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The mean of a binomial distribution is $6$ and its standard deviation is $3$. Is this statement true or false?comment?

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Toolbox:
  • A random variable $X$ is said to follow a binomial distribution of its probability mass function is given by
  • $P(X=x)=p(x)=\left\{\begin{array}{1 1}nC_xP^xq^{n-x},&x=0,1.......n\\0,&otherwise\end{array}\right.$
  • Constants of Binomial Distribution :
  • Mean = np
  • Variance = npq
  • Standard deviation =$\sqrt{ variance }=\sqrt{ npq}$
  • In a bionomial distribution Mean > Variance.
  • The parameters of the distribution are $n,p\quad X\sim B(n,p)$
Step 1:
It is stated that the mean=6 and the standard deviation =3
$\therefore np=6$
$\sqrt{npq}=3$
$npq=9$
Step 2:
From the above $\large\frac{npq}{np}=\frac{9}{6}=\frac{3}{2}$
$\Rightarrow q=\large\frac{3}{2}$ which is not possible
Since $0 < q < 1$
$\therefore$ the statement is false.
answered Sep 18, 2013 by sreemathi.v
 

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