In a Binomial distribution, if ' n ' is the number of trials and the mean and variance are 4 and 3 respectively, then 2³² p (X= n 2 )=

Step-by-step solution

Let X be the binomial variate for which mean =4 and variance =3 , then n p=4 and n p q=3 q= 3 4 P=(1-q)= (1- 3 4 )= 1 4 and n p=4 n= 4 1 4=16 Thus, n=16, p= 1 4 and q= 3 4 Hence, the binomial distribution 2³² P (X= n 2 )=2³² 16 16 2 ( 1 4 )^ 16 2 ( 3 4 )^ 16- 16 2 =2³² ¹⁶ C₈ ( 1 4 )^8 ( 3 4 )^8=2³² ¹⁶ C₈ (3)^8 (4)¹⁶ = ¹⁶ C₈(3)^8 [ (4)¹⁶= (2)³² ] .