Q3
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Probability
└── Homework
└── W8
└── Q3.tex
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\fancyhead[l]{Li Yifeng}
\fancyhead[c]{Homework \#8}
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\begin{document}
\section*{Question 3}
\noindent In the NBA $3$-point contest the player shoots $20$ shots: $16$ normal balls ($1$ pt each), and $4$ money balls ($2$ pts each). Assume shot success probability is $0.43$.
\bigskip
\begin{enumerate}[start=1,label={\bfseries Part \arabic*:},leftmargin=0in]
\bigskip\item What is the expected value of points scored?
\subsection*{Solution}
Let $X_1\sim\mathrm{Binomial}(16,0.43)$ be hits on normal balls and $X_2\sim\mathrm{Binomial}(4,0.43)$ hits on money balls. Total points $T = X_1 + 2X_2$, so
\[
\bE[T] = 16(0.43) + 2\cdot4(0.43) = 6.88 + 3.44 = 10.32
\]
\subsection*{Answer}
\[\boxed{\bE[T] = 10.32}\]
\bigskip\item What is the standard deviation of the total points scored?
\subsection*{Solution}
We have
\[
\begin{aligned}
\mathrm{Var}(X_1)&=16\cdot0.43\cdot0.57\\
\mathrm{Var}(X_2)&=4\cdot0.43\cdot0.57
\end{aligned}
\]
and since $\mathrm{Var}(2X_2)=4\,\mathrm{Var}(X_2)$ and $X_1, X_2$ independent,
\[
\begin{aligned}
\mathrm{Var}(T)
&= 16\cdot0.43\cdot0.57 + 4\times(4\cdot0.43\cdot0.57)\\
&= 2\times16\cdot0.43\cdot0.57\\
&= 32\cdot0.43\cdot0.57\\
&\approx7.843
\end{aligned}
\]
so
\[
\sigma_T = \sqrt{32\cdot0.43\cdot0.57} \approx 2.80
\]
\subsection*{Answer}
\[\boxed{\sigma_T = \sqrt{32\cdot0.43\cdot0.57}\approx2.80}\]
\end{enumerate}
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