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

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.6: The Inverse Trigonometric Functions
Problem 91E
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Could you solve the part d, please?

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?
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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