Question

Let X represent a binomial random variable with n = 110 and p = 0.19. Find the following probabilities. (Do not round intermediate calculations. Round your final answers to 4 decimal places.)

a.	P(X ≤ 20)
b.	P(X = 10)
c.	P(X > 30)
d.	P(X ≥ 25)

Answer 1

Mean = n * P = 110 * 0.19 = 20.9

Standard deviation =

Using Normal approximation to Binomial

Part a)

P ( X <= 20 )

Using continuity correction

P ( X < 20 + 0.5 ) = P ( X < 20.5 )

P ( X < 20.5 )
Standardizing the value

Z = ( 20.5 - 20.9 ) / 4.1145
Z = -0.1

P ( X < 20.5 ) = P ( Z < -0.1 )
P ( X < 20.5 ) = 0.4602

Part b)

P ( X = 10 )

Using continuity correction

P ( 10 - 0.5 < X < 10 + 0.5 ) = P ( 9.5 < X < 10.5 )

P ( 9.5 < X < 10.5 )
Standardizing the value

Z = ( 9.5 - 20.9 ) / 4.1145
Z = -2.77
Z = ( 10.5 - 20.9 ) / 4.1145
Z = -2.53
P ( -2.77 < Z < -2.53 )
P ( 9.5 < X < 10.5 ) = P ( Z < -2.53 ) - P ( Z < -2.77 )
P ( 9.5 < X < 10.5 ) = 0.0057 - 0.0028
P ( 9.5 < X < 10.5 ) = 0.0029

Part c)

P ( X > 30 )

Using continuity correction

P ( X > 30 + 0.5 ) = P ( X > 30.5 )

P ( X > 30.5 ) = 1 - P ( X < 30.5 )
Standardizing the value

Z = ( 30.5 - 20.9 ) / 4.1145
Z = 2.33

P ( Z > 2.33 )
P ( X > 30.5 ) = 1 - P ( Z < 2.33 )
P ( X > 30.5 ) = 1 - 0.9901
P ( X > 30.5 ) = 0.0099

Part d)

P ( X >= 25 )

Using continuity correction

P ( X > 25 - 0.5 ) = P ( X > 25.5 )

P ( X > 25.5 ) = 1 - P ( X < 25.5 )
Standardizing the value

Z = ( 25.5 - 20.9 ) / 4.1145
Z = 1.12

P ( Z > 1.12 )
P ( X > 25.5 ) = 1 - P ( Z < 1.12 )
P ( X > 25.5 ) = 1 - 0.8686
P ( X > 25.5 ) = 0.1314