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CHAPTER 3 Section 3-1 3-1. The range of X is { } 1000,...,2,1,0 3-2. The range of X is { } 0 12 50, , ,..., 3-3. The range of X is { } 0 12 99999, , ,..., 3-4. The range of X is { } 0 12 3 4 5, , , , , 3-5. The range of X is { . Because 490 parts are conforming, a nonconforming part must be selected in 491 selections.
}
}
491,...,2,1
3-6. The range of X is { . Although the range actually obtained from lots typically might not exceed 10%.
0 12 100, , ,...,
3-7. The range of X is conveniently modeled as all nonnegative integers. That is, the range of X is { }0 12, , ,... 3-8. The range of X is conveniently modeled as all nonnegative integers. That is, the range of X is { }0 12, , ,... 3-9. The range of X is { } 15,...,2,1,0 3-10. The possible totals for two orders are 1/8 + 1/8 = 1/4, 1/8 + 1/4 = 3/8, 1/8 + 3/8 = 1/2, 1/4 + 1/4 = 1/2, 1/4 + 3/8 = 5/8, 3/8 + 3/8 = 6/8.
Therefore the range of X is 14
38
12
58
68
, , , ,⎧⎨⎩
⎫⎬⎭
3-11. The range of X is }10000,,2,1,0{ … 3-12.The range of X is {100, 101, …, 150} 3-13.The range of X is {0,1,2,…, 40000) Section 3-2 3-14.
6/1)3(6/1)2(
3/1)5.1()5.1(3/16/16/1)0()0(
==
====+===
X
X
X
X
ff
XPfXPf
a) P(X = 1.5) = 1/3 b) P(0.5< X < 2.7) = P(X = 1.5) +P(X = 2) = 1/3 + 1/6 = 1/2 c) P(X > 3) = 0 d) (0 2) ( 0) ( 1.5) 1/ 3 1/ 3 2 / 3P X P X P X≤ < = = + = = + =e) P(X = 0 or X = 2) = 1/3 + 1/6 = 1/2
3-15. All probabilities are greater than or equal to zero and sum to one.
3-1
CHAPTER 3 Section 3-1 3-1. The range of X is { } 1000,...,2,1,0 3-2. The range of X is { } 0 12 50, , ,..., 3-3. The range of X is { } 0 12 99999, , ,..., 3-4. The range of X is { } 0 12 3 4 5, , , , , 3-5. The range of X is { . Because 490 parts are conforming, a nonconforming part must be selected in 491 selections.
}
}
491,...,2,1
3-6. The range of X is { . Although the range actually obtained from lots typically might not exceed 10%.
0 12 100, , ,...,
3-7. The range of X is conveniently modeled as all nonnegative integers. That is, the range of X is { }0 12, , ,... 3-8. The range of X is conveniently modeled as all nonnegative integers. That is, the range of X is { }0 12, , ,... 3-9. The range of X is { } 15,...,2,1,0 3-10. The possible totals for two orders are 1/8 + 1/8 = 1/4, 1/8 + 1/4 = 3/8, 1/8 + 3/8 = 1/2, 1/4 + 1/4 = 1/2, 1/4 + 3/8 = 5/8, 3/8 + 3/8 = 6/8.
Therefore the range of X is 14
38
12
58
68
, , , ,⎧⎨⎩
⎫⎬⎭
3-11. The range of X is }10000,,2,1,0{ … 3-12.The range of X is {100, 101, …, 150} 3-13.The range of X is {0,1,2,…, 40000) Section 3-2 3-14.
6/1)3(6/1)2(
3/1)5.1()5.1(3/16/16/1)0()0(
==
====+===
X
X
X
X
ff
XPfXPf
a) P(X = 1.5) = 1/3 b) P(0.5< X < 2.7) = P(X = 1.5) +P(X = 2) = 1/3 + 1/6 = 1/2 c) P(X > 3) = 0 d) (0 2) ( 0) ( 1.5) 1/ 3 1/ 3 2 / 3P X P X P X≤ < = = + = = + =e) P(X = 0 or X = 2) = 1/3 + 1/6 = 1/2
3-15. All probabilities are greater than or equal to zero and sum to one.
3-1
a) P(X ≤ 2)=1/8 + 2/8 + 2/8 + 2/8 + 1/8 = 1 b) P(X > - 2) = 2/8 + 2/8 + 2/8 + 1/8 = 7/8 c) P(-1 ≤ X ≤ 1) = 2/8 + 2/8 + 2/8 =6/8 = 3/4 d) P(X ≤ -1 or X=2) = 1/8 + 2/8 +1/8 = 4/8 =1/2 3-16. All probabilities are greater than or equal to zero and sum to one.
a) P(X≤ 1)=P(X=1)=0.5714 b) P(X>1)= 1-P(X=1)=1-0.5714=0.4286 c) P(2<X<6)=P(X=3)=0.1429 d) P(X≤1 or X>1)= P(X=1)+ P(X=2)+P(X=3)=1 3-17. Probabilities are nonnegative and sum to one. a) P(X = 4) = 9/25 b) P(X ≤ 1) = 1/25 + 3/25 = 4/25 c) P(2 ≤ X < 4) = 5/25 + 7/25 = 12/25 d) P(X > −10) = 1 3-18. Probabilities are nonnegative and sum to one. a) P(X = 2) = 3/4(1/4)2 = 3/64 b) P(X ≤ 2) = 3/4[1+1/4+(1/4)2] = 63/64 c) P(X > 2) = 1 − P(X ≤ 2) = 1/64 d) P(X ≥ 1) = 1 − P(X ≤ 0) = 1 − (3/4) = 1/4 3-19. X = number of successful surgeries. P(X=0)=0.1(0.33)=0.033 P(X=1)=0.9(0.33)+0.1(0.67)=0.364 P(X=2)=0.9(0.67)=0.603 3-20. P(X = 0) = 0.023 = 8 x 10-6
P(X = 1) = 3[0.98(0.02)(0.02)]=0.0012 P(X = 2) = 3[0.98(0.98)(0.02)]=0.0576 P(X = 3) = 0.983 = 0.9412 3-21. X = number of wafers that pass P(X=0) = (0.2)3 = 0.008 P(X=1) = 3(0.2)2(0.8) = 0.096 P(X=2) = 3(0.2)(0.8)2 = 0.384 P(X=3) = (0.8)3 = 0.512 3-22. X: the number of computers that vote for a left roll when a right roll is appropriate.
p=0.0001. P(X=0)=(1-p)4=0.99994=0.9996 P(X=1)=4*(1-p)3p=4*0.999930.0001=0.0003999 P(X=2)=C4
2(1-p)2p2=5.999*10-8
P(X=3)=C43(1-p)1p3=3.9996*10-12
P(X=4)=C40(1-p)0p4=1*10-16
3-23. P(X = 50 million) = 0.5, P(X = 25 million) = 0.3, P(X = 10 million) = 0.2 3-24. P(X = 10 million) = 0.3, P(X = 5 million) = 0.6, P(X = 1 million) = 0.1 3-25. P(X = 15 million) = 0.6, P(X = 5 million) = 0.3, P(X = -0.5 million) = 0.1 3-26. X = number of components that meet specifications P(X=0) = (0.05)(0.02) = 0.001 P(X=1) = (0.05)(0.98) + (0.95)(0.02) = 0.068 P(X=2) = (0.95)(0.98) = 0.931
3-2
3-27. X = number of components that meet specifications P(X=0) = (0.05)(0.02)(0.01) = 0.00001 P(X=1) = (0.95)(0.02)(0.01) + (0.05)(0.98)(0.01)+(0.05)(0.02)(0.99) = 0.00167 P(X=2) = (0.95)(0.98)(0.01) + (0.95)(0.02)(0.99) + (0.05)(0.98)(0.99) = 0.07663 P(X=3) = (0.95)(0.98)(0.99) = 0.92169 Section 3-3
3-28. where
⎪⎪⎪
⎭
⎪⎪⎪
⎬
⎫
⎪⎪⎪
⎩
⎪⎪⎪
⎨
⎧
≤<≤<≤
<≤<
=
xxx
xx
xF
31326/525.13/25.103/1
0,0
)(
6/1)3(6/1)2(
3/1)5.1()5.1(3/16/16/1)0()0(
==
====+===
X
X
X
X
ff
XPfXPf
3-29.
where
⎪⎪⎪⎪
⎭
⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧
≤<≤<≤<≤−−<≤−
−<
=
xxxxx
x
xF
21218/7108/5018/3128/1
2,0
)(
8/1)2(8/2)1(8/2)0(8/2)1(8/1)2(
====−=−
X
X
X
X
X
fffff
a) P(X ≤ 1.25) = 7/8 b) P(X ≤ 2.2) = 1 c) P(-1.1 < X ≤ 1) = 7/8 − 1/8 = 3/4 d) P(X > 0) = 1 − P(X ≤ 0) = 1 − 5/8 = 3/8
3-30. where
⎪⎪⎪⎪
⎭
⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧
≤<≤<≤<≤<≤
<
=
xxxxx
x
xF
414325/163225/92125/41025/1
0,0
)(
25/9)4(25/7)3(25/5)2(25/3)1(25/1)0(
=====
X
X
X
X
X
fffff
a) P(X < 1.5) = 4/25 b) P(X ≤ 3) = 16/25 c) P(X > 2) = 1 − P(X ≤ 2) = 1 − 9/25 = 16/25 d) P(1 < X ≤ 2) = P(X ≤ 2) − P(X ≤ 1) = 9/25 − 4/25 = 5/25 = 1/5 3-31.
3-3
3-27. X = number of components that meet specifications P(X=0) = (0.05)(0.02)(0.01) = 0.00001 P(X=1) = (0.95)(0.02)(0.01) + (0.05)(0.98)(0.01)+(0.05)(0.02)(0.99) = 0.00167 P(X=2) = (0.95)(0.98)(0.01) + (0.95)(0.02)(0.99) + (0.05)(0.98)(0.99) = 0.07663 P(X=3) = (0.95)(0.98)(0.99) = 0.92169 Section 3-3
3-28. where
⎪⎪⎪
⎭
⎪⎪⎪
⎬
⎫
⎪⎪⎪
⎩
⎪⎪⎪
⎨
⎧
≤<≤<≤
<≤<
=
xxx
xx
xF
31326/525.13/25.103/1
0,0
)(
6/1)3(6/1)2(
3/1)5.1()5.1(3/16/16/1)0()0(
==
====+===
X
X
X
X
ff
XPfXPf
3-29.
where
⎪⎪⎪⎪
⎭
⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧
≤<≤<≤<≤−−<≤−
−<
=
xxxxx
x
xF
21218/7108/5018/3128/1
2,0
)(
8/1)2(8/2)1(8/2)0(8/2)1(8/1)2(
====−=−
X
X
X
X
X
fffff
a) P(X ≤ 1.25) = 7/8 b) P(X ≤ 2.2) = 1 c) P(-1.1 < X ≤ 1) = 7/8 − 1/8 = 3/4 d) P(X > 0) = 1 − P(X ≤ 0) = 1 − 5/8 = 3/8
3-30. where
⎪⎪⎪⎪
⎭
⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧
≤<≤<≤<≤<≤
<
=
xxxxx
x
xF
414325/163225/92125/41025/1
0,0
)(
25/9)4(25/7)3(25/5)2(25/3)1(25/1)0(
=====
X
X
X
X
X
fffff
a) P(X < 1.5) = 4/25 b) P(X ≤ 3) = 16/25 c) P(X > 2) = 1 − P(X ≤ 2) = 1 − 9/25 = 16/25 d) P(1 < X ≤ 2) = P(X ≤ 2) − P(X ≤ 1) = 9/25 − 4/25 = 5/25 = 1/5 3-31.
