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(D) For a Binomial distribution with $n$ trials,the mean is $\mu = np$ and the variance is $\sigma^2 = npq$,where $q = 1 - p$.Given $n = 5$ and $\mu +

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$0.2$

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(D) For a Binomial distribution with $n$ trials,the mean is $\mu = np$ and the variance is $\sigma^2 = npq$,where $q = 1 - p$.Given $n = 5$ and $\mu + \sigma^2 = 1.8$.Substituting the values,we get:$np + npq = 1.8$$5p + 5p(1 - p) = 1.8$$5p + 5p - 5p^2 = 1.8$$10p - 5p^2 = 1.8$$5p^2 - 10p + 1.8 = 0$Multiplying by $10$ to clear the decimal:$50p^2 - 100p + 18 = 0$Dividing by $2$:$25p^2 - 50p + 9 = 0$Using the quadratic formula $p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$p = \frac{50 \pm \sqrt{2500 - 4(25)(9)}}{50} = \frac{50 \pm \sqrt{2500 - 900}}{50} = \frac{50 \pm \sqrt{1600}}{50} = \frac{50 \pm 40}{50}$$p_1 = \frac{90}{50} = 1.8$ (Not possible since $0 \le p \le 1$)$p_2 = \frac{10}{50} = 0.2$Thus,$p = 0.2$.

If the sum of the mean and the variance of a Binomial distribution for $5$ trials is $1.8$,then the value of $p$ is

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