A)+ Pr(B) + Pr(AB)
0.8 = 0.2 + Pr(B) – 0.2Pr(B)
0.6 = 0.8 Pr(B)
Pr(B) = 0.75
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Pr(A/B) =
Pr 0.2
I4. 57
a) Pr(YY)= 0.318
Or use
!
! !
21
12!2! 10!
66
∴ 0.318
b) Pr(GY) + Pr(YG) = 0.53
I5. ... day is 0.35 and the probability that it will be windy (the
winds will exceed 30 km/hour) is 0.25 and the probability that
it will be windy given that it doesn’t rain is 0.35.
Pr(W) = 0.25
Pr(R) =0.35
Pr(W/ ′ 0.35
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a) Pr( ′ Pr 0.65 0.25 0.90 b) Pr(W) = Pr(RW)+Pr( ′
0.25 0.35 0.65 0.35
0.25 0.35 0.2275
0.0225 0.35
0.0643
Pr(RW) =
. .
.0.09002
I6. 0.004
Pr(D /+) =
. .
. . . .0.164
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I7. Pr( ′ 1 0.5 0.50
Pr(A / ′.
. .
.0.40
Pr(A or B′ Pr Pr ′ Pr ′
Pr Pr Pr Pr
0.4 0.5 0.4 0.5 0.70
I8. Pr(A / C′
. .
. . . .0.529
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I9. 5CB, 4CC, 3HK
a) Pr(CC / B′ .
. . .0.295
b) Pr(CC or B′ Pr Pr Pr
0.8 0.7 0.9 0.7
0.642
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I10. Pr(B) = Pr(WB)+Pr(MB)
= 0.3(0.6) + 0.7(0.3)
=0.39
I11. Pr(W / B )=
. .
.0.46
I12. Pr(C′/. .
. . . .0.781
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I13. B = Brittany wins
Pr(R / B)=
. .
. . . .0.56
I14. Pr(S / P) =
. .
. . . .0.413
S = use steroids S′ doesn’t use steroids
P = positive test
P′= negative test
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Section J
J1. a) Pr (HD and C′ 0.10
to find
0.2 0.10 0.5
b)To find 0.60 0.6 0.75 0.2 0.5 0.2
0.6 0.45 0.1 0.2
0.05 0.2
0.25 Pr(C′/F)= from tree 0.75
J2. a) Pr (2 red, 2 green, 2 blue) = 0.195
b)Pr( one person gets all green) =
choose which person gets all green
all green
8 left – 4 to each of 2 people
c) 6 even, 6 odd
red #1‐5 green #6‐9 blue #10‐12
Pr (1st and last both odd or both red) =
Pr(both odd) + Pr(both red) – Pr(odd and red)
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=
J3. a) 15
555756756
b) Pr(Megan gets all hearts) =
0.000333
All hearts to Megan 10 cards left – 5 to
each of 2 people
OR
55
105
55
15555
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Section K
Example 2.
2 3 4 5 6 7
8 9 10 11 12
7
Pr 2 1
36
3 236
4 336
5 436
6 536
7 636
8 536
9 436
10 336
11 236
12
136
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Example 3.
X Pr(X) 1 12000/20000=0.6 2 6000/20000=0.3 3 0.1
Sample Pr( 1,1 1 0.6(0.6)=0.36 1,2 1.5 0.6(0.3)=0.18 1,3 2 0.6(0.1)=0.06 2,1 1.5 0.18 2,3 2.5 0.3x0.1=0.03 2,2 2 0.3x0.3=0.09 3,1 2 0.06 3,2 2.5 0.03 3,3 3 0.1x0.1=0.01
Pr( 1 0.36 1,5 0.36 2 0.21 2,5 0.06 3 0.01
Example 4.
X Pr(x) 0
0.559
1 0.382
2 0.059
Pr 0 0.559 1 0.382 2 0.059 0.5
0 0.5 2 0.559 1 0.5 2 0.382 2 0.5 2 0.059 0.368 0.61
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Example 5.
X Pr(X) 1 1/4 2 3/4 x 1/3 = 3/12= 1/4 3 3/4 x 2/3 x 1/2= 1/4 4 3/4 x2/3 x1/2 x1= 1/4
Example 6. 3P 5R
X Pr(X) 1 R 5/8 = 35/56 2 PR 3/8 x 5/7 = 15/56 3 PPR 3/8 x 2/7 x 5/6 = 5/56 4 PPPR 3/8x 2/7x 1/6 x1=1/56
b) 135
562
15
563
5
564
1
56
84
561.5
c) 2 1 1.5 2 35
562 1.5 2 15
563 1.5 2 5
564 1.5 2 1
56
2 0.54
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Example 7.
