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Probability and Statistics I
45-733, Spring 2002
Homework #2
Solution
Pp 141-4
3(a)
Proportion of Packages
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
46
47
48
49
50
51
Number of Clips
3(b)
Fx (47) = 0.04
Fx (48) = 0.17
Fx (49) = 0.38
Fx (50) = 0.67
Fx (51) = 0.87
Fx (52) = 0.97
Fx (53) = 1.00
52
53
54
3(c)
P(49  X  51)  0.21  0.29  0.20  0.70
3(d)
1 – [P(X < 50)]2 = 1 – 0.1444 = 0.8556
Pp 141-4
5(a)
Px (0) = (0.90)(0.90) = 0.81
Px (1) = (0.90)(0.10) + (0.10)(0.90) = 0.18
Px (2) = (0.10)(0.10) = 0.01
5(b)
P(Y = 0) = 18/20  17/19 = 153/190
P(Y = 1) = (2/20  18/19) + (18/20  2/19) = 36/190
P(Y = 2) = 2/20  1/19 = 1/190
The answer is part (b) is different from that of part (a) because in part (b) the probability
of picking a defective part on the second draw depends upon the result of the first draw.
Pp 141-4
11(a)
  0  1(0.12)  2(0.23)  3(0.31)  4(0.19)  5(0.18)  6(0.03)  7(0.02)  2.99
 x P ( x)  10.93
2
x
  10.93  (2.99) 2  1.9899  1.4106
11(b)
Revenue: r = 0.50X
E(r) = 0.50(2.99) = 1.495
 r | 0.50 | (1.4106)  0.7053
Pp 141-4
13
“One and One” E(x) = 1(0.75)(0.25) + 2(0.75)2 = 1.3125
“Two-shot foul” E(x) = 1((0.75)(0.25) + (0.25)(0.75)) + 2(0.75)2 = 1.50
The “Two-shot foul” has a higher expected value.
Pp 154-6
19(a)
Px(0) = 0.07 + 0.07 + 0.06 + 0.02 = 0.22
Px(1) = 0.09 + 0.06 + 0.07 + 0.04 = 0.26
Px(2) = 0.06 + 0.07 + 0.14 + 0.16 = 0.43
Px(3) = 0.01 + 0.01 + 0.03 + 0.04 = 0.09
 X  0  0.21  2(0.30)  3(0.26)  1.39
19(b)
Py(0) = 0.07 + 0.09 + 0.06 + 0.01 = 0.23
Py(1) = 0.07 + 0.06 + 0.07 + 0.01 = 0.21
Py(2) = 0.06 + 0.07 + 0.14 + 0.03 = 0.30
Py(3) = 0.02 + 0.04 + 0.16 + 0.40 = 0.26
 y  0  0.21  2(0.30)  3(0.26)  1.59
19(c)
PY|X (0|3) = 0.01/0.09 = 1/9
PY|X (1|3) = 0.01/0.09 = 1/9
PY|X (2|3) = 0.03/0.09 = 3/9
PY|X (3|3) = 0.04/0.09 = 4/9
19(d)
Cov(X,Y) = E(XY) -  X Y
E(XY) = 0 + (1)(1)(0.06) + (1)(2)(0.07) + (1)(3)(0.04) + (1)(3)(0.04) + (2)(1)(0.07) +
(2)(2)(0.14) + (2)(3)(0.16) + (3)(1)(0.01) + (3)(2)(0.03) + (3)(3)(0.04) = 2.55
Cov(X,Y) = 2.55 – (1.39)(1.59) = 0.3399
19(e)
No, because Cov(X,Y)  0
Pp 154-6
23(a)
Y\X
0
1
Total
0
0.704
0.096
0.8
1
0.168
0.032
0.2
23(b)
PY|X (0|0) = 0.704 / 0.8 = 0.88
PY|X (1|0) = 0.096 / 0.8 = 0.12
23(c)
Px(0) = 0.8
Px(1) = 0.2
Py(0) = 0.872
Py(1) = 0.128
Total
0.872
0.128
1
23(d)
E(XY) = 0.032;  X  0.20; Y  0.128
Cov(X,Y) = 0.032 – (0.20)(0.128) = 0.0064
There is a positive relationship between X & Y, as expected by the book-seller.
Professors are more likely to be away from the office on Friday than the during the other
days.
Pp 171-7
29(a)
Px(0)  (5 C 0)(0.25)0 (0.75)5  0.2373
P( X  1)  1  Px(0)  1  0.2373  0.7627
29(b)
P( X  3)
 Px (3)  Px (4)  Px(5)
 (5 C 3)(0.25)3 (0.75) 2  (5 C 4)(0.25) 4 (0.75)1  (5 C 5)(0.25) 5 (0.75)0
 10(0.25)3 (0.75) 2  5(0.25) 4 (0.75)  (0.25)5
 0.08791  0.01465  0.00098
 0.1035
Pp 171-7
33(a)
Px(5)  (5 C 5)(0.4)5 (0.6)0  (0.4)5  0.01024
33(b)
P( X  3)
 Px (3)  Px (4)  Px (5)
 (5 C 3)(0.4)3 (0.6)2  (5 C 4)(0.4)4 (0.6)1  (5 C 5)(0.4)5 (0.6)0
 0.2304  0.0768  0.0102
 0.3174
33(c)
Now n = 4 and p = 0.4
P( X  2)
 Px (2)  Px(3)  Px(4)
 6(0.4) 2 (0.6) 2  4(0.4)3 (0.6)1  (0.4) 4
 0.3456  0.1536  0.0256
 0.5248
33(d)
E(X) = np = 2
33(e)
E(X) = 1 + (4)(0.4) = 2.6
Pp 171-7
49.
P( X  3)
 1  P( X  2)
 1  e 4.2  e 4.2 (4.2) 
e 4.2 (4.2) 2
2!
 0.7898
59(a)
P(X<3) = 0.05 + 0.2 = 0.25
59(b)
E(X) = 0.05 + 2(0.02) + 3(0.35) + 4(0.3) + 5(0.1) = 3.2
59(c)
 x2 Px( x)  11.3
 x  11.3  (3.2)2  1.02956
59(d)
Let Z = 20000 + 2000X
E(Z) = $26,400
z = |2000|(1.02956) = $2,059.12
59(e)
P(X  4) = 0.40
Let Y = Projects which take at least 4 days
P(Y  2) = 3(0.4)2(0.6) + (0.4)3 = 0.352