Question

Suppose that $p(x)=\frac{1}{5}, x=1,2,3,4,5$, zero elsewhere, is the pmf of the discrete-type random variable $X$. Compute $E(X)$ and $E\left(X^{2}\right)$. Use these two results to find $E\left[(X+2)^{2}\right]$ by writing $(X+2)^{2}=X^{2}+4 X+4$.

          Suppose that $p(x)=\frac{1}{5}, x=1,2,3,4,5$, zero elsewhere, is the pmf of the discrete-type random variable $X$. Compute $E(X)$ and $E\left(X^{2}\right)$. Use these two results to find $E\left[(X+2)^{2}\right]$ by writing $(X+2)^{2}=X^{2}+4 X+4$.

Added by Lorenzo B.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Ben Blakesley

06/15/2022

Step 1

Given that $p(x) = \frac{1}{5}$ for $x=1,2,3,4,5$, we have: $E(X) = 1 \cdot \frac{1}{5} + 2 \cdot \frac{1}{5} + 3 \cdot \frac{1}{5} + 4 \cdot \frac{1}{5} + 5 \cdot \frac{1}{5}$ $E(X) = \frac{1}{5} + \frac{2}{5} + \frac{3}{5} + \frac{4}{5} + \frac{5}{5}$ $E(X) = 3$ Show more…

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Suppose that $p(x)=\frac{1}{5}, x=1,2,3,4,5$, zero elsewhere, is the pmf of the discrete-type random variable $X$. Compute $E(X)$ and $E\left(X^{2}\right)$. Use these two results to find $E\left[(X+2)^{2}\right]$ by writing $(X+2)^{2}=X^{2}+4 X+4$.