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CEE 110 Probability and Statistics for Engineers

PS 3 Solution Set

3-2. The range of X is{ }0 12 50, , ,...,

3-4. The range of X is 0! 1! 2! 3! 4! "#

3-$. The range of X is{ }0 12 100, , ,...,

. %lthough the range actually obtained fro& lots ty'ically &ight not e(ceed 10).

3-*. The range of X is con+eniently &odeled as all nonnegati+e integers. That is! the range of X is 0! 1! 2! ,#

3-1*. %ll 'robabilities are greater than or eual to ero and su& to one.

a/ PX≤ 1/PX1/0."14

 b/ PX1/ 1-PX1/1-0."140.42*$

c/ P2X$/PX3/0.1425d/ PX≤1 or X1/ PX1/6 PX2/6PX3/1

3-20. Probabilities are nonnegati+e and su& to one.a/ PX 2/ 374174/2  37$4

 b/ PX ≤ 2/ 3748161746174/29 $37$4

c/ PX 2/ 1 − PX ≤ 2/ 17$4

d/ PX ≥ 1/ 1 − PX ≤ 0/ 1 − 374/ 174

3-34. X days until changePX1."/ 0.0"

PX3/ 0.2"

PX4."/ 0.3"PX"/ 0.20

PX/ 0.1"

=====

=

!1".0

"!20.0

".4!3".0

3!2".0

".1!0".0

/

 x

 x

 x

 x

 x

 x f  

3-40.

 

F ( x) =

0   x

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c/ P X   2/ 1 : P X  ≤ 2/ 1 : $7 17

d/ P1  X  ≤ 2/ P X  ≤ 2/ : P X  ≤ 1/ $7 : 47 27

3-$4.

$

/ 0 0/ 1 1/ 2 2/ 3 3/

0* 10 / 10.0012/ 20.0"$/ 30.5412/

2.54000*

 E X f f f f   µ 

−

= = + + +

= × + + +=

2 2 2 2 2 / 0 0/ 1 1/ 2 2/ 3 3/

0.0"*$05$

V X f f f f      µ = + + + −

=

3-$. EX/ 0 6 55/72 45."! ;X/ 855 : 0 6 1/2 : 19712 *33.2"

3-54. a/

24$1.0/".0".0"

10/"   "" =  

  

  == X  P 

 b/

*251100 ".0".02

10".0".0

1

10".0".0

0

10/2   

 

  

 +  

 

  

 +  

 

  

 =≤ X  P 

 0"4.0/".04"/".010".0

  101010 =++=

c/

010.0/".0".010

10/".0".0

5

10/5   01015 =  

 

  

 +  

 

  

 =≥ X  P 

d/

$43 ".0".04

10".0".03

10/"3     

  +  

  

  =

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 b/

 

( )( )( )

"425.04"1.01/1

4"1.0A140A11"

A120A13"/0

/01/1

A120A20A140

A11"A20A13"

140

20

13"

20

"

0

=−=≥

=====

=−=≥

 X  P 

 X  P 

 X  P  X  P 

3-1$4.

a/ @et X denote the nu&ber of fla