Unit 3 Lesson 7- Determining an Equilibrium Constant Lab Chemistry OVS - SCIENCE 101

EECS 55 – Engineering Probability
Homework #5
Solutions
(1)
1
(a) c  (1  x
2
 
1
)dx = 1 => c x  x 1 = 1 => c (1-1/3) - c (-1+1/3) = 1 => 4c/3 = 1
1
=> c = 3/4

x
(b) F(x) = 3/4  (1  y 2 )dy =3/4( x  1  x  ) = ¾(  x  x  ) -1 < x < 1
1
F(x) = 0 for x ≤ -1, F(x) = 1 for x ≥ 1

(c) ( )=∫ ( ) × 3 4 (1 −
=∫ ) = [3 4 ( − )] = 0
(d) ( )= ( )− ( ) = ( )=∫ ( ) =
1 1
= × 3 4 (1 − ) = [3 4 ( − )] = 1 5
3 5

(2)
 
10 10 
(a) 20 x 2 dx = - x  20 = 1/2
x x
10 10  10
(b) F(x) =  2 dy = -  = -  1 x > 10. F(x) = 0 for x ≤ 10.
10 y
y  10 x

(c) We are assuming independence of the events that the devices exceed 15 hours.
Probability that a device will function at least 15 hours is 1-F(15) = 10/15=2/3.
The probability that out of 6 devices, at least 3 will function for at least 15 hours
i 6i
6
 6  2   1 
is going to be       .
i  3  i  3   3 



(3)

1 2 x / 2
E[X]= x e dx =
4 0
Let’s assume u(x) = x2 and v’(x) =e-x/2 => v(x) = -2 e-x/2
 
So E[X] = u(x)v(x) 0 -  u' ( x)v( x)dx =1/4( -2 x2e-x/2 0 - (- 4 x e  x / 2 dx ))=
0 0




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