In a binomial distribution,the difference bet | vedclass

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(C) Let the mean be $\mu = np$ and standard deviation be $\sigma = \sqrt{npq}$,where $q = 1-p$.Given,$\mu - \sigma = 3 \Rightarrow \mu - 3 = \sigma$.S

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$1 : 3$

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(C) Let the mean be $\mu = np$ and standard deviation be $\sigma = \sqrt{npq}$,where $q = 1-p$.Given,$\mu - \sigma = 3 \Rightarrow \mu - 3 = \sigma$.Squaring both sides,$(\mu - 3)^2 = \sigma^2 = npq$.Also given,$\mu^2 - \sigma^2 = 21$.Substituting $\sigma^2 = \mu^2 - 21$ into the first equation:$(\mu - 3)^2 = \mu^2 - 21$$\mu^2 - 6\mu + 9 = \mu^2 - 21$$-6\mu = -30 \Rightarrow \mu = 5$.Since $\mu = np = 5$,then $\sigma^2 = 5^2 - 21 = 25 - 21 = 4$.We know $\sigma^2 = npq = 5q = 4$,so $q = \frac{4}{5}$ and $p = 1 - \frac{4}{5} = \frac{1}{5}$.Since $np = 5$,$n(\frac{1}{5}) = 5 \Rightarrow n = 25$.The probability mass function is $P(X=k) = {}^{n}C_k p^k q^{n-k}$.$\frac{P(X=1)}{P(X=2)} = \frac{{}^{25}C_1 p^1 q^{24}}{{}^{25}C_2 p^2 q^{23}} = \frac{25 \cdot q}{300 \cdot p} = \frac{1}{12} \cdot \frac{4/5}{1/5} = \frac{1}{12} \cdot 4 = \frac{1}{3}$.

In a binomial distribution,the difference between the mean and standard deviation is $3$ and the difference between their squares is $21$,then $P(x=1) : P(x=2) =$

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