If the sum of mean and variance of a Binomial Distribution is $\frac{15}{2}$ for 10 trials, then the…

If the sum of mean and variance of a Binomial Distribution is $\frac{15}{2}$ for 10 trials, then the variance is
  1. $1.5$
  2. $2.5$
  3. $4.5$
  4. $3.5$

Solution

$\begin{array}{ll} & \text { Mean }+ \text { variance }=\frac{15}{2} \\ & \Rightarrow n p+n p q=\frac{15}{2} \\ & \Rightarrow n p+n p(1-p)=\frac{15}{2} \quad \ldots[\because p+q=1] \\ & \Rightarrow n\left(2 p-p^2\right)=\frac{15}{2} \\ & \Rightarrow 2 p-p^2=\frac{15}{2 \times 10} \quad \ldots[n=10] \\ & \Rightarrow 4 p^2-8 p+3=0 \\ & \Rightarrow(2 p-3)(2 p-1)=0 \\ & \Rightarrow p=\frac{3}{2} \text { or } p=\frac{1}{2} \\ & p=\frac{1}{2}...[0 < p < 1] \\ & q=1-\frac{1}{2}=\frac{1}{2} \\ & \text { Variance }=n p q=10 \times \frac{1}{2} \times \frac{1}{2}=2.5\end{array}$

Asked in: MHT CET 2023 (10 May Shift 2)

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