2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper (Held On 11-Jan-2019 Evening)

  • question_answer
    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 \[\left( \frac{Mean\,of\,X}{S\tan dard\,Deviation\,of\,X} \right)\] is equal to [JEE  Main Online Paper (Held on 11-jan-2019 Evening)]

    A) \[4\sqrt{3}\]                                          

    B) \[\frac{4\sqrt{3}}{3}\]

    C) \[3\sqrt{2}\]                              

    D)   4

    Correct Answer: A

    Solution :

    Here, p(Probability of getting white ball)\[=\frac{30}{40}=\frac{3}{4}\] \[\therefore \]\[q=\frac{1}{4}\]and\[n=16\] Now, mean\[=np=16.\frac{3}{4}=12\] Standard deviation \[=\sqrt{npq}=\sqrt{16.\frac{3}{4}.\frac{1}{4}}=\sqrt{3}\] \[\therefore \]\[\frac{Mean\,of\,X}{Standard\text{ }deviation\text{ }of}=\frac{12}{\sqrt{3}}=4\sqrt{3}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner