If the sum of mean and variance of a binomial | vedclass

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(A) For a binomial distribution,the mean is $\mu = np$ and the variance is $\sigma^2 = npq$,where $q = 1 - p$.Given $n = 5$ and $np + npq = 1.8$.Subst

Vedclass · Accepted Answer

$0.2$

Vedclass · Answer

(A) For a binomial distribution,the mean is $\mu = np$ and the variance is $\sigma^2 = npq$,where $q = 1 - p$.Given $n = 5$ and $np + npq = 1.8$.Substituting $n = 5$ and $q = 1 - p$:$5p + 5p(1 - p) = 1.8$$5p + 5p - 5p^2 = 1.8$$10p - 5p^2 = 1.8$$5p^2 - 10p + 1.8 = 0$Multiply by $5$ to clear decimals: $25p^2 - 50p + 9 = 0$.Using the quadratic formula $p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$p = \frac{50 \pm \sqrt{2500 - 900}}{50} = \frac{50 \pm \sqrt{1600}}{50} = \frac{50 \pm 40}{50}$.$p = \frac{90}{50} = 1.8$ (not possible as $p \leq 1$) or $p = \frac{10}{50} = 0.2$.Thus,the probability of success is $0.2$.

If the sum of mean and variance of a binomial distribution for $5$ trials is $1.8$,then the probability of success is:

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