If the sum and the product of the mean and va | vedclass

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(C) For a binomial distribution,the mean is $\mu = np$ and the variance is $\sigma^2 = npq$,where $p+q=1$.Given $np + npq = 24$ and $(np)(npq) = 128$.

Vedclass · Accepted Answer

$\frac{33}{2^{28}}$

Vedclass · Answer

(C) For a binomial distribution,the mean is $\mu = np$ and the variance is $\sigma^2 = npq$,where $p+q=1$.Given $np + npq = 24$ and $(np)(npq) = 128$.Let $A = np$ and $B = npq$. Then $A+B=24$ and $AB=128$.The quadratic equation $t^2 - 24t + 128 = 0$ has roots $t = 8$ and $t = 16$.Case $1$: $np = 16$ and $npq = 8$. Then $q = \frac{8}{16} = \frac{1}{2}$,so $p = \frac{1}{2}$. Thus $n(\frac{1}{2}) = 16 \implies n = 32$.Case $2$: $np = 8$ and $npq = 16$. Then $q = \frac{16}{8} = 2$,which is impossible since $q \leq 1$.So,$n=32, p=\frac{1}{2}, q=\frac{1}{2}$.The probability of one or two successes is $P(X=1) + P(X=2) = {}^{32}C_1 (\frac{1}{2})^1 (\frac{1}{2})^{31} + {}^{32}C_2 (\frac{1}{2})^2 (\frac{1}{2})^{30}$.$= 32 \cdot (\frac{1}{2})^{32} + \frac{32 \cdot 31}{2} \cdot (\frac{1}{2})^{32} = (32 + 496) \cdot \frac{1}{2^{32}} = 528 \cdot \frac{1}{2^{32}} = \frac{33 \cdot 16}{2^{32}} = \frac{33}{2^{28}}$.

If the sum and the product of the mean and variance of a binomial distribution are $24$ and $128$ respectively,then the probability of one or two successes is:

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