Because has the properties of a distribution function, its

Chapter 4, Problem 193SE

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QUESTION:

Because

                  \(P(Y \leq y \mid Y \geq c)=\frac{F(y)-F(c)}{1-F(c)}\)

has the properties of a distribution function, its derivative will have the properties of a probability

density function. This derivative is given by

                  \(\frac{f(y)}{1-F(c)}, \quad y \geq c.\)

We can thus find the expected value of 𝑌, given that 𝑌 is greater than 𝑐, by using

                  \(E(Y\mid Y\ge c)=\frac{1}{1-F(c)}\int_c^{\infty}yf\ (y)\ dy.\)

If 𝑌, the length of life of an electronic component, has an exponential distribution with mean

100 hours, find the expected value of 𝑌, given that this component already has been in use for

50 hours.

Equation Transcription:

Text Transcription:

P(Y</=y|Y>/=c)=F(y)-F(c) over 1-F(c)

f(y) over 1-F(c), y>/=c.

E(Y|Y>/=c)=1 over 1-F(c) integral c to infinity yf(y)dy.

Questions & Answers

QUESTION:

Because

                  \(P(Y \leq y \mid Y \geq c)=\frac{F(y)-F(c)}{1-F(c)}\)

has the properties of a distribution function, its derivative will have the properties of a probability

density function. This derivative is given by

                  \(\frac{f(y)}{1-F(c)}, \quad y \geq c.\)

We can thus find the expected value of 𝑌, given that 𝑌 is greater than 𝑐, by using

                  \(E(Y\mid Y\ge c)=\frac{1}{1-F(c)}\int_c^{\infty}yf\ (y)\ dy.\)

If 𝑌, the length of life of an electronic component, has an exponential distribution with mean

100 hours, find the expected value of 𝑌, given that this component already has been in use for

50 hours.

Equation Transcription:

Text Transcription:

P(Y</=y|Y>/=c)=F(y)-F(c) over 1-F(c)

f(y) over 1-F(c), y>/=c.

E(Y|Y>/=c)=1 over 1-F(c) integral c to infinity yf(y)dy.

ANSWER:

Solution:

Step 1 of 2:

We have

E(Y/ Yc) = j f(y) dy

An Exponential distribution with mean 100 hours

The claim is to find the value of Y, given that the component already has in use for 50 hours.


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