Applied Statistics and Probability for Engineers, 7th edition
CHAPTER 4
Section 4.1
4.1.1 a)
3679.0)()1(1
1
1
==−==−
−
−
eedxeXPxx
b)
2858.0)()5.21(5.21
5.2
1
5.2
1
=−=−==−−−−
eeedxeXPxx
c)
d)
9817.01)()4( 4
4
0
4
0
=−=−==−−−
eedxeXPxx
e)
0498.0)()3( 3
3
3
==−==−
−
−
eedxeXPxx
f)
10.0)()( ==−==−
−
−
x
x
x
x
xeedxeXxP
.
Then, x = −ln(0.10) = 2.3
g)
10.01)()( 0
0
=−=−==−−−
x
x
x
x
xeedxexXP
.
Then, x = −ln(0.9) = 0.1054
4.1.3 a)𝑃(𝑋<0)=∫0.5𝑐𝑜𝑠𝑥
0
−𝜋/2𝑑𝑥=(0.5𝑠𝑖𝑛𝑥)|−𝜋/2
0=0−(−0.5)=0.5
b)𝑃(𝑋<−𝜋/4)=∫0.5𝑐𝑜𝑠𝑥
−𝜋/4
−𝜋/2 𝑑𝑥=(0.5𝑠𝑖𝑛𝑥)|−𝜋/2
−𝜋/4=−0.3536−(−0.5)=0.1464
c)𝑃(−𝜋/4<𝑋<𝜋/4)=∫0.5𝑐𝑜𝑠𝑥
𝜋/4
−𝜋/4𝑑𝑥=(0.5𝑠𝑖𝑛𝑥)|−𝜋/4
𝜋/4=0.3536−(−0.3536)=
0.7072
d)𝑃(𝑋>−𝜋/4)=∫0.5𝑐𝑜𝑠𝑥
𝜋/2
−𝜋/4𝑑𝑥=(0.5𝑠𝑖𝑛𝑥)|−𝜋/4
𝜋/2=0.5−(−0.3536)=0.8536
e) 𝑃(𝑋<𝑥)=∫0.5𝑐𝑜𝑠𝑥
𝑥
−𝜋/2 𝑑𝑥=(0.5𝑠𝑖𝑛𝑥)|−𝜋/2
𝑥=(0.5𝑠𝑖𝑛𝑥)−(−0.5)=0.95
Then, sin x = 0.9, and x = 1.1198 radians
4.1.4 a) 𝑃(𝑋<4)=∫𝑥
8𝑑𝑥
4
3=𝑥2
16|3
4=42−32
16=0.4375, because 𝑓(𝑥)=0 for x < 3.
b) 𝑃(𝑋>3.5)=∫𝑥
8𝑑𝑥
5
3.5=𝑥2
16|3.5
5=52−3.52
16=0.7969 , because 𝑓(𝑥)=0 for x > 5.
c) 𝑃(4<𝑋<5)=∫𝑥
8𝑑𝑥
5
4=𝑥2
16|4
5=52−42
16=0.5625