Download Engineering 323 Beautiful Homework Set # 8 1 of 4 Khaleeli

Engineering 323
Khaleeli
Beautiful Homework Set # 8
Problem 4.8
1 of 4
4.8 In commuting to work. I must first get on a bus near my house and then transfer to a
second bus. If waiting time (in minutes) at each stop has a uniform distribution with A =
0 and B = 5, then it can be shown that my total waiting time Y has the pdf:
1
 25 y

2 1
f (y) =  −
y
 5 25
0


0≤ y≤ 5
5 ≤ y ≤ 10
y < 0 or y > 10
A) Sketch a graph of the pdf of Y.
Using the definition for the Continuous Random Variables and Probability
Distributions, which can be found on page 138 thru 141, will help you solve this
problem.
0.25
0.2
f(x)
0.15
0.1
0.05
0
-5
0
5
X=waiting time(min)
Figure 1. PDF for problem 4.8 a)
10
15
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B) Verify that
∫
∞
−∞
f ( y )dy = 1.
5
10
10
1
 1 y2 
2
1 y2 
1 1
2 1 
∫0 25 ydy + ∫5  5 − 25 y dy =  25 ⋅ 2  +  5 ⋅ y − 25 ⋅ 2  = 2 + 2 = 1
0
5
5
Since the answer is 1, this proves the definition found on page 140.
∞
∫ f (x )dx = area under the entire graph of f(x) = 1
−∞
C) What is the probability that total waiting time is at most 3 min?
0.25
0.2
0.15
f(x)
P( X ≤ 3) =
0.1
9
50
0.05
0
-5
0
5
10
15
X=waiting time(min)
Figure 2. PDF for problem 4.8 c)
Using the proposition on the bottom of page 141, will help you solve this problem.
3
 1 y2 
1
9
P( X ≤ 3) = ∫
ydy =  ⋅
 =
25
 25 2  0 50
0
3
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D) What is the probability that total waiting time is at most 8 min?
0.25
0.2
P( X ≤ 8) =
f(x)
0.15
23
25
0.1
0.05
0
-5
0
5
10
15
X=waiting time(min)
Figure 3. PDF for problem 4.8 d)
2 1
P( X ≤ 8) = ∫ ydy+ ∫  −
25
5 25
0
5 
4
1
8
 1 y2 


y dy =  ⋅



 25 2 
4
0
2
2
1 y
+ ⋅ y− ⋅
5
25 2





8
=
5
23
25
E) What is the probability that total waiting time is between 3 and 8 min?
0.25
0.2
P(3 ≤ X ≤ 8) =
37
50
f(x)
0.15
0.1
0.05
0
-5
0
5
X=waiting time(min)
Figure 4. PDF for problem 4.8 e)
10
15
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P(3 ≤ X ≤ 8) = P ( X ≤ 8) − P( X ≤ 3)
Taking the answers from (C) and (D), the answer would be 46/50 – 9/50 = 37/50.
You can also solve by:
5
8
1
37
2 1 
ydy + ∫  −
y dy =
25
5 25 
50
3
5
P(3 ≤ X ≤ 8) = ∫
F) What is the probability that total waiting time is either less than 2 min or more
than 6 min?
0.25
P( X < 2 or X > 6) =
0.2
2
5
f(x)
0.15
0.1
0.05
0
-5
0
5
X=waiting time(min)
Figure 5. PDF for problem 4.8 f)
2
10
1
2
2 1 
P( X < 2 or X > 6 = ∫
ydy + ∫  −
y dy =
25
5 25 
5
0
6
10
15