3-3
⎪⎪⎪
⎭
⎪⎪⎪
⎬
⎫
⎪⎪⎪
⎩
⎪⎪⎪
⎨
⎧
≤<≤<≤<≤
<
=
xxxx
x
xF
3,132,488.021,104.010,008.0
0,0
)(
where
,512.0)8.0()3(,384.0)8.0)(8.0)(2.0(3)2(,096.0)8.0)(2.0)(2.0(3)1(
,008.02.0)0(.
3
3
======
==
fff
f
3-32.
0, 0
0.9996, 0 1( ) 0.9999, 1 3
0.99999, 3 41, 4
xx
F x xx
x
<⎧ ⎫⎪ ⎪≤ <⎪ ⎪⎪ ⎪= ≤ <⎨ ⎬⎪ ⎪≤ <⎪ ⎪
≤⎪ ⎪⎩ ⎭
where
4
3
8
12
16
.(0) 0.9999 0.9996,
(1) 4(0.9999 )(0.0001) 0.0003999,(2) 5.999*10 ,
(3) 3.9996*10 ,(4) 1*10
ff
fff
−
−
−
= == =
=
=
=
3-33.
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
≤<≤<≤
<
=
xxx
x
xF
50,15025,5.02510,2.0
10,0
)(
where P(X = 50 million) = 0.5, P(X = 25 million) = 0.3, P(X = 10 million) = 0.2
3-4
3-34.
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
≤<≤<≤
<
=
xxx
x
xF
10,1105,7.051,1.0
1,0
)(
where P(X = 10 million) = 0.3, P(X = 5 million) = 0.6, P(X = 1 million) = 0.1
3-35. The sum of the probabilities is 1 and all probabilities are greater than or equal to zero;
pmf: f(1) = 0.5, f(3) = 0.5 a) P(X ≤ 3) = 1
b) P(X ≤ 2) = 0.5 c) P(1 ≤ X ≤ 2) = P(X=1) = 0.5 d) P(X>2) = 1 − P(X≤2) = 0.5 3-36. The sum of the probabilities is 1 and all probabilities are greater than or equal to zero; pmf: f(1) = 0.7, f(4) = 0.2, f(7) = 0.1
a) P(X ≤ 4) = 0.9 b) P(X > 7) = 0 c) P(X ≤ 5) = 0.9 d) P(X>4) = 0.1 e) P(X≤2) = 0.7 3-37. The sum of the probabilities is 1 and all probabilities are greater than or equal to zero; pmf: f(-10) = 0.25, f(30) = 0.5, f(50) = 0.25
a) P(X≤50) = 1 b) P(X≤40) = 0.75 c) P(40 ≤ X ≤ 60) = P(X=50)=0.25 d) P(X<0) = 0.25 e) P(0≤X<10) = 0 f) P(−10<X<10) = 0
3-5
3-38. The sum of the probabilities is 1 and all probabilities are greater than or equal to zero; pmf: f1/8) = 0.2, f(1/4) = 0.7, f(3/8) = 0.1
a) P(X≤1/18) = 0 b) P(X≤1/4) = 0.9 c) P(X≤5/16) = 0.9 d) P(X>1/4) = 0.1 e) P(X≤1/2) = 1 Section 3-4 3-39. Mean and Variance
2)2.0(4)2.0(3)2.0(2)2.0(1)2.0(0)4(4)3(3)2(2)1(1)0(0)(
=++++=++++== fffffXEμ
22)2.0(16)2.0(9)2.0(4)2.0(1)2.0(0)4(4)3(3)2(2)1(1)0(0)(2
222222
=−++++=
−++++= μfffffXV
3- 40. Mean and Variance for random variable in exercise 3-14
333.1)6/1(3)6/1(2)3/1(5.1)3/1(0)3(3)2(2)5.1(5.1)0(0)(
=+++=+++== ffffXEμ
139.1333.1)6/1(9)6/1(4)3/1(25.2)3/1(0)3(3)2(2)1(5.1)0(0)(
2
22222
=−+++=
−+++= μffffXV
3-41. Determine E(X) and V(X) for random variable in exercise 3-15
. 0)8/1(2)8/2(1)8/2(0)8/2(1)8/1(2
)2(2)1(1)0(0)1(1)2(2)(=+++−−=+++−−−−== fffffXEμ
5.10)8/1(4)8/2(1)8/2(0)8/2(1)8/1(4)2(2)1(1)0(0)1(1)2(2)(
2
222222
=−++++=
−+++−−−−= μfffffXV
3-42. Determine E(X) and V(X) for random variable in exercise 3-16
( ) 1 (1) 2 (2) 3 (3)1(0.5714286) 2(0.2857143) 3(0.1428571)1.571429
E X f f fμ = = + += + +=
2 2 2( ) 1 (1) 2 (2) 3 (3)1.428571
V X f f f 2μ= + + + −=
3-43. Mean and variance for exercise 3-17
8.2)36.0(4)28.0(3)2.0(2)12.0(1)04.0(0)4(4)3(3)2(2)1(1)0(0)(=++++=
++++== fffffXEμ
36.18.2)36.0(16)28.0(9)2.0(4)12.0(1)04.0(0)4(4)3(3)2(2)1(1)0(0)(
2
222222
=−++++=
−++++= μfffffXV
3-6
3-38. The sum of the probabilities is 1 and all probabilities are greater than or equal to zero; pmf: f1/8) = 0.2, f(1/4) = 0.7, f(3/8) = 0.1
a) P(X≤1/18) = 0 b) P(X≤1/4) = 0.9 c) P(X≤5/16) = 0.9 d) P(X>1/4) = 0.1 e) P(X≤1/2) = 1 Section 3-4 3-39. Mean and Variance
2)2.0(4)2.0(3)2.0(2)2.0(1)2.0(0)4(4)3(3)2(2)1(1)0(0)(
=++++=++++== fffffXEμ
22)2.0(16)2.0(9)2.0(4)2.0(1)2.0(0)4(4)3(3)2(2)1(1)0(0)(2
222222
=−++++=
−++++= μfffffXV
3- 40. Mean and Variance for random variable in exercise 3-14
333.1)6/1(3)6/1(2)3/1(5.1)3/1(0)3(3)2(2)5.1(5.1)0(0)(
=+++=+++== ffffXEμ
139.1333.1)6/1(9)6/1(4)3/1(25.2)3/1(0)3(3)2(2)1(5.1)0(0)(
2
22222
=−+++=
−+++= μffffXV
3-41. Determine E(X) and V(X) for random variable in exercise 3-15
. 0)8/1(2)8/2(1)8/2(0)8/2(1)8/1(2
)2(2)1(1)0(0)1(1)2(2)(=+++−−=+++−−−−== fffffXEμ
5.10)8/1(4)8/2(1)8/2(0)8/2(1)8/1(4)2(2)1(1)0(0)1(1)2(2)(
2
222222
=−++++=
−+++−−−−= μfffffXV
3-42. Determine E(X) and V(X) for random variable in exercise 3-16
( ) 1 (1) 2 (2) 3 (3)1(0.5714286) 2(0.2857143) 3(0.1428571)1.571429
E X f f fμ = = + += + +=
2 2 2( ) 1 (1) 2 (2) 3 (3)1.428571
V X f f f 2μ= + + + −=
3-43. Mean and variance for exercise 3-17
8.2)36.0(4)28.0(3)2.0(2)12.0(1)04.0(0)4(4)3(3)2(2)1(1)0(0)(=++++=
++++== fffffXEμ
36.18.2)36.0(16)28.0(9)2.0(4)12.0(1)04.0(0)4(4)3(3)2(2)1(1)0(0)(
2
222222
=−++++=
−++++= μfffffXV
3-6
3-44. Mean and variance for exercise 3-18
31
41
43
41
43)(
10=⎟
⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛= ∑∑
∞
=
∞
= x
x
x
x
xxXE
The result uses a formula for the sum of an infinite series. The formula can be derived from the fact that the
series to sum is the derivative of a
aaahx
x
−== ∑
∞
= 1)(
1
with respect to a.
For the variance, another formula can be derived from the second derivative of h(a) with respect to a. Calculate from this formula
95
41
43
41
43)(
1
2
0
22 =⎟⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛= ∑∑
∞
=
∞
= x
x
x
x
xxXE
Then [ ]94
91
95)()()( 22 =−=−= XEXEXV
3-45. Mean and variance for random variable in exercise 3-19
( ) 0 (0) 1 (1) 2 (2)
0(0.033) 1(0.364) 2(0.603)1.57
E X f f fμ = = + += + +=
2 2 2
2
( ) 0 (0) 1 (1) 2 (2)0(0.033) 1(0.364) 4(0.603) 1.570.3111
V X f f f 2μ= + + −
= + + −=
3-46. Mean and variance for exercise 3-20
6
( ) 0 (0) 1 (1) 2 (2) 3 (3)0(8 10 ) 1(0.0012) 2(0.0576) 3(0.9412)2.940008
E X f f f fμ−
= = + + +
= × + + +=
2 2 2 2( ) 0 (0) 1 (1) 2 (2) 3 (3)0.05876096
V X f f f f 2μ= + + + −=
3-47. Determine x where range is [0,1,2,3,x] and mean is 6.
242.08.4
2.02.16)2.0()2.0(3)2.0(2)2.0(1)2.0(06
)()3(3)2(2)1(1)0(06)(
==
+=++++=
++++===
xx
xx
xxfffffXEμ
3-7
3-48. (a) F(0)=0.17
Nickel Charge: X CDF 0 0.17 2 0.17+0.35=0.52 3 0.17+0.35+0.33=0.85 4 0.17+0.35+0.33+0.15=1
(b)E(X) = 0*0.17+2*0.35+3*0.33+4*0.15=2.29
V(X) =4
2
1( )( )i i
if x x μ
=
−∑ = 1.5259
3-49. X = number of computers that vote for a left roll when a right roll is appropriate.
