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Entered 0.451188 (1-e^(-0.6))/(0.25) (y-3) Answer Preview Result Message 1-e^{-0.6} incorrect This answer is equivalent to the one you just submitted. incorrect Variable 'y' is not defined in this context At least one of the answers above is NOT correct. Y is an exponential random variable with parameter \(\lambda = 0.2\). Define the event \(A = \{Y < 3\}\). (a) Find the conditional PDF. \(f_{Y|A}(y) = \begin{cases} 1 - e^{-0.6} & 0 \leq y \leq 3, \ 0 & \text{otherwise} \end{cases}\) (b) Calculate the conditional expected value. \(E[Y|A] = \frac{(1 - e^{-0.6})}{0.25} (y - 3)\)

          Entered
0.451188
(1-e^(-0.6))/(0.25) (y-3)
Answer Preview
Result
Message
1-e^{-0.6}
incorrect
This answer is equivalent to the one you just submitted.
incorrect
Variable 'y' is not defined in this context
At least one of the answers above is NOT correct.
Y is an exponential random variable with parameter \(\lambda = 0.2\). Define the event \(A = \{Y < 3\}\).
(a) Find the conditional PDF.
\(f_{Y|A}(y) = \begin{cases} 1 - e^{-0.6} & 0 \leq y \leq 3, \
0 & \text{otherwise} \end{cases}\)
(b) Calculate the conditional expected value.
\(E[Y|A] = \frac{(1 - e^{-0.6})}{0.25} (y - 3)\)

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$Entered 0.451188 (1-e^(-0.6))/(0.25) (y-3) Answer Preview Result Message 1-e^-0.6 incorrect This answer is equivalent to the one you just submitted. incorrect Variable 'y' is not defined in this context At least one of the answers above is NOT correct. Y is an exponential random variable with parameter λ = 0.2. Define the event A = {Y < 3}. (a) Find the conditional PDF. fY|A(y) = 1 - e^-0.6 0 ≤ y ≤ 3, 0 otherwise (b) Calculate the conditional expected value. E[Y|A] = ((1 - e^-0.6))/(0.25) (y - 3)$

Added by Joseph V.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Maitreya E

06/14/2022

Step 1

2 and event A={Y<3}, we know that the conditional PDF is: f(Y|A) = f(Y and A) / P(A) Since Y is an exponential random variable with parameter λ=0.2, the PDF of Y is f(Y) = 0.2 * e^(-0.2y) for y >= 0. To find the joint PDF f(Y and A), we need to find the Show more…

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Y is an exponential random variable with parameter lambda =0.2. Define the event A={Y<3}. (a) Find the conditional PDF. f_(Y|A)(y)={(,0<=y<=3),(0 otherwise ):} (b) Calculate the conditional expected value. E[Y|A]= Entered Answer Preview Result Message 0.451188 1-e0.6 incorrect This answer is equivalent to the one you just submitted 1-e^-0.6))/0.25(y-3) incorrect Variable'y'is not defined in this context At least one of the answers above is NOT correct Y is an exponential random variable with parameter X=0.2.Define the event A={Y<3} aFind the conditional PDF. s>f>0 fY|Ay) otherwise (b)Calculate the conditional expected value E[Y|A]= 1-e-0.6) y-3 0.25

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Transcript

00:01 Hello everyone.

00:02 You are given f of x of x is equal to c x to the part minus 2 in the interval 1 .2 and 0 otherwise.

00:11 First thing is we need to find c.

00:13 And the second thing is if a is even such that x is greater than 1 .5, we need to calculate the probability of a and the conditional probability density function of x given that a is a curve.

00:26 So first of all, let's calculate the constant.

00:28 Since the given pdf, is continuous, we have the formula here to be, we know that when we add all the probability is going to be equal to 1.

00:40 Therefore, we have from 1 to 2, c x to the power minus 2, dx will be equal to 1.

00:48 So let's calculate this integral.

00:49 This is going to be c, x to the power minus 1.

00:57 From 1 to 2 is equal to 1.

01:00 So when i substitute the limits, i'm going to get the value here as, c minus 1 divided by 2 plus 1 is equal to 1 or in other words c will be equal to 2.

01:14 So now we have calculated the constant c.

01:19 Next we are asked to find p of a.

01:22 So in order to find p of a, we are given that it is an event such that x is greater than 1 .5.

01:29 So what happens is p of a is going to be not from 1 .1, it is going to be, not from 1, it is going to be, from 1 .5 to 2 only because we have said that up to 2 only it is defined and 0 otherwise.

01:44 So we have to end it with 2.

01:45 And the function, what is the function? the constant c x to the power minus 2 d x...

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