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188 CHAPTER 4
5. (a) ln I∼N(1,0.2), ln R∼N(4,0.1), and Iand Rare independent. Therefore ln V∼N(5,0.3).
6. The probability that a given circuit has a voltage less than 200 volts is P(V < 200) = 0.7054.
Let Xbe the number of circuits whose voltage is less than 200 volts. Then X∼Bin(10,0.7054).
7. Let Xrepresent the withdrawal strength for a randomly chosen annularly threaded nail, and let Y
represent the withdrawal strength for a randomly chosen helically threaded nail.
(a) E(X) = e3.82+(0.219)2/2= 46.711 N/mm
(d) First find the median strength for annularly threaded nails.
Let mbe the median of X. Then P(X≤m) = P(ln X≤ln m) = 0.5.
Since ln X∼N(3.82,0.2192), P(ln Y < 3.82) = 0.5.
Page 189
190 CHAPTER 4
8. i. Since the lognormal population is skewed to the right, the mean will always be larger than the
9. Let Xrepresent the price of a share of company A one year from now. Let Yrepresent the price of a
share of company B one year from now.
10. (a) P(X < 20) = P(ln X < ln 20) = P(ln X < 2.996).
Page 190
SECTION 4.7 191
(c) Yes, because 20 MPa is unusually small if the claim is true.
Section 4.7
1. (a) µT= 1/0.45 = 2.2222
2. (a) λ= 1/µ = 1/0.5 = 2 seconds−1.
Page 191
192 CHAPTER 4
(b) Let Tbe the time between requests, and let mbe the median of T. Then P(T≤m) = 0.5.
P(T≤m) = 1 −e−2m= 0.5, so e−2m= 0.5.
3. Let Xbe the diameter in microns.
(a) µX= 1/λ = 1/0.25 = 4 microns
Page 192
SECTION 4.7 193
(g) Let x99 be the 99th percentile. Then P(T≤x99) = 0.75.
4. Let Tbe the distance between flaws, in meters.
(a) P(T > 15) = e−(1/12)(15) = 0.2865
5. (a) Let Xbe the number of pores with diameters less than 3 microns. The probability that a diameter is
less than 3 microns is 0.5276.
Page 193
194 CHAPTER 4
6. (a) P(X≥5) = 1 −P(X < 5) = 1 −P(X≤5) = 1 −(1 −e−1(5)) = 0.0067
7. No. If the lifetimes were exponentially distributed, the proportion of used components lasting longer
8. (a) T∼Exp(2), so µT= 1/2 = 0.5 seconds.
Page 194
SECTION 4.7 195
9. Let Tbe the waiting time between accidents. Then T∼Exp(3).
(a) µT= 1/3 year
10. (a) λ= 1/µX= 1/3
11. X1, ..., X5are each exponentially distributed with λ= 1/200 = 0.005.
i=1
Page 195
196 CHAPTER 4
= (e−0.5)5
=e−2.5
= 0.0821
(c) The time of the first replacement will be greater than 100 hours if and only if each of the bulbs lasts
longer than 100 hours.
Section 4.8
1. Let Tbe the waiting time.
Page 196
2. Let Xbe the resistance of a resistor.
(a) µX= (95 + 103)/2 = 99 mm
3. (a) µT= 4/0.5 = 8
(b) σT=p4/0.52= 4
4−1
X
4. µT=r/λ and σ2
T=r/λ2. It follows that µT=σ2
Tλ.
5. µT=r/λ and σ2
T=r/λ2. It follows that λ=r/µT.
6. Let X1, ..., X6be the lifetimes of the first 6 motors. Let T=X1+···+X6be the time that the sixth
j=0
j=0
7. α= 0.5, β= 3, 1/α = 2.
