Assignment 5
4.46
(a)
(b)
(c)
(d)
(e)
4.49
Age
18-25
26-40
Total
Getting Rich
405
310
715
Goals
Other
95
190
285
Total
500
500
1000
Simple event: “Has a goal of getting rich”.
Joint event: “Has a goal of getting rich and is between 18-25 years old.
P(Has a goal of getting rich) = 715/1000 = 0.715
P(Has a goal of getting rich and is in the 26-40 year old group) = 310/1000 =
0.31
P(Has a goal of getting rich | in the 26-40 year old group) = 310/500 = 0.62
Since P(Has a goal of getting rich | in the 26-40 year old group)  P(Has a goal
of getting rich), the events “age group” and “has getting rich as a goal” are not
statistically independent.
P(engage) = .13, P(not engage) = .67, P(actively disengaged) = .20
P(strongly agreed | engaged) = .48, P(strongly agreed | not engaged) = .20, P(strongly
agreed | actively disengaged) = .03
P(engaged | strongly agree) = (.48*.13)/(.48*.13 + .20*.67 + .03*.20) = 0.3083
4.51
P(HIV is present | ELISA has given a positive result )
= (0.995)(0.015)/((0.995)(0.015)+(0.01)(0.985) = 0.6024
5.1
cont.
(a)
(b)
Distribution A
Distribution B
X
P(X) X*P(X) X
P(X) X*P(X)
0
0.50
0.00
0
0.05
0.00
1
0.20
0.20
1
0.10
0.10
2
0.15
0.30
2
0.15
0.30
3
0.10
0.30
3
0.20
0.60
4
0.05
0.20
4
0.50
2.00
1.00
1.00
1.00
3.00
 = 1.00
 = 3.00
Distribution A
X
0
1
2
3
4
(X–  )2
(–1)2
(0)2
(1)2
(2)2
(3)2
P(X)
0.50
0.20
0.15
0.10
0.05
 2=
(X–  )2*P(X)
0.50
0.00
0.15
0.40
0.45
1.50
  ( X – ) 2  P(X ) = 1.22
(b)
Distribution B
(X–  )2
(–3)2
(–2)2
(–1)2
(0)2
(1)2
X
0
1
2
3
4
P(X)
0.05
0.10
0.15
0.20
0.50
 2=
(X–  )2*P(X)
0.45
0.40
0.15
0.00
0.50
1.50
  ( X – ) 2  P(X ) = 1.22
5.3
(c)
Distribution A: Because the mean of 1 is greater than the median of 0, the
distribution is right-skewed.
Distribution B: Because the mean of 3 is less than the median of 4, the
distribution is left-skewed.
The means are different but the variances are the same.
(a)
(b)
Based on the fact that the odds of winning are expressed out with a base of
31,478, you will think that the automobile dealership sent out 31,478 fliers.
   iN1 X i P X i  = $ 5.49
(c)
   iN1 X i  E  X i 2 P X i  = $ 84.56
(d)
5.6
The total cost of the prizes is $15,000 + $500 + 31,476 * $5 = $172,880.
Assuming that the cost of producing the fliers is negligible, the cost of reaching a
single customer is $172,880/31478 = $5.49. The effectiveness of the promotion
will depend on how many customers will show up in the show room.
PHStat output:
Probabilities & Outcomes:
P
X
0.125
0.240385
0.307692
0.163462
0.086538
0.057692
0.009615
0.009615
Statistics
E(X)
Variance(X)
Standard Deviation(X)
(a)
(b)
E(X) = 2.1058
2.105769
2.152274
1.467063
0
1
2
3
4
5
6
7