A manufacturer of window frames knows from long experience that 5% of the production will have some type of minor defect that will require an adjustment. What is the probability that in a sample of 20 window frames?

a) 0.3585

b) 0.6415

c) 0.07548

Step-by-step explanation:

This is a binomial distribution problem

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

p = probability of window frame with minor defect = 0.05

q = probability of window frame with no defect = 1 - 0.05 = 0.95

a) Probability that none of the frames picked will have defects

x = 0

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

P(X = 0) = ²⁰C₀ 0.05⁰ 0.95²⁰⁻⁰ = 0.3585

b) Probability that At least one will need adjustment = P(X ≥ 1) = 1 - P(X < 1) = 1 - P(X = 0) = 1 - 0.3585 = 0.6415

c) Probability that More than two will need adjustment = P(X > 2)

P(X > 2) = 1 - P(X ≤ 2) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)]

So, we evaluate each of that using the binomial distribution formula, slotting in the appropriate variables, and sum it all up.

P(X > 2) = 1 - 0.9245 = 0.07548

i dont know sorry

step-by-step explanation:

total amount of money that he should pay if he has not used his duty free exemption is $98

step-by-step explanation:

given a united states citizen returning to the states declares the following items at the customs office: 3 shirts at $8.50 each 2 dresses at $27.50 each 1 pair of gold cuff links at $17.50 per pair. we have to find how much he pay

cost of 3 shirts if one cost at $8.50 is

cost of 2 dresses if one cost at $27.50 is  

cost of 1 pair of gold cuff at $17.50

so, total amount of money that he should pay if he has not used his duty free exemption is   $25.5+$55+$17.50=$98