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The mean and variance of a binomial distribution are 8 and 4 respectively. What is $P(X=1)$ equal to?
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Verified Answer
The correct answer is:
$\frac{1}{2^{12}}$
We know that mean and variance of Binomial distribution are $n p$ and $n p q$ respectively therefore $n p=$ 8 and $n p q=4$ On dividing we get
$q=\frac{n p q}{n p}=\frac{4}{8}=\frac{1}{2}$ and $p+q=1 \Rightarrow p=1-\frac{1}{2}=\frac{1}{2}$
$\Rightarrow n\left(\frac{1}{2}\right)=8 \Rightarrow n=16$
We know that $P(X=r)=n_{C_{r}} p^{r} q^{n-r}$
therefore
$P(X=1)={ }^{16} C_{1}\left(\frac{1}{2}\right)^{16-1}\left(\frac{1}{2}\right)^{1}$
$=\frac{16}{2^{15} \cdot 2}=\frac{1}{2^{12}}$
$q=\frac{n p q}{n p}=\frac{4}{8}=\frac{1}{2}$ and $p+q=1 \Rightarrow p=1-\frac{1}{2}=\frac{1}{2}$
$\Rightarrow n\left(\frac{1}{2}\right)=8 \Rightarrow n=16$
We know that $P(X=r)=n_{C_{r}} p^{r} q^{n-r}$
therefore
$P(X=1)={ }^{16} C_{1}\left(\frac{1}{2}\right)^{16-1}\left(\frac{1}{2}\right)^{1}$
$=\frac{16}{2^{15} \cdot 2}=\frac{1}{2^{12}}$
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