A bag contains 30 white balls and 10 red balls. 16 balls are drawn one by one randomly from the bag with replacement. If X be the number of white balls drawn, then \left( {{{mean\,\,of\,X} \over {s an dard\,\,deviation\,\,of\,X}}} ight) is equal to :

Question

A bag contains 30 white balls and 10 red balls. 16 balls are drawn one by one randomly from the bag with replacement. If X be the number of white balls drawn, then \left( {{{mean\,\,of\,X} \over {s	an dard\,\,deviation\,\,of\,X}}} ight) is equal to :

Accepted Answer

Answer: D. 4\sqrt 3 p (probability of getting white ball) = {{30} \over {40}} q = {1 \over 4} and n = 16 mean = np = 16.{3 \over 4} = 12 and standard diviation = \sqrt {npq} = \sqrt {16.{3 \over 4}.{1 \over 4}} = \sqrt 3 \because mean \over standard\,deviation = 12\over{\sqrt3} = 4{\sqrt3}

A bag contains 30 white balls and 10 red balls. 16 balls are drawn one by one randomly from the bag with replacement. If X be the number of white balls drawn, then $$\left( {{{mean\,\,of\,X} \over {s\tan dard\,\,deviation\,\,of\,X}}} \right)$$ is equal to :

Answer

Explanation

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