X Pr(X) 3 3-2/10 = 1/10 = 0.1 4 2/10 = 0.2 5 3/10 = 0.3 6 4/10 = 0.4
Sample Pr( 3,3 3 0.1x0.1=0.01 3,4 3.5 0.1x0.2=0.02 3,5 4 0.1x0.3=0.03 3,6 4.5 0.04 4,3 3.5 0.2x0.1=0.02 4,4 4 0.2x0.2=0.04 4,5 4.5 0.06 4,6 5 0.08 5,3 4 0.03 5,4 4.5 0.06 5,5 5 0.09 5,6 5.5 0.12 6,3 4.5 0.04 6,4 5 0.08 6,5 5.5 0.12 6,6 6 0.16
K1.
2R, 4B #
Pr BB 0 4
6
3
5
12
30
1 16
30
RR 2 2
6
1
5
2
30
Pr 116
300.53
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K2.a) 8 - #0, 2 - #1, 4 - #0, 6 - #1
Pr 0 0, 0 8
10410
32100 0.32
0.5 0, 1 810
610
48100 0.48
0.5 1, 0 210
410
8100 0.08
1 1, 1 210
610
12100 0.12
b) 0 0.32 0.5 0.48 0.5 0.08 1 0.12 0.40
0 0.4 2 0.32 0.5 0.4 2 0.48 0.5 0.4 2 0.08 1 0.4 2 0.12
0.32
K3.a)
X Roll Pr(x) 1 (even 2, 4, or 6) 3
6
-3 (roll 1 or 3) 26
4 (roll 5) 16
b) 1 36 3 2
6 4 16
3 6 4
6
1
6$0.17
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c)
Sample Pr( ) Even, even 2 1
2
1
20.25
Even, 1 or 3 -2 1
2
2
60.1675
Even, 5 5 1
2
1
6
1
120.083
1 or 3, even -2 0.1675 1 or 3, 1 or 3 -6 2
6
2
6
4
360.11
1 or 3, 5 1 2
6
1
6
2
360.056
5, even 5 1
6
3
6
3
360.083
5, 1 or 3 1 0.056 5, 5 8 1
360.028
K4. A bag contains 4 purple, 3 red and 6 green marbles. Two marbles are drawn from the bag without replacement. Let X count the number of red marbles.
#
Pr 0 0.58 1 0.381 2 0.039 Total=1
Pr 0 Pr ′ ′ 10
13
9
120.58
Pr 2 Pr3
13
2
120.039
Pr 1 1 0.58 0.039 0.381
E(x)=0(0.58)+1(0.381) + 2(0.039)=0.459
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K5.
Pr -1 0.3 1 1 0.4 1 3 0.3 9
∑ 1 0.3 1 0.4 3 0.3 1
∑ 2 Pr 2 1 0.3 1 0.4 9 0.3 12
√2.4
1.55
K6. 3G 4P
Pr 1 4
72035
2 3
7
4
6
12
42
2
7
10
35
3 3
7
2
6
4
5
4
35
4 3
7
2
6
1
51
1
35
120
35210
3534
3541
35
56
351.6
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K7. 6R 4B → 3 # 0,1,2,
Pr BB 0 4
10
3
9
12
90
1 48
90
RR 2 6
10
5
9
30
90
012
90148
90230
90
1.2
2 ∑ 2 Pr 0 1.2 2 12
901 1.2 2 48
902 1.2 2 30
900.43
√0.43 0.656
K8. 3R 4P → 2 #
Pr RR 0 3
7
2
6
1
7
RP,PR 1 4
7
PP 2 4
7
3
6
12
42
2
7
01
714
722
7
8
7
1.14
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K9.a)
X Pr(x) 2 0.133 3 0.2 4 0.667
Sample Pr 2,2 2 0.0177 2,3 2.5 0.0266 2,4 3 0.0887 3,2 2.5 0.0266 3,3 3 0.04 3,4 3.5 0.1334 4,2 3 0.0887 4,3 3.5 0.1334 4,4 4 0.4449 b) Pr 3 Pr 4 0.667
K10.
a) x = # matches she wins
Sample Pr L 0 0.3 WL 1 0.7x0.4=0.28 WWL 2 0.7x0.6x0.5=0.21 WWWL 3 0.7x0.6x0.5x0.6=0.126 WWWW 4 0.7x0.6x0.5x0.4=0.084
b) 0 0.3 1 0.28 2 0.21 3 0.126 4 0.084 1.41
c)
0 1.4 2 0.3 1 1.4 2 0.28 2 1.4 2 0.21 3 1.4 2 0.126 4 1.4 2 0.0.084
1.5986 1.26
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Section L
Example 1. 20 0.5 0.5
Pr 8 208
0.5 0.5 !
