Question

Let X be a binomial random variable with n = 11 and p = 0.3. Find the following values. (Round your answers to three decimal places.)

(a)    

P(X = 5)

(b)    

P(X ≥ 5)

(c)    

P(X > 5)

(d)    

P(X ≤ 5)

(e)    

μ = np

μ =

(f)    σ =

npq

σ =

Answer 1

X ~ bin ( n , p)

Where n = 11 , p = 0.3

P(X) = nCx * p^x * ( 1 - p)^n-x

a)

P(X = 5) = ¹¹C₅ * 0.3⁵ * ( 1 - 0.3)⁶

= 0.132

b)

P(X >= 5) = 1 - P(X <= 4)

= 1 - [ P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) ]

= 1 - [ ¹¹C₀ * 0.3⁰ * ( 1 - 0.3)¹¹ +¹¹C₁ * 0.3¹ * ( 1 - 0.3)¹⁰ +¹¹C₂ * 0.3² * ( 1 - 0.3)⁹ +¹¹C₃ * 0.3³ * ( 1 - 0.3)⁸

+¹¹C₄ * 0.3⁴ * ( 1 - 0.3)⁷ ]

= 0.210

c)

P(X > 5) = 1 - P(X <= 5)

= 1 - [ P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) ]

= 1 - [ ¹¹C₀ * 0.3⁰ * ( 1 - 0.3)¹¹ +¹¹C₁ * 0.3¹ * ( 1 - 0.3)¹⁰ +¹¹C₂ * 0.3² * ( 1 - 0.3)⁹ +¹¹C₃ * 0.3³ * ( 1 - 0.3)⁸

+¹¹C₄ * 0.3⁴ * ( 1 - 0.3)⁷ + ¹¹C₅ * 0.3⁵ * ( 1 - 0.3)⁶ ]

= 0.078

d)

P(X <= 5) =  P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

= ¹¹C₀ * 0.3⁰ * ( 1 - 0.3)¹¹ +¹¹C₁ * 0.3¹ * ( 1 - 0.3)¹⁰ +¹¹C₂ * 0.3² * ( 1 - 0.3)⁹ +¹¹C₃ * 0.3³ * ( 1 - 0.3)⁸

+¹¹C₄ * 0.3⁴ * ( 1 - 0.3)⁷ + ¹¹C₅ * 0.3⁵ * ( 1 - 0.3)⁶

= 0.922

e)

= n p = 11 * 0.3 = 3.3

= sqrt [ n p ( 1 - p) ]

= sqrt [ 11 * 0.3 * ( 1 - 0.3) ]

= 1.520