 Hello and welcome to the session. In this session we discussed the following question which says what is the binomial distribution with mean 16 and standard deviation 2 root 3. Let's recall the formula to find out the mean and the standard deviation. We have mean is equal to NP standard deviation is equal to square root of NPQ where we have N is the number of trials, P is the probability of getting success and Q is the probability of getting failure. This is the key idea that we use for this question. Let's proceed with the solution now. We take let X be a binomial variant for which mean is equal to 16 and standard deviation is equal to 2 root 3. Now here since we have mean equal to 16 so this means we have NP equal to 16 since we know that mean is equal to NP. So let this be equation 1 then we have standard deviation equal to 2 root 3. So this means square root of NPQ equal to 2 root 3 since we know standard deviation is equal to square root NPQ. Now square in both sides where we get NPQ is equal to 12. Let this be equation 2. Now using equation 1 in equation 2 we get 16 into Q equal to 12 that is in equation 2 in place of NP we put 16 so we get 16 into Q equal to 12. This means Q is equal to 12 upon 16. Now 3 times is 12 and 4 times is 16 so we get Q equal to 3 upon 4 as we know P plus Q is equal to 1. So from here we have P would be equal to 1 minus Q that means P is equal to 1 minus 3 upon 4 which gives us P equal to 1 upon 4. So we have got the values for Q and P. Now substituting the value of P in equation 1 we get N into 1 upon 4 is equal to 16 since we have equation 1 as NP equal to 16. So in place of P we put 1 upon 4 that becomes N into 1 upon 4 equal to 16 which gives us N equal to 64. The binomial distribution is given by probability of the random variate or the binomial variate X equal to R this equal to NCR into P to the power R into Q to the power N minus R where we have R is equal to 0, 1, 2 and so on up to N. Now putting the values for N, P and Q in this binomial distribution we get the binomial distribution as probability of the binomial variate X equal to R equal to 64 CR into 1 upon 4 whole to the power R into 3 upon 4 whole to the power 64 minus R where this R is equal to 0, 1, 2 and so on up to 64. So this is required by the binomial distribution. This completes the session. Hope you have understood the solution of this question.