Question

# The mean and variance of a binomial distribution are $$12$$ and $$3$$ respectively. Find the probability distribution.

Solution

## Let the parameters of the distribution be $$n, p$$. Give the mean is $$12 \Rightarrow np=12$$ Variance is $$3\Rightarrow npq=3$$So, $$q=\dfrac{3}{12}=\dfrac{1}{4}$$$$P=1-q\Rightarrow P=1-\dfrac{1}{4}=\dfrac{3}{4}$$$$np=12\Rightarrow \dfrac{3}{4}n=12\Rightarrow n=16$$$$\therefore$$ The binomial distribution is ,$$P(x=r)=\, ^{16}C_r\left(\dfrac{3}{4}\right)^r \left(\dfrac{1}{4}\right)^{16-r}$$ for $$r=0,1,2,....16$$         $$=\, ^{16}C_r.\dfrac{3^r}{4^{16}}.$$ for $$r=0, 1,2,...,16$$Maths

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