Let X refers the time to failure of a machine component. Suppose the pdf of X is f(x)=32/(x+4)^3 for x>0. The cdf is: F(x)=[6" 32 -dx (x+4)³ =6* 32 (x + 4)-3 dx (x+4)* )-3+1 =32
Let X refers the time to failure of a machine component. Suppose the pdf of X is f(x)=32/(x+4)^3 for x>0. The cdf is: F(x)=[6" 32 -dx (x+4)³ =6* 32 (x + 4)-3 dx (x+4)* )-3+1 =32
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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Could you solve the part d, please?
Transcribed Image Text:Let X refers the time to failure of a machine component. Suppose the pdf of X is
f(x)=32/(x+4)^3 for x>0.
The cdf is:
F(x)=[6°
dx
32
3
(x+4)³
=6 32 (x + 4)-3 dx
-3+1 jx
(x+4)
=32
-3+1
(x+4)-2
=32[
-2
1
=32
-2(x+4)²
=32 -2)(x+4)*
(-2)(0+4)?
1
=32
2(x+4)²
32
16
+ 1
(x+4)?
c. Use the previous result (part b) and calculate the probability that time to failure is
between 6 and 4 unit of time.
The CDF provide the probability of X <x. So to compute the probability between 4 and 6, the
probability of X <4 can be subtracted from probability of X < 6.
P(4<X<6)=P(X < 6) – P (X < 4)
=F(6) – F(4)
16
16
+
+
(6+4)²
16
+1
100
+ * - 1
=-0. 16 + 0. 25
=0. 09
d. What is the expected time until failure?
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