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{\bf Question}

Suppose that $F$ is a continuous cdf on the real line; and let
$\alpha_1$ and $\alpha_2$ be numbers such that $F(\alpha_1)=0.3$
and $F(\alpha_2)=0.8$.  If 25 observations are selected at random
from the distribution for which the cdf is $F$, what is the
probability that 6 of the observed values will be less than
$\alpha_1$, 10 of the observed values will be between $\alpha_1$
and $\alpha_2$, and 9 of the observed values will be greater than
$\alpha_2$?



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{\bf Answer}

\setlength{\unitlength}{.5in}

\begin{picture}(5,2)
\put(0,1){\line(1,0){5}} \put(2,.9){\line(0,1){.2}}
\put(4,.9){\line(0,1){.2}} \put(1,1.3){6} \put(3,1.3){10}
\put(4.5,1.3){9} \put(1.9,0.5){$\alpha_1$}
\put(3.9,0.5){$\alpha_2$}


\end{picture}

$F(\alpha_1)=0.3,\ \ \ F(\alpha_2)=0.8$
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\begin{tabular}{|c|c|l|}
\hline {Group 1 $\leq \alpha_1$} & {$p_1=0.3$} & {$x_1=6$}\\
\hline Group 2 $(\alpha_1,\alpha_2)$ & $p_2 = 0.8-0.3 = 0.5$ &
$x_2=10$\\ \hline Group 3 $(\alpha_3,\infty)$ & $p_3 = 1-0.8=0.2$
& $x_2=9$\\ \hline
\end{tabular}

Therefore required probability is: $$
\frac{n!}{x_1!x_2!x_3!}p_1^{x_1}p_2^{x_2}p_3^{x_3} =
\frac{25!}{6!10!9!}(0.3)^6(0.5)^10(0.2)^9$$



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