µ = E(X)=0*f(0)+1*f(1)+2*f(2)+3*f(3)+4*f(4) = 0+0.0003999+2*5.999*10-8+3*3.9996*10-12+4*1*10-16= 0.0004
V(X)=5
2
1
( )( )i ii
f x x μ=
−∑ = 0.00039996
3-50. µ=E(X)=350*0.06+450*0.1+550*0.47+650*0.37=565
V(X)= =6875 ∑=
−4
1
2))((i
i xxf μ
σ= )(XV =82.92 3-51. (a) Transaction Frequency Selects: X f(X)
New order 43 23 0.43 Payment 44 4.2 0.44 Order status 4 11.4 0.04 Delivery 5 130 0.05 Stock level 4 0 0.04 total 100
µ =
E(X) = 23*0.43+4.2*0.44+11.4*0.04+130*0.05+0*0.04 =18.694
V(X) = = 735.964 ∑=
−5
1
2))((i
i xxf μ 1287.27)( == XVσ
(b)
Transaction Frequency All operation: X f(X) New order 43 23+11+12=46 0.43 Payment 44 4.2+3+1+0.6=8.8 0.44 Order status 4 11.4+0.6=12 0.04 Delivery 5 130+120+10=260 0.05 Stock level 4 0+1=1 0.04 total 100
µ = E(X) = 46*0.43+8.8*0.44+12*0.04+260*0.05+1*0.04=37.172
3-8
V(X) = =2947.996 ∑=
−5
1
2))((i
i xxf μ 2955.54)( == XVσ
Section 3-5 3-52. E(X) = (0+100)/2 = 50, V(X) = [(100-0+1)2-1]/12 = 850 3-53. E(X) = (3+1)/2 = 2, V(X) = [(3-1+1)2 -1]/12 = 0.667 3-54. X=(1/100)Y, Y = 15, 16, 17, 18, 19.
E(X) = (1/100) E(Y) = 17.02
1915100
1=⎟
⎠⎞
⎜⎝⎛ +
mm
0002.012
1)11519(100
1)(22
=⎥⎦
⎤⎢⎣
⎡ −+−⎟⎠⎞
⎜⎝⎛=XV mm2
3-55. 3314
313
312)( =⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛=XE
in 100 codes the expected number of letters is 300
( ) ( ) ( ) ( )323
314
313
312)( 2222 =−⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛=XV
in 100 codes the variance is 6666.67 3-56. X = 590 + 0.1Y, Y = 0, 1, 2, ..., 9
E(X) = 45.5902
901.0590 =⎟⎠⎞
⎜⎝⎛ +
+ mm
0825.012
1)109()1.0()(2
2 =⎥⎦
⎤⎢⎣
⎡ −+−=XV mm2
3-57. a = 675, b = 700 (a) µ = E(X) = (a+b)/2= 687.5 V(X) = [(b – a +1)2 – 1]/12= 56.25 (b) a = 75, b = 100 µ = E(X) = (a+b)/2 = 87.5 V(X) = [(b – a + 1)2 – 1]/12= 56.25 The range of values is the same, so the mean shifts by the difference in the two minimums (or maximums) whereas the variance does not change.
3-58. X is a discrete random variable. X is discrete because it is the number of fields out of 28 that has an error. However, X is not uniform because P(X=0) ≠ P(X=1).
3-59. The range of Y is 0, 5, 10, ..., 45, E(X) = (0+9)/2 = 4.5 E(Y) = 0(1/10)+5(1/10)+...+45(1/10) = 5[0(0.1) +1(0.1)+ ... +9(0.1)] = 5E(X) = 5(4.5) = 22.5 V(X) = 8.25, V(Y) = 52(8.25) = 206.25, σY = 14.36
3-9
V(X) = =2947.996 ∑=
−5
1
2))((i
i xxf μ 2955.54)( == XVσ
Section 3-5 3-52. E(X) = (0+100)/2 = 50, V(X) = [(100-0+1)2-1]/12 = 850 3-53. E(X) = (3+1)/2 = 2, V(X) = [(3-1+1)2 -1]/12 = 0.667 3-54. X=(1/100)Y, Y = 15, 16, 17, 18, 19.
E(X) = (1/100) E(Y) = 17.02
1915100
1=⎟
⎠⎞
⎜⎝⎛ +
mm
0002.012
1)11519(100
1)(22
=⎥⎦
⎤⎢⎣
⎡ −+−⎟⎠⎞
⎜⎝⎛=XV mm2
3-55. 3314
313
312)( =⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛=XE
in 100 codes the expected number of letters is 300
( ) ( ) ( ) ( )323
314
313
312)( 2222 =−⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛=XV
in 100 codes the variance is 6666.67 3-56. X = 590 + 0.1Y, Y = 0, 1, 2, ..., 9
E(X) = 45.5902
901.0590 =⎟⎠⎞
⎜⎝⎛ +
+ mm
0825.012
1)109()1.0()(2
2 =⎥⎦
⎤⎢⎣
⎡ −+−=XV mm2
3-57. a = 675, b = 700 (a) µ = E(X) = (a+b)/2= 687.5 V(X) = [(b – a +1)2 – 1]/12= 56.25 (b) a = 75, b = 100 µ = E(X) = (a+b)/2 = 87.5 V(X) = [(b – a + 1)2 – 1]/12= 56.25 The range of values is the same, so the mean shifts by the difference in the two minimums (or maximums) whereas the variance does not change.
3-58. X is a discrete random variable. X is discrete because it is the number of fields out of 28 that has an error. However, X is not uniform because P(X=0) ≠ P(X=1).
3-59. The range of Y is 0, 5, 10, ..., 45, E(X) = (0+9)/2 = 4.5 E(Y) = 0(1/10)+5(1/10)+...+45(1/10) = 5[0(0.1) +1(0.1)+ ... +9(0.1)] = 5E(X) = 5(4.5) = 22.5 V(X) = 8.25, V(Y) = 52(8.25) = 206.25, σY = 14.36
3-9
3-60. , ∑∑ ===xx
XcExxfcxcxfcXE )()()()(
∑∑ =−=−=xx
XcVxfxcxfccxcXV )()()()()()( 222 μμ
Section 3-6 3-61. A binomial distribution is based on independent trials with two outcomes and a constant probability of success on each trial. a) reasonable b) independence assumption not reasonable c) The probability that the second component fails depends on the failure time of the first component. The binomial distribution is not reasonable. d) not independent trials with constant probability e) probability of a correct answer not constant. f) reasonable g) probability of finding a defect not constant. h) if the fills are independent with a constant probability of an underfill, then the binomial distribution for the number packages underfilled is reasonable. i) because of the bursts, each trial (that consists of sending a bit) is not independent j) not independent trials with constant probability 3-62. (a) P(X≤3) = 0.411
(b) P(X>10) = 1 – 0.9994 = 0.0006 (c) P(X=6) = 0.1091 (d) P(6 ≤X ≤11) = 0.9999 – 0.8042 = 0.1957
3-63. (a) P(X≤2) = 0.9298
(b) P(X>8) = 0 (c) P(X=4) = 0.0112 (d) P(5≤X≤7) = 1 - 0.9984 = 0.0016
3-64. a) 2461.0)5.0(5.05
10)5( 55 =⎟⎟
⎠
⎞⎜⎜⎝
⎛==XP
b) 8291100 5.05.02
105.05.0
110
5.05.00
10)2( ⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=≤XP
0547.0)5.0(45)5.0(105.0 101010 =++=
c) 0107.0)5.0(5.01010
)5.0(5.09
10)9( 01019 =⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=≥XP
d) 6473 5.05.04
105.05.0
310
)53( ⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=<≤ XP
3223.0)5.0(210)5.0(120 1010 =+=
3-65. a) ( ) 855 1040.299.001.05
10)5( −×=⎟⎟
⎠
⎞⎜⎜⎝
⎛==XP
3-10
3-60. , ∑∑ ===xx
XcExxfcxcxfcXE )()()()(
∑∑ =−=−=xx
XcVxfxcxfccxcXV )()()()()()( 222 μμ
Section 3-6 3-61. A binomial distribution is based on independent trials with two outcomes and a constant probability of success on each trial. a) reasonable b) independence assumption not reasonable c) The probability that the second component fails depends on the failure time of the first component. The binomial distribution is not reasonable. d) not independent trials with constant probability e) probability of a correct answer not constant. f) reasonable g) probability of finding a defect not constant. h) if the fills are independent with a constant probability of an underfill, then the binomial distribution for the number packages underfilled is reasonable. i) because of the bursts, each trial (that consists of sending a bit) is not independent j) not independent trials with constant probability 3-62. (a) P(X≤3) = 0.411
(b) P(X>10) = 1 – 0.9994 = 0.0006 (c) P(X=6) = 0.1091 (d) P(6 ≤X ≤11) = 0.9999 – 0.8042 = 0.1957
3-63. (a) P(X≤2) = 0.9298
(b) P(X>8) = 0 (c) P(X=4) = 0.0112 (d) P(5≤X≤7) = 1 - 0.9984 = 0.0016
3-64. a) 2461.0)5.0(5.05
10)5( 55 =⎟⎟
⎠
⎞⎜⎜⎝
⎛==XP
b) 8291100 5.05.02
105.05.0
110
5.05.00
10)2( ⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=≤XP
0547.0)5.0(45)5.0(105.0 101010 =++=
c) 0107.0)5.0(5.01010
)5.0(5.09
10)9( 01019 =⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=≥XP
d) 6473 5.05.04
105.05.0
310
)53( ⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=<≤ XP
3223.0)5.0(210)5.0(120 1010 =+=
3-65. a) ( ) 855 1040.299.001.05
10)5( −×=⎟⎟
⎠
⎞⎜⎜⎝
⎛==XP
3-10
( ) ( ) ( )
( ) ( )
( ) 46473
1801019
8291100
10138.1)99.0(01.04
1099.001.0
310
)53()
1091.999.001.01010
99.001.09
10)9()
9999.0
99.001.02
1099.001.0
110
99.001.00
10)2()
−
−
×=⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=<≤
×=⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=≥
=
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=≤
XPd
XPc
XPb
3-66.
1050
0.25
0.20
0.15
0.10
0.05
0.00
x
f(x)
a) , X= 5 is most likely, also ( 5) 0.9999P X = = 5)5.0(10)( === npXE
b) Values X=0 and X=10 are the least likely, the extreme values
3-67.
