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, the (( textmean textof X/ textstandard textdeviation textof X )) 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, the (( textmean textof X/ textstandard textdeviation textof X )) is equal to : (A) 4 (B) (4 √3/3) (C)  4 √3  (D)  3 √2

Tardigrade Admin · Accepted Answer

p (probability of getting white ball) = (30/40) q = (1/4) and n = 16 mean =np = 16 . (3/4) = 12 and standard diviation = √npq = √ 16 . ( 3/4) . (1/4) = √3

Q. 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, the $(\frac{mean of X}{standard deviation of X})$ is equal to :

$4$

$\frac{4 3}{3}$

$43$

$32$

p (probability of getting white ball) = $\frac{30}{40}$
$q = \frac{1}{4}$ and n = 16
mean =np = $16. \frac{3}{4} = 12$
and standard diviation
$= n pq = 16. \frac{3}{4} . \frac{1}{4} = 3$

Q. 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, the (standarddeviationofXmeanofX​) is equal to :

4

343​​

43​

32​

p (probability of getting white ball) = 4030​ q=41​ and n = 16 mean =np = 16.43​=12 and standard diviation =npq​=16.43​.41​​=3​

Q. 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, the $(\frac{mean of X}{standard deviation of X})$ is equal to :

$4$

$\frac{4 3}{3}$

$43$

$32$

p (probability of getting white ball) = $\frac{30}{40}$
$q = \frac{1}{4}$ and n = 16
mean =np = $16. \frac{3}{4} = 12$
and standard diviation
$= n pq = 16. \frac{3}{4} . \frac{1}{4} = 3$