A courier service claims that 5% of all of its deliveries arrive late. Assuming the claim is true and deliveries are independent, a sample of 10 deliveries is randomly selected. What is the probability that more than 2 of the sample deliveries arrive late

The probability that more than 2 of the sample deliveries arrive late = 0.0115

Step-by-step 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!!!

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.

lockers 1, 4, 9, 16, 25, 40, and 49 are open

i ran through it by the other answer is wrong