Binomial probability #1 - Questions and Answers

Question 1 Binomial experiment with 𝑛 = 20 n=20 and 𝑝 = 0.70 p=0.70: a) Compute 𝑓 ( 12 ) f(12) b) Compute 𝑓 ( 16 ) f(16) c) Compute 𝑃 ( 𝑥 ≥ 16 ) P(x≥16) d) Compute 𝑃 ( 𝑥 ≤ 15 ) P(x≤15) e) Compute 𝐸 ( 𝑥 ) E(x) f) Compute var ( 𝑥 ) var(x) and 𝜎 σ Answer 1 Given: 𝑛 = 20 n=20, 𝑝 = 0.70 p=0.70 a) Compute 𝑓 ( 12 ) f(12): The binomial probability formula is: 𝑓 ( 𝑥 ) = ( 𝑛 𝑥 ) 𝑝 𝑥 ( 1 − 𝑝 ) 𝑛 − 𝑥 f(x)=( x n ​ )p x (1−p) n−x For 𝑥 = 12 x=12: 𝑓 ( 12 ) = ( 20 12 ) ( 0.70 ) 12 ( 1 − 0.70 ) 8 f(12)=( 12 20 ​ )(0.70) 12 (1−0.70) 8 𝑓 ( 12 ) = 20 ! 12 ! ( 20 − 12 ) ! ( 0.70 ) 12 ( 0.30 ) 8 = 0.1144 f(12)= 12!(20−12)! 20! ​ (0.70) 12 (0.30) 8 =0.1144 b) Compute 𝑓 ( 16 ) f(16): For 𝑥 = 16 x=16: 𝑓 ( 16 ) = ( 20 16 ) ( 0.70 ) 16 ( 1 − 0.70 ) 4 f(16)=( 16 20 ​ )(0.70) 16 (1−0.70) 4 𝑓 ( 16 ) = 20 ! 16 ! ( 20 − 16 ) ! ( 0.70 ) 16 ( 0.30 ) 4 = 0.1304 f(16)= 16!(20−16)! 20! ​ (0.70) 16 (0.30) 4 =0.1304 c) Compute 𝑃 ( 𝑥 ≥ 16 ) P(x≥16): Add the probabilities from 𝑥 = 16 x=16 to 𝑥 = 20 x=20: 𝑃 ( 𝑥 ≥ 16 ) = 𝑓 ( 16 ) + 𝑓 ( 17 ) + 𝑓 ( 18 ) + 𝑓 ( 19 ) + 𝑓 ( 20 ) P(x≥16)=f(16)+f(17)+f(18)+f(19)+f(20) 𝑃 ( 𝑥 ≥ 16 ) = 0.1304 + ⋯ = 0.2375 P(x≥16)=0.1304+⋯=0.2375 d) Compute 𝑃 ( 𝑥 ≤ 15 ) P(x≤15): Using the complement rule: 𝑃 ( 𝑥 ≤ 15 ) = 1 − 𝑃 ( 𝑥 ≥ 16 ) P(x≤15)=1−P(x≥16) 𝑃 ( 𝑥 ≤ 15 ) = 1 − 0.2375 = 0.7625 P(x≤15)=1−0.2375=0.7625 e) Compute 𝐸 ( 𝑥 ) E(x): The mean of the binomial distribution is: 𝐸 ( 𝑥 ) = 𝑛 𝑝 E(x)=np 𝐸 ( 𝑥 ) = 20 × 0.70 = 14 E(x)=20×0.70=14 f) Compute var ( 𝑥 ) var(x) and 𝜎 σ: Variance is given by: var ( 𝑥 ) = 𝑛 𝑝 ( 1 − 𝑝 ) var(x)=np(1−p) var ( 𝑥 ) = 20 × 0.70 × ( 1 − 0.70 ) = 4.2 var(x)=20×0.70×(1−0.70)=4.2 Standard deviation ( 𝜎 σ) is the square root of variance: 𝜎 = var ( 𝑥 ) = 4.2 ≈ 2.05 σ= var(x) ​ = 4.2 ​ ≈2.05 Answer Summary: a) 𝑓 ( 12 ) = 0.1144 f(12)=0.1144 b) 𝑓 ( 16 ) = 0.1304 f(16)=0.1304 c) 𝑃 ( 𝑥 ≥ 16 ) = 0.2375 P(x≥16)=0.2375 d) 𝑃 ( 𝑥 ≤ 15 ) = 0.7625 P(x≤15)=0.7625 e) 𝐸 ( 𝑥 ) = 14 E(x)=14 f) var ( 𝑥 ) = 4.2 var(x)=4.2, 𝜎 ≈ 2.05 σ≈2.05 Question 2 Which of the following is not a requirement of the binomial probability distribution? Answer: B. The trials must be dependent. Explanation: For a binomial distribution: Each trial must be independent. There are two outcomes: success or failure. The probability of success is constant for each trial. The number of trials 𝑛 n is fixed. Question 3 What are the four conditions necessary for 𝑋 X to have Binomial Distribution? Answer: For a Binomial Distribution: The probability of success is constant for every trial. There are only two outcomes (success and failure). The trials must be independent. The number of trials 𝑛 n is fixed. Question 4 Find 𝑃 ( 𝑋 = 12 ) P(X=12) for 𝑛 = 13 n=13, 𝑝 = 0.83 p=0.83: Answer: The binomial probability formula is: 𝑃 ( 𝑋 = 𝑘 ) = ( 𝑛 𝑘 ) 𝑝 𝑘 ( 1 − 𝑝 ) 𝑛 − 𝑘 P(X=k)=( k n ​ )p k (1−p) n−k For 𝑘 = 12 k=12, 𝑛 = 13 n=13, 𝑝 = 0.83 p=0.83: 𝑃 ( 𝑋 = 12 ) = ( 13 12 ) ( 0.83 ) 12 ( 1 − 0.83 ) 1 P(X=12)=( 12 13 ​ )(0.83) 12 (1−0.83) 1 𝑃 ( 𝑋 = 12 ) = 13 × ( 0.83 ) 12 × ( 0.17 ) ≈ 0.2362 P(X=12)=13×(0.83) 12 ×(0.17)≈0.2362 Question 5 Compute 𝑃 ( 𝑋 = 3 ) P(X=3) for 𝑛 = 5 n=5, 𝑝 = 0.85 p=0.85: Answer: Using the binomial formula: 𝑃 ( 𝑋 = 𝑘 ) = ( 𝑛 𝑘 ) 𝑝 𝑘 ( 1 − 𝑝 ) 𝑛 − 𝑘 P(X=k)=( k n ​ )p k (1−p) n−k For 𝑘 = 3 k=3, 𝑛 = 5 n=5, 𝑝 = 0.85 p=0.85: 𝑃 ( 𝑋 = 3 ) = ( 5 3 ) ( 0.85 ) 3 ( 1 − 0.85 ) 5 − 3 P(X=3)=( 3 5 ​ )(0.85) 3 (1−0.85) 5−3 𝑃 ( 𝑋 = 3 ) = 10 × ( 0.85 ) 3 × ( 0.15 ) 2 ≈ 0.1382 P(X=3)=10×(0.85) 3 ×(0.15) 2 ≈0.1382 Thus, 𝑃 ( 𝑋 = 3 ) = 0.1382 P(X=3)=0.1382.