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If the sum of the mean and the variance of a Binomial distribution for 5 trials is 1.8 , then the value of $\mathrm{p}$ is
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Verified Answer
The correct answer is:
0.2
According to the given condition,
mean + variance $=1.8$
$\begin{aligned} & \Rightarrow n p+n p q=1.8 \\ & \Rightarrow 5 p+5 p q=1.8 \\ & \Rightarrow 5 p+5 p(1-p)=1.8 \\ & \Rightarrow 5 p+5 p-5 p^2=1.8 \\ & \Rightarrow 5 p^2-10 p+1.8=0 \\ & \Rightarrow 50 p^2-100 p+18=0 \\ & \Rightarrow(10 p-2)(5 p-9)=0 \\ & \Rightarrow p=\frac{2}{10}=0.2 \text { or } p=\frac{9}{5}=1.8\end{aligned}$
Since $0 < \mathrm{p} < 1$,
$\mathrm{p}=0.2$
mean + variance $=1.8$
$\begin{aligned} & \Rightarrow n p+n p q=1.8 \\ & \Rightarrow 5 p+5 p q=1.8 \\ & \Rightarrow 5 p+5 p(1-p)=1.8 \\ & \Rightarrow 5 p+5 p-5 p^2=1.8 \\ & \Rightarrow 5 p^2-10 p+1.8=0 \\ & \Rightarrow 50 p^2-100 p+18=0 \\ & \Rightarrow(10 p-2)(5 p-9)=0 \\ & \Rightarrow p=\frac{2}{10}=0.2 \text { or } p=\frac{9}{5}=1.8\end{aligned}$
Since $0 < \mathrm{p} < 1$,
$\mathrm{p}=0.2$
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