Question

Consider a binomial experiment with and . a. Compute (to 4 decimals). .1144 b. Compute (to 4 decimals). .1304 c. Compute (to 4 decimals). .2374 d. Compute (to 4 decimals). .7626 e. Compute . 14 f. Compute (to 1 decimal) and (to 2 decimals). 4.2 1.74

Answer 1

Answer:

a) f(12) = P(X=12) = 0.1144

b) f(16) = P(X=16) = 0.1304

c) P(x≥16) = 0.2375

d) P(x≤15) = 0.7625

e) E(X) = 14

f) Var(X) = 4.2

Standard deviation = 2.05

Step-by-step explanation:

Binomial distribution formula is given as

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 20

x = Number of successes required = variable

p = probability of success = 0.7

q = probability of failure = 1 - 0.7 = 0.3

a) f(12) = P(X=12)

x = 12

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

P(X = 12) = ²⁰C₁₂ (0.7)¹² (0.3)²⁰⁻¹²

P(X=12) = 0.1144

b) f(16) = P(X=16)

x = 16

P(X = 16) = ²⁰C₁₆ (0.7)¹⁶ (0.3)²⁰⁻¹⁶

P(X=16) = 0.1304

c) P(x≥16)

This is a sum of probabilities from x = 16 to x = 20. (Total number of sample space = 20)

P(x≥16) = P(X=16) + P(X=17) + P(X=18) + P(X=19) + P(X=20) = 0.2375 (computed using calculator)

d) P(x≤15)

This is a sum of probabilities from 0 to 15.

But it can be rewritten as

P(X≤15) = 1 - P(X > 15) = 1 - [P(X=16) + P(X=17) + P(X=18) + P(X=19) + P(X=20)]

= 1 - 0.2375 = 0.7625

e) Expected value = E(X) = Mean = np = (20)(0.7) = 14

f) Variance = np(1-p) = (20×0.7×0.3) = 4.2

Standard deviation = σ = √(variance) = √(4.2) = 2.05

Hope this Helps!!!