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Let $X$ be the number of heads out of 4 flips of a fair coin, let $Y_{i}$ be the $i$ th roll of a 6 -sided die. Find the mean and variance of:

(a) $X+Y_{1}$
(b) $Y_{1}-Y_{2}$
(c) $2 X+3 Y_{2}+1.52$
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We need the pmf of $X$ first.

\begin{array}{c|c}
x & f(x) \\
\hline 0 & \frac{1}{2^{4}}=\frac{1}{16} \\
1 & \left(\begin{array}{l}
4 \\
1
\end{array}\right) \frac{1}{16}=\frac{4}{16} \\
2 & \left(\begin{array}{l}
4 \\
2
\end{array}\right) \frac{1}{16}=\frac{6}{16} \\
3 & \left(\begin{array}{l}
4 \\
3
\end{array}\right) \frac{1}{16}=\frac{4}{16} \\
4 & \left(\begin{array}{c}
1 \\
16
\end{array}\right)
\end{array}

$E(X)=0\left(\frac{1}{16}\right)+1\left(\frac{4}{16}\right)+2\left(\frac{6}{16}\right)+3\left(\frac{4}{16}\right)+4\left(\frac{1}{16}\right)$

$=\dfrac{4+12+12+4}{16}=2$

$E\left(X^{2}\right)=0^{2}\left(\frac{1}{16}\right)+1^{2}\left(\frac{4}{16}\right)+2^{2}\left(\frac{6}{16}\right)+3^{2}\left(\frac{4}{16}\right)+4^{2}\left(\frac{1}{16}\right)$

$=\dfrac{4+24+36+16}{16}=5$

$\operatorname{Var}(X)=E\left(X^{2}\right)-E(X)^{2}=5-2^{2}=1$

$E\left(Y_{i}\right)=1\left(\frac{1}{6}\right)+2\left(\frac{1}{6}\right)+3\left(\frac{1}{6}\right)+4\left(\frac{1}{6}\right)+5\left(\frac{1}{6}\right)+6\left(\frac{1}{6}\right)=\frac{21}{6}=3.5$

$E\left(Y_{i}^{2}\right) 1^{2}\left(\frac{1}{6}\right)+2^{2}\left(\frac{1}{6}\right)+3^{2}\left(\frac{1}{6}\right)+4^{2}\left(\frac{1}{6}\right)+5^{2}\left(\frac{1}{6}\right)+6^{2}\left(\frac{1}{6}\right)$

$=\dfrac{1+4+9+16+25+36}{6}$

$=\frac{91}{6}$

$\operatorname{Var}\left(Y_{i}\right)=E\left(Y_{i}^{2}\right)-E(Y)^{2}=\frac{91}{6}-\frac{21^{2}}{6^{2}}=\frac{108}{36}=3$

(a) $E\left(X+Y_{1}\right)=E(X)+E\left(Y_{1}\right)=2+3.5=5.5$,
$\operatorname{Var}\left(X+Y_{1}\right)=\operatorname{Var}(X)+\operatorname{Var}\left(Y_{1}\right)=5+3=8$
(b) $E\left(Y_{1}-Y_{2}\right)=E\left(Y_{1}\right)-E\left(Y_{2}\right)=3.5-3.5=0$,
$\operatorname{Var}\left(Y_{1}-Y_{2}\right)=\operatorname{Var}\left(Y_{1}\right)+(-1)^{2} \operatorname{Var}\left(Y_{2}\right)=3+3=6$

(c) $E\left(2 X+3 Y_{2}+1.52\right)=2 E(X)+3 E\left(Y_{2}\right)+1.52$

$=2(2)+3(3.5)+1.52=16.02$

$\operatorname{Var}\left(2 X+3 Y_{2}+1.52\right)=2^{2} \operatorname{Var}(X)+3^{2} \operatorname{Var}(Y)=4(1)+9(3) 4+18=22$
by Diamond (40,719 points)
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