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The mean and standard deviation of a binomial variate x are 4 and $\sqrt 3$ respectively. Then $P( X \geq 1)$ is equal to

\[\begin {array} {1 1} (a)\;1-\bigg(\frac{1}{4}\bigg)^{16} & \quad (b)\;1-\bigg(\frac{3}{4}\bigg)^{16} \\ (c)\;1-\bigg(\frac{2}{3}\bigg)^{16} & \quad (d)\;1-\bigg(\frac{1}{3}\bigg)^{16} \end {array}\]

1 Answer

$(b)\;1-\bigg(\frac{3}{4}\bigg)^{16}$
answered Nov 7, 2013 by pady_1
 
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