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If the mean and variance of a binomial variate $X$ are 8 and 4 respectively, then $P(X < 3)$ equals to
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Verified Answer
The correct answer is:
$\frac{137}{2^{16}}$
Given, mean of binomial variable, $n p=8$ and variance of binomial variable, $n p q=4$ $\therefore \quad q=\frac{1}{2}$
and
$$
\begin{aligned}
p & =1-q \\
& =1-\frac{1}{2}=\frac{1}{2}
\end{aligned}
$$
and $\quad n\left(\frac{1}{2}\right)=8 \Rightarrow n=16$
$$
\begin{aligned}
& \therefore P(X < 3)=P(X=0)+P(X=1)+P(X=2) \\
& ={ }^{16} C_0\left(\frac{1}{2}\right)^0\left(\frac{1}{2}\right)^{16-0}+{ }^{16} C_1\left(\frac{1}{2}\right)^1\left(\frac{1}{2}\right)^{16-1} \\
& +{ }^{16} C_2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^{16-2} \\
& =1\left(\frac{1}{2}\right)^{16}+16\left(\frac{1}{2}\right)^{16}+120\left(\frac{1}{2}\right)^{16}=\frac{137}{2^{16}} \\
&
\end{aligned}
$$
and
$$
\begin{aligned}
p & =1-q \\
& =1-\frac{1}{2}=\frac{1}{2}
\end{aligned}
$$
and $\quad n\left(\frac{1}{2}\right)=8 \Rightarrow n=16$
$$
\begin{aligned}
& \therefore P(X < 3)=P(X=0)+P(X=1)+P(X=2) \\
& ={ }^{16} C_0\left(\frac{1}{2}\right)^0\left(\frac{1}{2}\right)^{16-0}+{ }^{16} C_1\left(\frac{1}{2}\right)^1\left(\frac{1}{2}\right)^{16-1} \\
& +{ }^{16} C_2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^{16-2} \\
& =1\left(\frac{1}{2}\right)^{16}+16\left(\frac{1}{2}\right)^{16}+120\left(\frac{1}{2}\right)^{16}=\frac{137}{2^{16}} \\
&
\end{aligned}
$$
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