psych 2810 november exam a - uwo math...

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

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

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

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

. 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

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

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

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

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

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

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

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.

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

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

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

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

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


chart!

∆ ∆ 25.5 ∴ 50 25.5 75.5

∴ 75.5 ∴ 100 75.524.5% 135

number of scores=0.245x150=36 people

c) 120 20 100 ← 16


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

∴

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

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

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

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

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

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

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

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

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

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

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}

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.

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

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

∴ 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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)

=

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

psych 2810 november exam a - uwo math...

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