3-11
109876543210
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
x
pro
b of
x
Binomal (10, 0.01)
P(X = 0) = 0.904, P(X = 1) = 0.091, P(X = 2) = 0.004, P(X = 3) = 0. P(X = 4) = 0 and so forth. Distribution is skewed with 1.0)01.0(10)( === npXE a) The most-likely value of X is 0.
b) The least-likely value of X is 10. 3-68. n=3 and p=0.5
⎪⎪⎪
⎭
⎪⎪⎪
⎬
⎫
⎪⎪⎪
⎩
⎪⎪⎪
⎨
⎧
≤<≤<≤<≤
<
=
xxxx
x
xF
3132875.0215.010125.0
00
)( where
81
41)3(
83
43
413)2(
83
21
213)1(
81
21)0(
3
2
2
3
=⎟⎠⎞
⎜⎝⎛=
=⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛=
=⎟⎠⎞
⎜⎝⎛⎟⎠⎞
⎜⎝⎛=
=⎟⎠⎞
⎜⎝⎛=
f
f
f
f
3-12
3-69. n=3 and p=0.25
where
⎪⎪⎪
⎭
⎪⎪⎪
⎬
⎫
⎪⎪⎪
⎩
⎪⎪⎪
⎨
⎧
≤<≤<≤<≤
<
=
xxxx
x
xF
31329844.0218438.0104219.0
00
)(
641
41)3(
649
43
413)2(
6427
43
413)1(
6427
43)0(
3
2
2
3
=⎟⎠⎞
⎜⎝⎛=
=⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛=
=⎟⎠⎞
⎜⎝⎛⎟⎠⎞
⎜⎝⎛=
=⎟⎠⎞
⎜⎝⎛=
f
f
f
f
3-70. Let X denote the number of defective circuits. Then, X has a binomial distribution with n = 40 and
p = 0.01. Then, P(X = 0) = ( ) 6690.099.001.0 400400 = .
3-71. Let X denote the number of times the line is occupied. Then, X has a binomial distribution with n = 10 and p = 0.4
a.) P X ( ) . ( . ) .= =⎛⎝⎜
⎞⎠⎟ =3
103
0 4 0 6 0 2153 7
b.) ( ) 994.06.04.01)0(1)1( 100100 =−==−=≥ XPXP
c.) 4)4.0(10)( ==XE 3-72. Let X denote the number of questions answered correctly. Then, X is binomial with n = 25 and p = 0.25.
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) 2137.075.025.0425
75.025.0325
75.025.0225
75.025.0125
75.025.0025
)5()
10677.975.025.02525
75.025.02425
75.025.02325
75.025.02225
75.025.02125
75.025.02025
)20()
214223
232241250
10025124223
322421520
=⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=<
×=⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=≥
−
XPb
XPa
3-73. Let X denote the number of mornings the light is green.
( )( )
370.0630.01)4(1)4()
218.08.02.0)4()
410.08.02.0)1()
164204
4151
=−=≤−=>
===
===
XPXPc
XPb
XPa
3-13
3-74. p=0.01, n=15 X: the number of sample mutated
(a) P(X=0) = = 0.86 150 )1(015
pp −⎟⎟⎠
⎞⎜⎜⎝
⎛
(b) P(X≤1)=P(X=0)+P(X=1)= 0.99 (c) P(X>7)=P(X=8)+P(X=9)+…+P(X=15)= 0
3-75. (a) n=20, p=0.6122,
P(X≥1) = 1-P(X=0) = 1
(b)P(X≥3) = 1- P(X<3)= 0.999997 (c) µ=E(X)= np=20*0.6122=12.244 V(X)=np(1-p) = 4.748
σ= )(XV =2.179
3-76. n=20,p=0.13
(a) P(X=3) = =0.235 173 )1(320
pp −⎟⎟⎠
⎞⎜⎜⎝
⎛
(b) P(X≥3) = 1-P(X<3)=0.492 (c) µ = E(X) = np = 20*0.13 = 2.6 V(X) = np(1-p) = 2.262
σ = )(XV = 1.504 3-77. (a) Binomial distribution, p =104/369 = 4.59394E-06, n = 1E09
(b) P(X=0) = = 0 0910 )1(0
091 EppE
−⎟⎟⎠
⎞⎜⎜⎝
⎛
(c) µ = E(X) = np =1E09*0.45939E-06 = 4593.9 V(X) = np(1-p) = 4593.9
3-78. E(X) = 20 (0.01) = 0.2 V(X) = 20 (0.01) (0.99) = 0.198 μ σX X+ = + =3 0 2 3 0198 153. . . a) ) X is binomial with n = 20 and p = 0.01
( ) ( )[ ] 0169.099.001.099.001.01)1(1)2()53.1(
191201
200200 =+−=
≤−=≥=> XPXPXP
b) X is binomial with n = 20 and p = 0.04
( ) ( )[ ] 1897.096.004.096.004.01
)1(1)1(19120
120020
0 =+−=
≤−=> XPXP
c) Let Y denote the number of times X exceeds 1 in the next five samples. Then, Y is binomial with n = 5 and p = 0.190 from part b.
( )[ ] 651.0810.0190.01)0(1)1( 5050 =−==−=≥ YPYP
3-14
The probability is 0.651 that at least one sample from the next five will contain more than one defective 3-79. Let X denote the passengers with tickets that do not show up for the flight. Then, X is binomial with n = 125 and p = 0.1.
( ) ( ) ( )
( ) ( )
9886.0)5(1)5()9961.0
9.01.04
1259.01.0
3125
9.01.02
1259.01.0
1125
9.01.00
125
1
)4(1)5()
12141223
123212411250
=≤−=>=
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛
−=
≤−=≥
XPXPb
XPXPa
3-80. Let X denote the number of defective components among those stocked.
( )( ) ( ) ( )
981.0)5()
666.098.002.098.002.098.002.0)2()
133.098.002.0)0()
10021022
10111021
10201020
10001000
=≤
=++=≤
===
XPc
XPb
XPa
Section 3-7
3-81 . a) 5.05.0)5.01()1( 0 =−==XPb) 0625.05.05.0)5.01()4( 43 ==−==XPc) 0039.05.05.0)5.01()8( 87 ==−==XPd) 5.0)5.01(5.0)5.01()2()1()2( 10 −+−==+==≤ XPXPXP 75.05.05.0 2 =+=e.) 25.075.01)2(1)2( =−=≤−=> XPXP
3-82. E(X) = 2.5 = 1/p giving p = 0.4
a) 4.04.0)4.01()1( 0 =−==XPb) 0864.04.0)4.01()4( 3 =−==XPc) 05184.05.0)5.01()5( 4 =−==XPd) )3()2()1()3( =+=+==≤ XPXPXPXP
7840.04.0)4.01(4.0)4.01(4.0)4.01( 210 =−+−+−=e) 2160.07840.01)3(1)3( =−=≤−=> XPXP
3-83. Let X denote the number of trials to obtain the first success. a) E(X) = 1/0.2 = 5 b) Because of the lack of memory property, the expected value is still 5. 3-84. a) E(X) = 4/0.2 = 20
b) P(X=20) = 0436.02.0)80.0(3
19 416 =⎟⎟⎠
⎞⎜⎜⎝
⎛
3-15
The probability is 0.651 that at least one sample from the next five will contain more than one defective 3-79. Let X denote the passengers with tickets that do not show up for the flight. Then, X is binomial with n = 125 and p = 0.1.
( ) ( ) ( )
( ) ( )
9886.0)5(1)5()9961.0
9.01.04
1259.01.0
3125
9.01.02
1259.01.0
1125
9.01.00
125
1
)4(1)5()
12141223
123212411250
=≤−=>=
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛
−=
≤−=≥
XPXPb
XPXPa
3-80. Let X denote the number of defective components among those stocked.
( )( ) ( ) ( )
981.0)5()
666.098.002.098.002.098.002.0)2()
133.098.002.0)0()
10021022
10111021
10201020
10001000
=≤
=++=≤
===
XPc
XPb
XPa
Section 3-7
3-81 . a) 5.05.0)5.01()1( 0 =−==XPb) 0625.05.05.0)5.01()4( 43 ==−==XPc) 0039.05.05.0)5.01()8( 87 ==−==XPd) 5.0)5.01(5.0)5.01()2()1()2( 10 −+−==+==≤ XPXPXP 75.05.05.0 2 =+=e.) 25.075.01)2(1)2( =−=≤−=> XPXP
3-82. E(X) = 2.5 = 1/p giving p = 0.4
a) 4.04.0)4.01()1( 0 =−==XPb) 0864.04.0)4.01()4( 3 =−==XPc) 05184.05.0)5.01()5( 4 =−==XPd) )3()2()1()3( =+=+==≤ XPXPXPXP
7840.04.0)4.01(4.0)4.01(4.0)4.01( 210 =−+−+−=e) 2160.07840.01)3(1)3( =−=≤−=> XPXP
3-83. Let X denote the number of trials to obtain the first success. a) E(X) = 1/0.2 = 5 b) Because of the lack of memory property, the expected value is still 5. 3-84. a) E(X) = 4/0.2 = 20
b) P(X=20) = 0436.02.0)80.0(3
19 416 =⎟⎟⎠
⎞⎜⎜⎝
⎛
3-15
c) P(X=19) = 0459.02.0)80.0(3
18 415 =⎟⎟⎠
⎞⎜⎜⎝
⎛
d) P(X=21) = 0411.02.0)80.0(3
20 417 =⎟⎟⎠
⎞⎜⎜⎝
⎛
e) The most likely value for X should be near μX. By trying several cases, the most likely value is x = 19. 3-85. Let X denote the number of trials to obtain the first successful alignment. Then X is a geometric random variable with p = 0.8
a) 0064.08.02.08.0)8.01()4( 33 ==−==XP b) )4()3()2()1()4( =+=+=+==≤ XPXPXPXPXP
8.0)8.01(8.0)8.01(8.0)8.01(8.0)8.01( 3210 −+−+−+−= 9984.08.02.0)8.0(2.0)8.0(2.08.0 32 =+++=
c) )]3()2()1([1)3(1)4( =+=+=−=≤−=≥ XPXPXPXPXP
]8.0)8.01(8.0)8.01(8.0)8.01[(1 210 −+−+−−= 008.0992.01)]8.0(2.0)8.0(2.08.0[1 2 =−=++−=
3-86. Let X denote the number of people who carry the gene. Then X is a negative binomial random variable with r=2 and p = 0.1
a) )]3()2([1)4(1)4( =+=−=<−=≥ XPXPXPXP
972.0)018.001.0(11.0)1.01(12
1.0)1.01(11
1 2120 =+−=⎥⎦
⎤⎢⎣
⎡−⎟⎟
⎠
⎞⎜⎜⎝
⎛+−⎟⎟
⎠
⎞⎜⎜⎝
⎛−=
b) 201.0/2/)( === prXE
3-87. Let X denote the number of calls needed to obtain a connection. Then, X is a geometric random variable with p = 0.02.
a) 0167.002.098.002.0)02.01()10( 99 ==−==XP b) )]5()4()3()2()1([1)4(1)5( =+=+=+=+=−=≤−=> XPXPXPXPXPXPXP
)]02.0(98.0)02.0(98.0)02.0(98.0)02.0(98.0)02.0(98.002.0[1 4332 +++++−=
9039.00961.01 =−= May also use the fact that P(X > 5) is the probability of no connections in 5 trials. That is,
9039.098.002.005
)5( 50 =⎟⎟⎠
⎞⎜⎜⎝
⎛=>XP
c) E(X) = 1/0.02 = 50 3-88. X: # of opponents until the player is defeated.
p=0.8, the probability of the opponent defeating the player. (a) f(x) = (1 – p)x – 1p = 0.8(x – 1)*0.2 (b) P(X>2) = 1 – P(X=1) – P(X=2) = 0.64 (c) µ = E(X) = 1/p = 5 (d) P(X≥4) = 1-P(X=1)-P(X=2)-P(X=3) = 0.512 (e) The probability that a player contests four or more opponents is obtained in part (d), which is po = 0.512. Let Y represent the number of game plays until a player contests four or more opponents. Then, f(y) = (1-po)y-1po.