! !0.5 0.5 0.12
Example 2. n = 10 p = get a 1=
Pr "1"atleastonce 1‐Pr getno"1"
1 100
≐ 0.84
Example 3.
0.4 0.6 10
a)Find the probability of 3 heads
Pr 3 103
0.4 0.6
!
! ! 0.4 0.6 0.21499
b) 10 0.4 4
10 0.4 0.6 √2.4 1.55
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Example 4.
20 0.20 0.8
Pr 2 Pr 2
1 Pr 0 Pr 1
1 200
0.2 0.8 201
0.2 0.8
1 0.01153 0.05765 ≐ 0.93
Example 5.
20 0.20
Pr 4 10 Pr 10 Pr 3
0.999 0.411 [TABLE]
0.588
Example 6. 2R 5B
Pr 1 0.71 B
2 0.24 RB
3 1 0.048 RRB
1 0.71 2 0.24 3 0.048
1.33
1 0.71 2 0.24 3 0.048 1.33 0.33
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b).
.5.1
c)Pr 3 Pr 1 0.048
Example 7. A store has found that 20% of the items sold at Christmas
are returned. On Saturday, a total of 25 items are sold by 5 different
clerks. Clerk A sold 10 items, clerk B sold 8 items, and the rest were sold
by the other 3 clerks.
a) Pr 0 100
0.15 0.85 0.197
b) Pr 2 8 Pr 8 Pr 1
0.953 0.027 0.972 25
c) Pr 5 10 Pr 10 Pr 4
0.994 0.421 0.578 25
Example 8. a) x = 3 damaged, p = 0.05
Pr 310
30.05 3 0.95 7
Pr 310!3! 7!
0.05 3 0.95 7 0.0105
b) n = 20 p = never returned = 0.02 Pr 1 1 Pr
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1 20
00.02 0 0.98 20
1 20!0! 20!
0.02 0 0.98 20 0.332
c) n =25 p = 0.05 + 0.02 = 0.07 (damaged or never returned) x = 5
Pr 5255
0.07 0.9325!5! 20!
0.07 0.93
0.0209
Example 9. a) n = 20, p = 0.30 made in Canada
Pr 6 10 Pr 10 Pr 5
=0.983 - 0.416 = 0.567 [table]
b) n = 15, x = 2
Pr
Pr Pr 0.30 0.20 0.06 (since independent Pr(AB) = Pr(A)xPr(B) )
Pr 215
20.06 2 0.94 13 15!
2! 13!0.06 0.94 0.169
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L1.
Pr 6 256
0.5 0.5 0.00528
L2.
Pr 1 1 Pr
1 250
0.5 0.5 0.9999
L3.
Pr 8 208
0.00841
↳ "6"
L4.
Binomial n=20 3R, 4P, 2B → 9
Pr 5 205
0.1457
↙↓↘
choose 5 red prob of red not red
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L5. Pr 0.40 Pr 0.60 10
P 6 Pr 4 104
0.4 0.6
= 0.2508
L6.
0.01 0.99 10
Pr 8
1 Pr Pr 1 ) Pr 2defective
1 100
0.01 0.99 101
0.01 0.99
102
0.01 0.99
0.0001138
L7. 0.20 0.8 25
25 0.20 5
L8. 100 50
100 √25 5
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L9. a) 200 0.2 40
b) 200 0.2 0.8 √32
5.7
L10.a)n=10p=0.30q=0.70
Pr 2102
0.30 0.7010!2! 8!
