Question

The random variable X has distribution: $f(x) = (\frac{1}{18})x^2$ $-3 < x < 3$. Consider a new random variable that is a function of X as shown below: where $Y = \begin{cases} X^{0.5} & X > 0 \\ 0 & \text{elsewhere} \end{cases}$ a. Find the expected value of Y. b. Find the variance of Y. c. Find the median of Y.

          The random variable X has distribution: $f(x) = (\frac{1}{18})x^2$  $-3 < x < 3$. Consider a new random variable that is a function of X as shown below:
where $Y = \begin{cases} X^{0.5} & X > 0 \\ 0 & \text{elsewhere} \end{cases}$
a. Find the expected value of Y.
b. Find the variance of Y.
c. Find the median of Y.

The random variable X has distribution: f(x) = ((1)/(18))x^2 -3 < x < 3. Consider a new random variable that is a function of X as shown below:
where Y = X^0.5 X > 0
0 elsewhere
a. Find the expected value of Y.
b. Find the variance of Y.
c. Find the median of Y.

Added by Shelly H.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Sri K

06/27/2022

Step 1

To find the expected value of Y, we need to integrate the product of Y and its probability density function (pdf) f(x) over the entire domain of x. Since Y is defined as x^(0.5) for x > 0 and 0 elsewhere, we can write the expected value as: E(Y) = ∫[Y * f(x)] dx Show more…

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The random variable x has distribution: xY={(x^(0.5),x>0),(0 elsewhere ):}Yf(x)=((1)/(18))x^(2),-3. Consider a new random variable that is a function of x as shown below: where Y={(x^(0.5),x>0),(0 elsewhere ):} a. Find the expected value of Y. b. Find the variance of Y. c. Find the median of Y. The random variable X has distribution: f(x) 2--3<x<3. Consider a new random variable that is a function of X as shown below: JX0.5 X >0 where Y= 0 elsewhere a. Find the expected value of Y. b. Find the variance of Y. c. Find the median of Y.