3-16
µY = E(Y) = 1/po = 1.95 3-89. p=0.13
(a) P(X=1) = (1-0.13)1-1*0.13=0.13. (b) P(X=3)=(1-0.13)3-1*0.13 =0.098 (c) µ=E(X)= 1/p=7.69≈8
3-90. X: # number of attempts before the hacker selects a user password.
(a) p=9900/366=0.0000045 µ=E(X) = 1/p= 219877 V(X)= (1-p)/p2 = 4.938*1010
σ= )(XV =222222 (b) p=100/363=0.00214 µ=E(X) = 1/p= 467 V(X)= (1-p)/p2 = 217892.39
σ= )(XV =466.78 Based on the answers to (a) and (b) above, it is clearly more secure to use a 6 character password. 3-91 . p = 0.005 , r = 8
a.) 198 1091.3005.0)8( −=== xXP
b). 200005.01)( === XEμ days
c) Mean number of days until all 8 computers fail. Now we use p=3.91x10-19
1891 1056.2
1091.31)( xx
YE === −μ days or 7.01 x1015 years
3-92. Let Y denote the number of samples needed to exceed 1 in Exercise 3-66. Then Y has a geometric distribution with p = 0.0169. a) P(Y = 10) = (1 − 0.0169)9(0.0169) = 0.0145 b) Y is a geometric random variable with p = 0.1897 from Exercise 3-66. P(Y = 10) = (1 − 0.1897)9(0.1897) = 0.0286 c) E(Y) = 1/0.1897 = 5.27 3-93. Let X denote the number of transactions until all computers have failed. Then, X is negative binomial random variable with p = 10-8 and r = 3. a) E(X) = 3 x 108
b) V(X) = [3(1−10-80]/(10-16) = 3.0 x 1016
3-94. (a) p6=0.6, p=0.918
(b) 0.6*p2=0.4, p=0.816
3-95 . Negative binomial random variable: f(x; p, r) = . rrx pprx −−⎟⎟
⎠
⎞⎜⎜⎝
⎛−−
)1(11
When r = 1, this reduces to f(x; p, r) = (1−p)x-1p, which is the pdf of a geometric random variable. Also, E(X) = r/p and V(X) = [r(1−p)]/p2 reduce to E(X) = 1/p and V(X) = (1−p)/p2, respectively.
3-17
3-96. Let X denote a geometric random variable with parameter p. Let q = 1-p.
1 12
1 1
1 1( ) (1 )x x
x x
E X x p p p xq pp p
∞ ∞− −
= =
⎛ ⎞= − = = =⎜ ⎟
⎝ ⎠∑ ∑
( )
2 2
2
2
2
2 2
2 1 21 1
1 1
2 1 1 11
1 1 1
2 1 2 1
1
2 1 1
1
2 3 1
2 1
1(1 )
( ) ( ) (1 ) 2 (1 )
2
2 3 ...
(1 2 3 ...)
2
1x xp p
x x
x x xp
x x x
xp p
x
xp
x
ddq p
ddq p
qddq q p
V X x p p px x p
p x q xq q
p x q
p x q
p q q q
p q q q
p
∞ ∞− −
= =
∞ ∞ ∞− − −
= = =
∞−
=
∞−
=
−
= − − = − + −
= − +
= − +
= −
⎡ ⎤= + + + −⎣ ⎦
⎡ ⎤= + + + −⎣ ⎦
⎡ ⎤= − =⎣ ⎦
∑ ∑
∑ ∑ ∑
∑
∑
[ ]2
3 2 1
2 2 2
(1 ) (1 )
2(1 ) 1 (1 )
ppq q p q
p p p qp p p
− −− + − −
− + − −= = =
Section 3-8 3-97. X has a hypergeometric distribution N=100, n=4, K=20
a.)( )( )( ) 4191.0
3921225)82160(20)1( 100
4
803
201 ====XP
b.) , the sample size is only 4 0)6( ==XP
c.) ( )( )( ) 001236.0
3921225)1(4845)4( 100
4
800
204 ====XP
d.) 8.0100204)( =⎟
⎠⎞
⎜⎝⎛===
NKnnpXE
6206.09996)8.0)(2.0(4
1)1()( =⎟
⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛
−−
−=N
nNpnpXV
3-18
3-96. Let X denote a geometric random variable with parameter p. Let q = 1-p.
1 12
1 1
1 1( ) (1 )x x
x x
E X x p p p xq pp p
∞ ∞− −
= =
⎛ ⎞= − = = =⎜ ⎟
⎝ ⎠∑ ∑
( )
2 2
2
2
2
2 2
2 1 21 1
1 1
2 1 1 11
1 1 1
2 1 2 1
1
2 1 1
1
2 3 1
2 1
1(1 )
( ) ( ) (1 ) 2 (1 )
2
2 3 ...
(1 2 3 ...)
2
1x xp p
x x
x x xp
x x x
xp p
x
xp
x
ddq p
ddq p
qddq q p
V X x p p px x p
p x q xq q
p x q
p x q
p q q q
p q q q
p
∞ ∞− −
= =
∞ ∞ ∞− − −
= = =
∞−
=
∞−
=
−
= − − = − + −
= − +
= − +
= −
⎡ ⎤= + + + −⎣ ⎦
⎡ ⎤= + + + −⎣ ⎦
⎡ ⎤= − =⎣ ⎦
∑ ∑
∑ ∑ ∑
∑
∑
[ ]2
3 2 1
2 2 2
(1 ) (1 )
2(1 ) 1 (1 )
ppq q p q
p p p qp p p
− −− + − −
− + − −= = =
Section 3-8 3-97. X has a hypergeometric distribution N=100, n=4, K=20
a.)( )( )( ) 4191.0
3921225)82160(20)1( 100
4
803
201 ====XP
b.) , the sample size is only 4 0)6( ==XP
c.) ( )( )( ) 001236.0
3921225)1(4845)4( 100
4
800
204 ====XP
d.) 8.0100204)( =⎟
⎠⎞
⎜⎝⎛===
NKnnpXE
6206.09996)8.0)(2.0(4
1)1()( =⎟
⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛
−−
−=N
nNpnpXV
3-18
3-98. a) ( )( )( ) 4623.0
24/)17181920(6/)1415164()1(
204
163
41 =
××××××
===XP
b) ( )( )( ) 00021.0
24/)17181920(1)4(
204
160
44 =
×××===XP
c)
( )( )( )
( )( )( )
( )( )( )
9866.0
)2()1()0()2(
2417181920
215166
61415164
2413141516
204
162
42
204
163
41
204
164
40
==
++=
=+=+==≤
⎟⎠⎞
⎜⎝⎛ ×××
⎟⎠⎞
⎜⎝⎛ ××
+×××
+×××
XPXPXPXP
d) E(X) = 4(4/20) = 0.8 V(X) = 4(0.2)(0.8)(16/19) = 0.539 3-99. N=10, n=3 and K=4
0 1 2 3
0.0
0.1
0.2
0.3
0.4
0.5
x
P(x
)
3-100. (a) f(x) = / ⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛xx 3
1224⎟⎟⎠
⎞⎜⎜⎝
⎛336
(b) µ=E(X) = np= 3*24/36=2 V(X)= np(1-p)(N-n)/(N-1) =2*(1-24/36)(36-3)/(36-1)=0.629 (c) P(X≤2) =1-P(X=3) =0.717
3-19
3-101. Let X denote the number of men who carry the marker on the male chromosome for an increased risk for high
blood pressure. N=800, K=240 n=10 a) n=10
( )( )( )
( )( )1201.0)1(
!790!10!800
!551!9!560
!239!1!240
80010
5609
2401 ====XP
b) n=10 )]1()0([1)1(1)1( =+=−=≤−=> XPXPXPXP
( )( )( )
( )( )0276.0)0(
!790!10!800
!550!10!560
!240!0!240
80010
56010
2400 ====XP
8523.0]1201.00276.0[1)1(1)1( =+−=≤−=> XPXP
3-102. Let X denote the number of cards in the sample that are defective.
a)
( )( )( )
9644.00356.01)1(
0356.0)0(
)0(1)1(
!120!20!140
!100!20!120
14020
12020
200
=−=≥
====
=−=≥
XP
XP
XPXP
b)
( )( )( )
5429.04571.01)1(
4571.0!140!115!120!135)0(
)0(1)1(
!120!20!140
!115!20!135
14020
13520
50
=−=≥
=====
=−=≥
XP
XP
XPXP
3-103. N=300
(a) K=243, n=3, P(X=1)=0.087 (b) P(X≥1)=0.9934 (c) K=26+13=39 P(X=1)=0.297 (d) K=300-18=282 P(X≥1)=0.9998
3-104. Let X denote the count of the numbers in the state's sample that match those in the player's sample. Then, X has a hypergeometric distribution with N = 40, n = 6, and K = 6.
a) ( )( )( )
71
406
340
66 1061.2
!34!6!40)6( −
−
×=⎟⎠⎞
⎜⎝⎛===XP
b) ( )( )( ) ( )
5406
406
341
65 1031.5346)5( −×=
×===XP
c) ( )( )( ) 00219.0)4( 40
6
342
64 ===XP
d) Let Y denote the number of weeks needed to match all six numbers. Then, Y has a geometric distribution with p =
3-20
380,838,3
1 and E(Y) = 1/p = weeks. This is more than 738 centuries! 380,838,3
3-105. Let X denote the number of blades in the sample that are dull.
a)
( )( )( )
7069.0)0(1)1(
2931.0!33!48!43!38)0(
)0(1)1(
!43!5!48!33!5
!38
485
385
100
==−=≥
=====
=−=≥
XPXP
XP
XPXP
b) Let Y denote the number of days needed to replace the assembly.
