Answer:
The probability that more than 2 of the sample deliveries arrive late = 0.0115
Explanation:
This is a binomial distribution problem
A binomial experiment is one in which the probability of success doesn't change with every run or number of trials.
It usually consists of a fixed number of runs/trials with only two possible outcomes, a success or a failure. The outcome of each trial/run of a binomial experiment is independent of one another.
The probability of each delivery arriving late = 5% = 0.05
- Each delivery is independent from the other.
- There is a fixed number of deliveries to investigate.
- Each delivery has only two possible outcomes, a success or a failure of arriving late.
Binomial distribution function is represented by
P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ
n = total number of sample spaces = number of deliveries we're considering = 10
x = Number of successes required = number of deliveries that we expect to arrive late = more than 2 = > 2
p = probability of success = probability of a delivery arriving late = 0.05
q = probability of failure = probability of a delivery NOT arriving late = 0.95
P(X > 2) = 1 - P(X ≤ 2)
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
= 0.59873693924 + 0.31512470486 + 0.07463479852
= 0.98849644262
P(X > 2) = 1 - P(X ≤ 2)
= 1 - 0.98849644262
= 0.01150355738
= 0.0115
Hope this Helps!!!