SECTION 4.8 199
9. Let Tbe the lifetime in hours of a bearing.
(a) P(T > 1000) = 1 −P(T≤1000) = 1 −(1 −e−[(0.0004474)(1000)]2.25 ) = e−[(0.0004474)(1000)]2.25 = 0.8490
10. Let Tbe the lifetime of a battery.
(a) P(T > 10) = 1 −P(T≤10) = 1 −(1 −e−[(0.1)(10)]2)=e−1= 0.3679
Page 199
11. Let Tbe the lifetime of a fan.
12. (a) P(T≤1) = 1 −e−[(0.01)(1)]3= 1.000 ×10−6
13. (a) P(X1>5) = 1 −P(X1≤5) = 1 −(1 −e−[(0.2)(5)]2) = e−1= 0.3679
Page 200
SECTION 4.8 201
15. µX2=Zb
a
x21
b−adx =a2+ab +b2
3
16. (a) Since P(X≤a) = 0, P(X≤x) = 0 if x≤a.
Page 201
17. (a) The cumulative distribution function of Uis
FU(x) =
0x≤0
x0< x ≤1
1x > 1
0x≤a
x−a
Section 4.9
1. iii. By definition, an estimator is unbiased if its mean is equal to the true value.
Page 202
SECTION 4.9 203
(c) We denote the mean of bµ3by E(bµ3) and the variance of bµ3by V(bµ3).
E(bµ3) = µX1+µX2
4=µ+µ
4=µ
2.
8+−µ
22=2µ2+ 1
8.
(d) bµ3has smaller mean squared error than bµ1whenever 2µ2+ 1
4. (a) We denote the mean of bσ2
kby E(bσ2
k).
E(bσ2
k) = E(n−1)s2
k=(n−1)µs2
k=(n−1)σ2
k
Page 203
5. The probability mass function of Xis f(x;p) = p(1 −p)x−1.
The MLE is the value of pthat maximizes f(x;p), or equivalently, ln f(x;p).
6. The joint probability mass function of X1, ..., Xnis
n
Y
i
i=1 xi
7. (a) The probability mass function of Xis f(x;p) = n!
x!(n−x)!(p)x(1 −p)n−x.
The MLE of pis the value of pthat maximizes f(x;p), or equivalently, ln f(x;p).
d
dp ln f(x;p) = d
dp[ln n!−ln x!−ln(n−x)! + xln p+ (n−x) ln(1 −p)] = x
p−n−x
1−p= 0.
(c) The MLE of λis b
λ=X. The MLE of e−λis therefore e−b
Page 204
8. The joint probability density function of X1, ..., Xnis
f(x1, ..., xn;µ) =
n
Y
1
√2πe−(xi−µ)2/2= (2π)−n/2e−Pn
i=1 (xi−µ)2/2.
9. The joint probability density function of X1, ..., Xnis
f(x1, ..., xn;σ) =
n
Y
i=1
1
σ√2πe−x2
i/2σ2= (2π)−n/2σ−ne−Pn
i=1 x2
i/2σ2
.
10. The joint probability density function of X1, ..., Xnis
n
i=1
Page 205
206 CHAPTER 4
Now we solve for µ:
Pn
i=1(xi−µ)
σ2= 0.
Pn
Now we substitute bµfor µin ln f(x,..., xn;µ, σ) and maximize over σ:
∂
∂σ ln f(x1, ..., xn;bµ, σ) = ∂
∂σ "−(n/2) ln 2π−nln σ−
n
X
i=1
(xi−bµ)2
2σ2#=−n
σ+Pn
i=1(xi−bµ)2
σ3= 0.
Section 4.10
1. (a) No
Page 206
SECTION 4.10 207
2.
12 14 16 18 20
0.001
0.01
0.05
0.1
0.25
0.5
0.75
0.9
0.95
0.99
0.999
3.
2 2.5 3 3.5 4 4.5
0.001
0.01
0.05
0.1
0.25
0.5
0.75
0.9
0.95
0.99
0.999
4.
50 60 70 80 90
0.001
0.01
0.05
0.1
0.25
0.5
0.75
0.9
0.95
0.99
0.999
5.
0 5 10 15 20 25
0.001
0.01
0.05
0.1
0.25
0.5
0.75
0.9
0.95
0.99
0.999
Page 207