P(Y = 3) = 0607.0)7069.0(2931.0 2 =
c) On the first day, ( )( )( ) 8005.0
!41!48!43!46)0(
!43!5!48!41!5
!46
485
465
20 =====XP
On the second day, ( )( )( ) 4968.0
!37!48!43!42)0(
!43!5!48!37!5
!42
485
425
60 =====XP
On the third day, P(X = 0) = 0.2931 from part a. Therefore, P(Y = 3) = 0.8005(0.4968)(1-0.2931) = 0.2811.
3-106. a) For Exercise 3-97, the finite population correction is 96/99. For Exercise 3-98, the finite population correction is 16/19. Because the finite population correction for Exercise 3-97 is closer to one, the binomial approximation to the distribution of X should be better in Exercise 3-97.
b) Assuming X has a binomial distribution with n = 4 and p = 0.2,
( )( ) 0016.08.02.0)4(
4096.08.02.0)1(044
4
3141
===
===
XP
XP
The results from the binomial approximation are close to the probabilities obtained in Exercise 3-97.
c) Assume X has a binomial distribution with n = 4 and p = 0.2. Consequently, P(X = 1) and P(X = 4) are the same as computed in part b. of this exercise. This binomial approximation is not as close to the true answer as the results obtained in part b. of this exercise.
d) From Exercise 3-102, X is approximately binomial with n = 20 and p = 20/140 = 1/7.
( )( ) ( ) 9542.00458.01)0(1)1( 20760
7120
0 =−===−=≥ XPXP finite population correction is 120/139=0.8633
From Exercise 3-92, X is approximately binomial with n = 20 and p = 5/140 =1/28
( )( ) ( ) 5168.04832.01)0(1)1( 2028270
28120
0 =−===−=≥ XPXP finite population correction is 120/139=0.8633
3-21
Section 3-9
3-107. a) P X e e( )!
.= = = =−
−0 40
0 01834 0
4
b) P X P X P X P X( ) ( ) ( ) (≤ = = + = )+ =2 0 1 2
= + +
=
−− −
e e e44 1 4 241
42
0 2381! !
.
c) P X e( )!
.= = =−
4 44
019544 4
d) P X e( )!
.= = =−
8 48
0 02984 8
3-108 a) P X e( ) ..= = =−0 0 67030 4
b) P X e e e( ) ( . )!
( . )!
... .
≤ = + + =−− −
2 0 41
0 42
0 99210 40 4 0 4 2
c) P X e( ) ( . )!
..
= = =−
4 0 44
0 0007150 4 4
d) P X e( ) ( . )!
..
= = = ×−
−8 0 48
109 100 4 8
8
3-109. . Therefore, λ = −ln(0.05) = 2.996. P X e( ) .= = =−0 0λ 05 Consequently, E(X) = V(X) = 2.996. 3-110 a) Let X denote the number of calls in one hour. Then, X is a Poisson random variable with λ = 10.
P X e( )!
.= = =−
5 105
0 037810 5
.
b) P X e e e e( )! ! !
.≤ = + + + =−− − −
3 101
102
103
0 01031010 10 2 10 3
c) Let Y denote the number of calls in two hours. Then, Y is a Poisson random variable with
λ = 20. P Y e( )!
.= = =−
15 2015
0 051620 15
d) Let W denote the number of calls in 30 minutes. Then W is a Poisson random variable with
λ = 5. P W e( )!
.= = =−
5 55
017555 5
3-111. λ=1, Poisson distribution. f(x) =e- λ λx/x!
(a) P(X≥2)= 0.264 (b) In order that P(X≥1) = 1-P(X=0)=1-e- λ exceed 0.95, we need λ=3.
Therefore 3*16=48 cubic light years of space must be studied. 3-112. (a) λ=14.4, P(X=0)=6*10-7
(b) λ=14.4/5=2.88, P(X=0)=0.056 (c) λ=14.4*7*28.35/225=12.7 P(X≥1)=0.999997
(d) P(X≥28.8) =1-P(X ≤ 28) = 0.00046. Unusual.
3-113. (a) λ=0.61. P(X≥1)=0.4566
(b) λ=0.61*5=3.05, P(X=0)= 0.047.
3-22
3-114. a) Let X denote the number of flaws in one square meter of cloth. Then, X is a Poisson random variable
with = 0.1. λ 0045.0!2
)1.0()2(21.0
===−eXP
b) Let Y denote the number of flaws in 10 square meters of cloth. Then, Y is a Poisson random variable
with = 1. λ 3679.0!11)1( 1
11
==== −−
eeYP
c) Let W denote the number of flaws in 20 square meters of cloth. Then, W is a Poisson random variable
with λ = 2. 1353.0)0( 2 === −eWP d) )1()0(1)1(1)2( =−=−=≤−=≥ YPYPYPYP
2642.0
1 11
=−−= −− ee
3-115.a) 2.0)( == λXE errors per test area
b) 9989.0!2
)2.0(!1
2.0)2(22.02.0
2.0 =++=≤−−
− eeeXP
99.89% of test areas 3-116. a) Let X denote the number of cracks in 5 miles of highway. Then, X is a Poisson random variable with
= 10. λ 510 1054.4)0( −− ×=== eXP b) Let Y denote the number of cracks in a half mile of highway. Then, Y is a Poisson random variable with
λ = 1. 6321.01)0(1)1( 1 =−==−=≥ −eYPYP c) The assumptions of a Poisson process require that the probability of a count is constant for all intervals. If the probability of a count depends on traffic load and the load varies, then the assumptions of a Poisson process are not valid. Separate Poisson random variables might be appropriate for the heavy and light load sections of the highway.
3-117. a) Let X denote the number of flaws in 10 square feet of plastic panel. Then, X is a Poisson random
variable with = 0.5. λ 6065.0)0( 5.0 === −eXP b) Let Y denote the number of cars with no flaws,
0067.0)3935.0()6065.0(1010
)10( 010 =⎟⎟⎠
⎞⎜⎜⎝
⎛==YP
c) Let W denote the number of cars with surface flaws. Because the number of flaws has a Poisson distribution, the occurrences of surface flaws in cars are independent events with constant probability. From part a., the probability a car contains surface flaws is 1−0.6065 = 0.3935. Consequently, W is binomial with n = 10 and p = 0.3935.
0 10
1 9
10( 0) (0.3935) (0.6065) 0.0067
0
10( 1) (0.3935) (0.6065) 0.0437
1( 1) 0.0067 0.0437 0.0504
P W
P W
P W
⎛ ⎞= = =⎜ ⎟
⎝ ⎠⎛ ⎞
= = =⎜ ⎟⎝ ⎠
≤ = + =
3-23
3-118. a) Let X denote the failures in 8 hours. Then, X has a Poisson distribution with λ = 0.16.
8521.0)0( 16.0 === −eXP b) Let Y denote the number of failure in 24 hours. Then, Y has a Poisson distribution with
λ = 0.48. 3812.01)0(1)1( 48 =−==−=≥ −eYPYP Supplemental Exercises
3-119. 41
31
83
31
41
31
81)( =⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛=XE ,
0104.041
31
83
31
41
31
81)(
2222
=⎟⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛=XV
3-120. a) 3681.0)999.0(001.01
1000)1( 9991 =⎟⎟
⎠
⎞⎜⎜⎝
⎛==XP
( )
( ) ( )
999.0)999.0)(001.0(1000)(1)001.0(1000)()d
9198.0
999.0001.02
1000999.0001.0
11000
999.0001.00
1000)2()c
6319.0999.0001.00
10001)0(1)1()b
9982999110000
9990
====
=
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=≤
=⎟⎟⎠
⎞⎜⎜⎝
⎛−==−=≥
XVXE
XP
XPXP
3-121. a) n = 50, p = 5/50 = 0.1, since E(X) = 5 = np.
b) ( ) ( ) ( ) 112.09.01.02
509.01.0
150
9.01.00
50)2( 482491500 =⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=≤XP
c) ( ) ( ) 48050149 1051.49.01.05050
9.01.04950
)49( −×=⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=≥XP
3-122. (a)Binomial distribution, p=0.01, n=12.
(b) P(X>1)=1-P(X≤1)= 1- - =0.0062 120 )1(012
pp −⎟⎟⎠
⎞⎜⎜⎝
⎛ 141 )1(112
pp −⎟⎟⎠
⎞⎜⎜⎝
⎛
(c) µ =E(X)= np =12*0.01 = 0.12
V(X)=np(1-p) = 0.1188 σ= )(XV = 0.3447
3-123. (a) 12(0.5) 0.000244=(b) = 0.2256 6 6
12 (0.5) (0.5)C 6
(c) 4189.0)5.0()5.0()5.0()5.0( 66126
75125 =+CC
3-124. (a) Binomial distribution, n=100, p=0.01.
(b) P(X≥1) = 0.634 (c) P(X≥2)= 0.264
3-24
3-118. a) Let X denote the failures in 8 hours. Then, X has a Poisson distribution with λ = 0.16.
8521.0)0( 16.0 === −eXP b) Let Y denote the number of failure in 24 hours. Then, Y has a Poisson distribution with
λ = 0.48. 3812.01)0(1)1( 48 =−==−=≥ −eYPYP Supplemental Exercises
3-119. 41
31
83
31
41
31
81)( =⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛=XE ,
0104.041
31
83
31
41
31
81)(
2222
=⎟⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛=XV
3-120. a) 3681.0)999.0(001.01
1000)1( 9991 =⎟⎟
⎠
⎞⎜⎜⎝
⎛==XP
( )
( ) ( )
999.0)999.0)(001.0(1000)(1)001.0(1000)()d
9198.0
999.0001.02
1000999.0001.0
11000
999.0001.00
1000)2()c
6319.0999.0001.00
10001)0(1)1()b
9982999110000
9990
====
=
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=≤
=⎟⎟⎠
⎞⎜⎜⎝
⎛−==−=≥
XVXE
XP
XPXP
3-121. a) n = 50, p = 5/50 = 0.1, since E(X) = 5 = np.
b) ( ) ( ) ( ) 112.09.01.02
509.01.0
150
9.01.00
50)2( 482491500 =⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=≤XP
c) ( ) ( ) 48050149 1051.49.01.05050
9.01.04950
)49( −×=⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=≥XP
3-122. (a)Binomial distribution, p=0.01, n=12.
