Let X ~ Bin(9, 0.4). Find

a. P(X > 6)

b. P(X ≥ 2)

c. P(2 ≤ X < 5)

d. P(2 < X ≤ 5)

e. P(X = 0)

f. P(X = 7)

g. μx

h.

Answer :

Step 1 of 9:

Let X follows binomial distribution with probability mass function is

P(x: n,p) = (1 - p , x = 0, 1, 2, ….., n.

We have binomial distribution with n = 9 and p = 0.4

Step 2 of 9:

The claim is to find the P(x > 6) = P(x=7) + P(x =8) + P(x = 9)The probability mass function is

P(x: n,p) = (1 - p , x = 0, 1, 2, ….., n.

Where, n = 9, p = 0.4 and x = 7, 8, 9

Therefore, P(x > 6) = (1 - 0.4+ (1 - 0.4+

(1 - 0.4

= (36) (0.00163) (0.36) + (9) (0.00065536) (0.6) +

(1) (0.0002621) (1)

= 0.02112 + 0.003539 + 0.0002621

= 0.025

Hence, P(x >6) = 0.025

Step 3 of 9:

The claim is to find the P(x 2) = 1 - ( P( x = 0) + P( x = 1) )The probability mass function is

P(x: n,p) = (1 - p , x = 0, 1, 2, ….., n.

Where, n = 9, p = 0.4 and x = 0 and 1

Therefore, P(x 2) = 1 - [ (1 - 0.4+ (1 - 0.4]

= 1 - [(1) (1) (0.0101) + (9) (0.4) (0.0168) ]

= 1 - [ 0.0101 + 0.06048]

= 1 - (0.07058)

= 0.9294

Hence, P(x 2) = 0.9294