Applied statistics and probability for engineers 7th edition solutions

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