(b) P(X>1)=1-P(X≤1)= 1- - =0.0062 120 )1(012
pp −⎟⎟⎠
⎞⎜⎜⎝
⎛ 141 )1(112
pp −⎟⎟⎠
⎞⎜⎜⎝
⎛
(c) µ =E(X)= np =12*0.01 = 0.12
V(X)=np(1-p) = 0.1188 σ= )(XV = 0.3447
3-123. (a) 12(0.5) 0.000244=(b) = 0.2256 6 6
12 (0.5) (0.5)C 6
(c) 4189.0)5.0()5.0()5.0()5.0( 66126
75125 =+CC
3-124. (a) Binomial distribution, n=100, p=0.01.
(b) P(X≥1) = 0.634 (c) P(X≥2)= 0.264
3-24
(d) µ=E(X)= np=100*0.01=1
V(X)=np(1-p) = 0.99
σ= )(XV =0.995 (e) Let pd= P(X≥2)= 0.264,
Y: # of messages requires two or more packets be resent. Y is binomial distributed with n=10, pm=pd*(1/10)=0.0264 P(Y≥1) = 0.235
3-125 Let X denote the number of mornings needed to obtain a green light. Then X is a geometric random variable with p = 0.20. a) P(X = 4) = (1-0.2)30.2= 0.1024 b) By independence, (0.8)10 = 0.1074. (Also, P(X > 10) = 0.1074) 3-126 Let X denote the number of attempts needed to obtain a calibration that conforms to specifications. Then, X is geometric with p = 0.6. P(X ≤ 3) = P(X=1) + P(X=2) + P(X=3) = 0.6 + 0.4(0.6) + 0.42(0.6) = 0.936. 3-127. Let X denote the number of fills needed to detect three underweight packages. Then X is a negative binomial random variable with p = 0.001 and r = 3. a) E(X) = 3/0.001 = 3000 b) V(X) = [3(0.999)/0.0012] = 2997000. Therefore, σX = 1731.18 3-128. Geometric with p=0.1
(a) f(x)=(1-p)x-1p=0.9(x-1)0.1 (b) P(X=5) = 0.94*0.1=0.0656 (c) µ=E(X)= 1/p=10 (d) P(X≤10)=0.651
3-129. (a) λ=6*0.5=3.
P(X=0) = 0.0498 (b) P(X≥3)=0.5768 (c) P(X≤x) ≥0.9, x=5 (d) σ2= λ=6. Not appropriate.
3-130. Let X denote the number of totes in the sample that do not conform to purity requirements. Then, X has a hypergeometric distribution with N = 15, n = 3, and K = 2.
3714.0!15!10!12!131
315
313
02
1)0(1)1( =−=
⎟⎟⎠
⎞⎜⎜⎝
⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛
−==−=≥ XPXP
3-131. Let X denote the number of calls that are answered in 30 seconds or less. Then, X is a binomial random variable with p = 0.75.
a) P(X = 9) = 1877.0)25.0()75.0(9
10 19 =⎟⎟⎠
⎞⎜⎜⎝
⎛
b) P(X ≥ 16) = P(X=16) +P(X=17) + P(X=18) + P(X=19) + P(X=20)
3-25
4148.0)25.0()75.0(2020
)25.0()75.0(1920
)25.0()75.0(1820
)25.0()75.0(1720
)25.0()75.0(1620
020119
218317416
=⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=
c) E(X) = 20(0.75) = 15 3-132. Let Y denote the number of calls needed to obtain an answer in less than 30 seconds.
a) 0117.075.025.075.0)75.01()4( 33 ==−==YP b) E(Y) = 1/p = 1/0.75 = 4/3 3-133. Let W denote the number of calls needed to obtain two answers in less than 30 seconds. Then, W has a negative binomial distribution with p = 0.75.
a) P(W=6) = 51
0 25 0 75 0 01104 2⎛⎝⎜⎞⎠⎟ =( . ) ( . ) .
b) E(W) = r/p = 2/0.75 = 8/3
3-134. a) Let X denote the number of messages sent in one hour. 1755.0!55)5(
55
===−eXP
b) Let Y denote the number of messages sent in 1.5 hours. Then, Y is a Poisson random variable with
λ =7.5. 0858.0!10
)5.7()10(105.7
===−eYP
c) Let W denote the number of messages sent in one-half hour. Then, W is a Poisson random variable with = 2.5. λ 2873.0)1()0()2( ==+==< WPWPWP 3-135X is a negative binomial with r=4 and p=0.0001
400000001.0/4/)( === prXE requests 3-136. X ∼ Poisson(λ = 0.01), X ∼ Poisson(λ = 1)
9810.0!3)1(
!2)1(
!1)1()3(
3121111 =+++=≤
−−−− eeeeYP
3-137. Let X denote the number of individuals that recover in one week. Assume the individuals are independent. Then, X is a binomial random variable with n = 20 and p = 0.1. P(X ≥ 4) = 1 − P(X ≤ 3) = 1 − 0.8670 = 0.1330. 3-138. a.) P(X=1) = 0 , P(X=2) = 0.0025, P(X=3) = 0.01, P(X=4) = 0.03, P(X=5) = 0.065
P(X=6) = 0.13, P(X=7) = 0.18, P(X=8) = 0.2225, P(X=9) = 0.2, P(X=10) = 0.16 b.) P(X=1) = 0.0025, P(X=1.5) = 0.01, P(X=2) = 0.03, P(X=2.5) = 0.065, P(X=3) = 0.13 P(X=3.5) = 0.18, P(X=4) = 0.2225, P(X=4.5) = 0.2, P(X=5) = 0.16
3-139. Let X denote the number of assemblies needed to obtain 5 defectives. Then, X is a negative binomial
random variable with p = 0.01 and r=5. a) E(X) = r/p = 500. b) V(X) =(5* 0.99)/0.012 = 49500 and σX = 222.49 3-140. Here n assemblies are checked. Let X denote the number of defective assemblies. If P(X ≥ 1) ≥ 0.95, then
3-26
P(X=0) ≤ 0.05. Now,
P(X=0) = and 0.99nnn99)99.0()01.0(
00 =⎟⎟
⎠
⎞⎜⎜⎝
⎛ n ≤ 0.05. Therefore,
07.298
)95.0ln()05.0ln(
)05.0ln())99.0(ln(
=≥
≤
n
n
This would require n = 299. 3-141. Require f(1) + f(2) + f(3) + f(4) = 1. Therefore, c(1+2+3+4) = 1. Therefore, c = 0.1. 3-142. Let X denote the number of products that fail during the warranty period. Assume the units are independent. Then, X is a binomial random variable with n = 500 and p = 0.02.
a) P(X = 0) = 4.1 x 10=⎟⎟⎠
⎞⎜⎜⎝
⎛ 5000 )98.0()02.0(0
500-5
b) E(X) = 500(0.02) = 10 c) P(X >2) = 1 − P(X ≤ 2) = 0.9995 3-143. 16.0)3.0)(3.0()7.0)(1.0()0( =+=Xf
19.0)3.0)(4.0()7.0)(1.0()1( =+=Xf
20.0)3.0)(2.0()7.0)(2.0()2( =+=Xf
31.0)3.0)(1.0()7.0)(4.0()3( =+=Xf
14.0)3.0)(0()7.0)(2.0()4( =+=Xf 3-144. a) P(X ≤ 3) = 0.2 + 0.4 = 0.6 b) P(X > 2.5) = 0.4 + 0.3 + 0.1 = 0.8 c) P(2.7 < X < 5.1) = 0.4 + 0.3 = 0.7 d) E(X) = 2(0.2) + 3(0.4) + 5(0.3) + 8(0.1) = 3.9 e) V(X) = 22(0.2) + 32(0.4) + 52(0.3) + 82(0.1) − (3.9)2 = 3.09 3-145.
x 2 5.7 6.5 8.5 f(x) 0.2 0.3 0.3 0.2
3-146. Let X denote the number of bolts in the sample from supplier 1 and let Y denote the number of bolts in the sample from supplier 2. Then, x is a hypergeometric random variable with N = 100, n = 4, and K = 30. Also, Y is a hypergeometric random variable with N = 100, n = 4, and K = 70. a) P(X=4 or Y=4) = P(X = 4) + P(Y = 4)
2408.04
1004
700
30
4100
070
430
=
⎟⎟⎠
⎞⎜⎜⎝
⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛
+
⎟⎟⎠
⎞⎜⎜⎝
⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛
=
3-27
b) P[(X=3 and Y=1) or (Y=3 and X = 1)]= 4913.0
4100
370
130
170
330
=
⎟⎟⎠
⎞⎜⎜⎝
⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛
=
3-147. Let X denote the number of errors in a sector. Then, X is a Poisson random variable with λ = 0.32768. a) P(X>1) = 1 − P(X≤1) = 1 − e-0.32768 − e-0.32768(0.32768) = 0.0433 b) Let Y denote the number of sectors until an error is found. Then, Y is a geometric random variable and P = P(X ≥ 1) = 1 − P(X=0) = 1 − e-0.32768 = 0.2794 E(Y) = 1/p = 3.58
3-148. Let X denote the number of orders placed in a week in a city of 800,000 people. Then X is a Poisson random variable with λ = 0.25(8) = 2. a) P(X ≥ 3) = 1 − P(X ≤ 2) = 1 − [e-2 + e-2(2) + (e-222)/2!] = 1 − 0.6767 = 0.3233. b) Let Y denote the number of orders in 2 weeks. Then, Y is a Poisson random variable with λ = 4, and P(Y>2) =1- P(Y ≤ 2) = e-4 + (e-441)/1!+ (e-442)/2! =1 - [0.01832+0.07326+0.1465] = 0.7619. 3-149.a) hypergeometric random variable with N = 500, n = 5, and K = 125
2357.0115524.2100164.6
5500
5375
0125
)0( ==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛
=EEf X
3971.0115525.2
)810855.8(125
5500
4375
1125
)1( ==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛
=E
Ef X
2647.0115524.2
)8718875(7750
5500
3375
2125
)2( ==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛
=E
f X
0873.0115524.2
)70125(317750
5500
2375
3125
)3( ==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛
=E
f X
01424.0115524.2
)375(9691375
5500
1375
4125
)4( ==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛
=E
f X
3-28
00092.0115524.283453.2
5500
0375
5125
)5( ==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛
=EEf X
b) x 0 1 2 3 4 5 6 7 8 9 10f(x) 0.0546 0.1866 0.2837 0.2528 0.1463 0.0574 0.0155 0.0028 0.0003 0.0000 0.0000 3-150. Let X denote the number of totes in the sample that exceed the moisture content. Then X is a binomial random variable with n = 30. We are to determine p.