0.30 0.70
0.233
b)n=10p=0.30
Pr 2 8
Pr 8 Pr 1 1 0.149 0.851
L11.a)n=15,p=0.30(rain)q=0.70
Pr 5 1 Pr 5 1 0.722 0.278(table)
b)n=10,p=0.60(construction)
Pr 7 1 Pr 6 1 0.618 0.382
c)n=10
Pr 2 3 Pr Pr Pr
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102
0.30 0.70103
0.60 0.40
102
0.30 0.70 ∙103
0.60 0.40
=0.2335+0.0425‐0.00992
M.HypothesisTesting
Rejection/Non‐rejection region
1. 012…8
910
a) Pr Pr / is true)
Pr /
Pr 9/p=0.5) 1 Pr 8
1 0.989 0.011
b) Pr
Pr / )
Pr / )
Pr 8/ p=0.8) 0.624
2. : 0.50
Step 1 Look up n=10 on binomial table and go across to p=0.5
Take the closest to 0.05but still less than 0.05
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Step 2 Exact 0.011 1 0.011 0.989 Step 3 Now go to bottom of table and any probability >0.989
is in the rejection region (we do this since : 0.5
Step 4 Rejection/Non‐rejection region
∴ 9 8
3. : 0.50 2
Step 1 Look up n = 20 and go across to p=0.5. Find the
closest to 0.05/2=0.025 ( 2 sided so divide alpha by 2) but less than 0.025
Step 2 Exact 0.021
1 – 0.021 = 0.979
Step 3 Since it is 2‐ sided, we will have one rejection region of all
values below and including 0.021 ( 5 Also, go to bottom and
all values greater than 0.979 are also in the rejection region
Step 4 Rejection/Non‐rejection region:
012345
67 … 1314 1516. . .20
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Example 1.
: 0.50 : 0.70 70%
8
a) Go to n= 10 across to 0.50
Pr Pr /
Pr /
Pr 8/p = 0.5) 1 Pr 7 1 0.945 0.055
b) Pr Pr /
Pr /
Pr 7/p = 0.70) 0.617 ∗ 0.70 7
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Example 2. 10 : 0.5
: 0.6
7
Pr Pr /
Pr /
Pr 7/p = 0.5 *go across n = 10 to
0.5 and down to X=6
1 Pr 6
1 0.828 0.172
a) Pr
Pr /
Pr /
Pr 6/p=0.60) 0.618 ∗ 10 0.6 6
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Example 3. a) Rejection region 16 0.5 n = 20 0.5 Pr Pr /
Pr /
Pr 16/p=0.5) 1 Pr 15
1 0.994 0.006
b)n = 25 0.5 0.8
0.05 0.022
Go across n=25 to 0.5 and find exact (largest number less
than 0.05)
1 1 0.022 0.978 ∴
0.978
∴ 18
Pr /
Pr 18/p=0.8)
Pr 17
1 0.109 0.891
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Practise Exam Questions on Hypothesis Testing
M1. 0.5 0.60 60%
10 0.5
a) 7 Pr Pr /
Pr /
Pr 7/p=0.5) 1 Pr 6 1 0.828 0.172
b) Pr Pr /
Pr /
Pr 7 /p=0.60) Pr 6
0.618 ∗ 10 0.6
6
M2. a) 0.5 0.8
10
Pr Pr /
Pr /
Pr 9/p=0.5)
1 Pr 8 1 0.989 0.011
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b) Pr Pr /
Pr /
Pr 9/p=0.8)
Pr 8/p=0.8
0.624 * go across n=10 0.8 8
c) 0.5 0.8 25 0.05 0.022 1 0.022 0.978 ∴ 0.978
∴ 18
Pr Pr /
Pr /
Pr 17/p=0.8)
0.109
M3. a) 0.5 0.3
20 reject if X<7 and do not reject if X 7
Pr Pr /
Pr /
Pr 7/p=0.5)
Pr 6 0.058
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b) Pr Pr /
Pr /
Pr 7/p=0.3)
1 Pr 6/p=0.3
1 0.608=0.392
0.5 0.8
25 0.05 0.022 1 0.022 0.978 ∴ 0.978
∴ 18
Pr Pr /
Pr /
Pr 17/p=0.8)
0.109
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M4. a) 0.5 0.7
20 reject if X>14 and do not reject if X 14
Pr Pr /
Pr /
Pr 14/p=0.5)
1 Pr 14 1 0.979 0.021
b) Pr Pr /
Pr /
Pr 14/p=0.7)
0.584
c)
25 0.05 0.022 1 0.022 0.978 ∴ 0.978
∴ 18
Pr Pr /
Pr /
Pr 17/p=0.7)
0.488