If X follow a binomial distribution with mean 4 and variance 2 find `P(X ge 5)`

Explanation

We know that mean = np and variance = npq
`:.` np=4 and npq =2
Now up =4 and npq = 2 `rArr 4q =2 rArr q =(1)/(2)`
`:. P=(1-q) =(1-(1)/(2)) =(1)/(2)`
Now np =4 and p`=(1)/(2) rArr (1)/(2) n= 4 rArr n=8`
So the binomial distributions is given by
`P(X =r)=.^(n)C_(r ).p^(r).q^((n-r)) =.^(8)C_(r ).((1)/(2))^(r) .((1)/(2))^((8-r)) =.^(8)C_( r) .((1)/(2))^(8)`
`:. P(Xge5) =P(X=5)+P(X=6)+P(X=7)+P(X=8)`
`=.^(8)C_( 5).((1)/(2))^(8)+.^(8)C_(6).((1)/(2))^(8)+.^(8)C_(7).((1)/(2))^(8)+.^(8)C_(8).((1)/(2))^(8)`
`=[.^(8)C_(3)+.^(8)C_(2)+.^(8)C_(2)+.^(8)C_(1)+1].((1)/(2))^(8)`
` =(56 +28 +8 +1).(1)/(256) =(93)/(256)`