If P(X ≥ 1) = 0.9, then P(X = 0) = 0.1. Then , giving 30ln(1−p)=ln(0.1),
which results in p = 0.0739.
1.0)1()(0
30 300 =−⎟⎟⎠
⎞⎜⎜⎝
⎛pp
3-151. Let t denote an interval of time in hours and let X denote the number of messages that arrive in time t. Then, X is a Poisson random variable with λ = 10t. Then, P(X=0) = 0.9 and e-10t = 0.9, resulting in t = 0.0105 hours = 37.8 seconds 3-152. a) Let X denote the number of flaws in 50 panels. Then, X is a Poisson random variable with λ = 50(0.02) = 1. P(X = 0) = e-1 = 0.3679. b) Let Y denote the number of flaws in one panel, then P(Y ≥ 1) = 1 − P(Y=0) = 1 − e-0.02 = 0.0198. Let W denote the number of panels that need to be inspected before a flaw is found. Then W is a geometric random variable with p = 0.0198 and E(W) = 1/0.0198 = 50.51 panels.
c) 0198.01)0(1)1( 02.0 =−==−=≥ −eYPYP Let V denote the number of panels with 1 or more flaws. Then V is a binomial random
variable with n=50 and p=0.0198
491500 )9802.0(0198.0150
)9802(.0198.00
50)2( ⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=≤VP
9234.0)9802.0(0198.02
50 482 =⎟⎟⎠
⎞⎜⎜⎝
⎛+
Mind Expanding Exercises 3-153. The binomial distribution
P(X=x) = ( )!xn!x!n−
px(1-p)n-x
If n is large, then the probability of the event could be expressed as λ/n, that is λ=np. We could re-write the probability mass function as:
P(X=x) = ( )!xn!x!n−
[λ/n]x[1 – (λ/n)]n-x
Where p = λ/n.
P(X=x)n
1)x(n3)......(n2)(n1)(nnx
+−×−×−×−×x!λx
(1 – (λ/n))n-x
Now we can re-express:
3-29
00092.0115524.283453.2
5500
0375
5125
)5( ==
⎟⎟⎠
⎞⎜⎜⎝
⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛
=EEf X
b) x 0 1 2 3 4 5 6 7 8 9 10f(x) 0.0546 0.1866 0.2837 0.2528 0.1463 0.0574 0.0155 0.0028 0.0003 0.0000 0.0000 3-150. Let X denote the number of totes in the sample that exceed the moisture content. Then X is a binomial random variable with n = 30. We are to determine p.
If P(X ≥ 1) = 0.9, then P(X = 0) = 0.1. Then , giving 30ln(1−p)=ln(0.1),
which results in p = 0.0739.
1.0)1()(0
30 300 =−⎟⎟⎠
⎞⎜⎜⎝
⎛pp
3-151. Let t denote an interval of time in hours and let X denote the number of messages that arrive in time t. Then, X is a Poisson random variable with λ = 10t. Then, P(X=0) = 0.9 and e-10t = 0.9, resulting in t = 0.0105 hours = 37.8 seconds 3-152. a) Let X denote the number of flaws in 50 panels. Then, X is a Poisson random variable with λ = 50(0.02) = 1. P(X = 0) = e-1 = 0.3679. b) Let Y denote the number of flaws in one panel, then P(Y ≥ 1) = 1 − P(Y=0) = 1 − e-0.02 = 0.0198. Let W denote the number of panels that need to be inspected before a flaw is found. Then W is a geometric random variable with p = 0.0198 and E(W) = 1/0.0198 = 50.51 panels.
c) 0198.01)0(1)1( 02.0 =−==−=≥ −eYPYP Let V denote the number of panels with 1 or more flaws. Then V is a binomial random
variable with n=50 and p=0.0198
491500 )9802.0(0198.0150
)9802(.0198.00
50)2( ⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=≤VP
9234.0)9802.0(0198.02
50 482 =⎟⎟⎠
⎞⎜⎜⎝
⎛+
Mind Expanding Exercises 3-153. The binomial distribution
P(X=x) = ( )!xn!x!n−
px(1-p)n-x
If n is large, then the probability of the event could be expressed as λ/n, that is λ=np. We could re-write the probability mass function as:
P(X=x) = ( )!xn!x!n−
[λ/n]x[1 – (λ/n)]n-x
Where p = λ/n.
P(X=x)n
1)x(n3)......(n2)(n1)(nnx
+−×−×−×−×x!λx
(1 – (λ/n))n-x
Now we can re-express:
3-29
[1 – (λ/n)]n-x = [1 – (λ/n)]n[1 – (λ/n)]-x And the first part of this can be re-expressed further as
[1 – (λ/n)]n = ( )( )nλ-1 n/λλ
−−
So:
P(X=x)=n
1)x(n3)......(n2)(n1)(nnx
+−×−×−×−×x!λx
( )( )nλ-1 n/λλ
−−
[1 – (λ/n)]-x
Now: In the limit as n → ∞
n1)x(n3)......(n2)(n1)(nn
x
+−×−×−×−× ≅ 1
In the limit as n → ∞ [1 – (λ/n)]-x ≅ 1 Thus:
P(X=x) = x!λx
( )( )nλ-1 n/λλ
−−
We know from exponential series that:
Limit z → ∞ ⎟⎠⎞
⎜⎝⎛ +
z11
z
= e ≅ 2.7183
In our case above –n/λ = z, so ( )nλ-1 n/λ− = e. Thus,
P(X=x) = x!λx
e λ−
It is the case, therefore, that
Limit n → ∞ ( )!xn!x!n−
px(1 – p)n-x = !x
e xλλ−
The distribution of the probability associated with this process is known as the Poisson distribution. The pmf can be expressed as:
f(x) = !x
e xλλ−
3-154. Sow that using an infinite sum. ∑∞
=
− =−1
1 1)1(i
i pp
To begin, , by definition of an infinite sum this can be rewritten
as
∑∑∞
=
−∞
=
− −=−1
1
1
1 )1()1(i
i
i
i pppp
1)1(1
)1(1
1 ==−−
=−∑∞
=
−
pp
pppp
i
i
3-30
3-155.
[ ]
1
1 1
2 2
2 22
( ) [( ( 1) ... ]( 1)
( 1) ( 1)2 2
( 1) ( 1)
( ) ( )( 1)2 2
( 1) ( 1)( )
2( 1)(( )
( )1
b a
i i
b bbb a
i a i ai a
E X a a b b a
b b a ai i
b a b a
b a b a b a b a
b a b ab a
b a b ai b a iiV X
b a
−
= =
+
= ==
= + + + + − +
⎡ ⎤ + −⎡ ⎤− −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦= =− + − +
⎡ ⎤− + + + − +⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦= =− + − ++
=
− + +− + +−= =
+ −
∑ ∑
∑ ∑∑2
2
2
)4
1( 1)(2 1) ( 1) (2 1) ( 1) ( 1) ( 1)( )( )
6 6 2 41
( 1) 112
b ab b b a a a b b a a b a b ab a
b ab a
⎡ ⎤⎢ ⎥⎣ ⎦
+ −+ + − − + − − − + +⎡ ⎤− − + +⎢ ⎥⎣ ⎦=
− +− + −
=
3-156. Let X denote the number of nonconforming products in the sample. Then, X is approximately binomial with p = 0.01 and n is to be determined. If , then 90.0)1( ≥≥XP 10.0)0( ≤=XP .
Now, P(X = 0) = ( ) nnn ppp )1()1(00 −=− . Consequently, , and
( ) .1 0− ≤p n 10
11.229)1ln(
10.0ln=
−≤
pn . Therefore, n = 230 is required
3-157. If the lot size is small, 10% of the lot might be insufficient to detect nonconforming product. For example, if the lot size is 10, then a sample of size one has a probability of only 0.2 of detecting a nonconforming product in a lot that is 20% nonconforming. If the lot size is large, 10% of the lot might be a larger sample size than is practical or necessary. For example, if the lot size is 5000, then a sample of 500 is required. Furthermore, the binomial approximation to the hypergeometric distribution can be used to show the following. If 5% of the lot of size 5000 is nonconforming, then the probability of zero nonconforming product in the sample is approximately
. Using a sample of 100, the same probability is still only 0.0059. The sample of size 500 might 7 10 12× −
be much larger than is needed.
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3-158. Let X denote the number of panels with flaws. Then, X is a binomial random variable with n =100 and p is
the probability of one or more flaws in a panel. That is, p = = 0.095. 1.01 −− e
( ) ( ) ( )( ) ( )
034.0
)1()1(
)1()1()1(
)4()3()2()1()0()4()5(
9641004
9731003
9821002
9911001
10001000
=
−+−+
−+−+−=
=+=+=+=+==≤=<
pppp
pppppp
XPXPXPXPXPXPXP
3-159. Let X denote the number of rolls produced.
Revenue at each demand 0 1000 2000 3000
0 1000≤ ≤x 0.05x 0.3x 0.3x 0.3x mean profit = 0.05x(0.3) + 0.3x(0.7) - 0.1x
1000 2000≤ ≤x 0.05x 0.3(1000) + 0.05(x-1000)
0.3x 0.3x
mean profit = 0.05x(0.3) + [0.3(1000) + 0.05(x-1000)](0.2) + 0.3x(0.5) - 0.1x 2000 3000≤ ≤x 0.05x 0.3(1000) +
0.05(x-1000) 0.3(2000) +
0.05(x-2000) 0.3x
mean profit = 0.05x(0.3) + [0.3(1000)+0.05(x-1000)](0.2) + [0.3(2000) + 0.05(x-2000)](0.3) + 0.3x(0.2) - 0.1x 3000 ≤ x 0.05x 0.3(1000) +
0.05(x-1000) 0.3(2000) +
0.05(x-2000) 0.3(3000)+
0.05(x-3000) mean profit = 0.05x(0.3) + [0.3(1000)+0.05(x-1000)](0.2) + [0.3(2000)+0.05(x-2000)]0.3 + [0.3(3000)+0.05(x-
3000)]0.2 - 0.1x
Profit Max. profit 0 1000≤ ≤x 0.125 x $ 125 at x = 1000
1000 2000≤ ≤x 0.075 x + 50 $ 200 at x = 2000 2000 3000≤ ≤x 200 $200 at x = 3000
3000 ≤ x -0.05 x + 350 $200 at x = 3000
The bakery can make anywhere from 2000 to 3000 and earn the same profit. 3-160.Let X denote the number of acceptable components. Then, X has a binomial distribution with p = 0.98 and n is to be determined such that P X . ( ) .≥ ≥100 0 95
n P X( )≥ 100 102 0.666 103 0.848 104 0.942 105 0.981
Therefore, 105 components are needed.
3-32
