Worksheet titled "Binomial Probability Worksheet" featuring exercises on calculating mean, standard deviation, and probabilities for binomial distributions, including real-world applications like quizzes, manufacturing, eggs, and sports.

1. Mean = 2.4, St. dev = 1.3856, P(X=k) for k=0 to 12 can be calculated using binomial formula.
2. Mean = 10, St. dev = 2.2361, P(X=k) for k=0 to 20 can be calculated using binomial formula.
3. P(3 failures) = P(8 successes) = C(11,8) * (0.05)^8 * (0.95)^3 ≈ 0.0000000037.
4. P(at least 3 successes) = 1 - [P(0) + P(1) + P(2)] = 1 - [C(6,0)(0.35)^0(0.65)^6 + C(6,1)(0.35)^1(0.65)^5 + C(6,2)(0.35)^2(0.65)^4] ≈ 0.3529.
5. a) P(Colin passes) = P(X≥6) where X~Bin(10,0.2) ≈ 0.0064. b) P(Diana passes) = P(X≥6) where X~Bin(10,0.75) ≈ 0.9219. c) Expected correct for Colin = 2. d) Expected correct for Diana = 7.5.
6. a) P(none defective) = (0.99)^50 ≈ 0.6050. b) P(at least one defective) = 1 - 0.6050 = 0.3950. c) P(at least two defective) = 1 - [P(0) + P(1)] = 1 - [(0.99)^50 + C(50,1)(0.01)^1(0.99)^49] ≈ 0.0894. d) Expected defective parts = 50 * 0.01 = 0.5.
7. a) P(none cracked) = (0.97)^24 ≈ 0.4814. b) P(at least one cracked) = 1 - 0.4814 = 0.5186. c) P(exactly two cracked) = C(24,2)(0.03)^2(0.97)^22 ≈ 0.1275. d) Expected cracked eggs = 24 * 0.03 = 0.72. e) Expected uncracked eggs = 24 * 0.97 = 23.28.
8. a) P(exactly 8) = C(10,8)(0.7)^8(0.3)^2 ≈ 0.2335. b) P(at least 8) = P(8)+P(9)+P(10) ≈ 0.3828. c) P(at most 2) = P(0)+P(1)+P(2) ≈ 0.0016. d) P(between 5 and 7 inclusive) = P(5)+P(6)+P(7) ≈ 0.5698. e) Expected sinks = 10 * 0.7 = 7.