A random variable X has the distribution Binomial 12p Given

A random variable X has the distribution Binomial (12,p) Given that p=0.25, find P(X10) The mean and variance of X Given that P(X=0)=0.05, find the value of p to 3 decimal places. Given that the variance of X is 1.92, find the possible values of p.

Solution

Binomial Distribution

PMF of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
Where   
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial

1)
a)
P( X < 3) = P(X=2) + P(X=1) + P(X=0)
= ( 12 2 ) * 0.25^2 * ( 1- 0.25 ) ^10 + ( 12 1 ) * 0.25^1 * ( 1- 0.25 ) ^11 + ( 12 0 ) * 0.25^0 * ( 1- 0.25 ) ^12
= 0.3907

b)
P( X = 11 ) = ( 12 11 ) * ( 0.25^11) * ( 1 - 0.25 )^1
= 0.0000021
P( X = 12 ) = ( 12 12 ) * ( 0.25^12) * ( 1 - 0.25 )^0
= 0.0000001

P( X > 10) = P( X = 11 ) + P( X = 12 ) = 0.0000021 + 0.0000001 = 0.0000022

c)
Mean ( np ) =3
Standard Deviation ( npq )= 12*0.25*0.75 = 1.5

2)
P( X = 0 ) = ( 12 0 ) * ( P^0) * ( 1 - P )^12
0.05 = 1 * ( P^0) * ( 1 - P )^12
0.05 = ( 1 - P )^12
Q^12 = 0.05
Q = 0.77907
P = 1 - Q = 1 - 0.77907 = 0.22093