A package delivery service claims that no more than 5 percent of all packages arrive at the address late. Assuming that the conditions for the binomial hold, if a sample of size 10 packages is randomly selected and the 5 percent rate holds, what is the probability that more than 2 packages will be delivered late?
A) 0.0115
B) 0.0105
C) 0.0862
D) 0.0746

Question

Quizplus · Accepted Answer

The probability of more than 2 packages being delivered late is the sum of the probabilities of 3, 4, 5, 6, 7, 8, 9, or 10 packages being delivered late. Using the binomial distribution, we can calculate these probabilities as follows:P(3 packages late) = (10 choose 3) * (0.05)^3 * (0.95)^7 = 0.0115P(4 packages late) = (10 choose 4) * (0.05)^4 * (0.95)^6 = 0.0029P(5 packages late) = (10 choose 5) * (0.05)^5 * (0.95)^5 = 0.0006P(6 packages late) = (10 choose 6) * (0.05)^6 * (0.95)^4 = 0.0001P(7 packages late) = (10 choose 7) * (0.05)^7 * (0.95)^3 = 0.0000P(8 packages late) = (10 choose 8) * (0.05)^8 * (0.95)^2 = 0.0000P(9 packages late) = (10 choose 9) * (0.05)^9 * (0.95)^1 = 0.0000P(10 packages late) = (10 choose 10) * (0.05)^10 * (0.95)^0 = 0.0000Therefore, the probability of more than 2 packages being delivered late is 0.0115 + 0.0029 + 0.0006 + 0.0001 + 0.0000 + 0.0000 + 0.0000 + 0.0000 = 0.0151.

A Package Delivery Service Claims That No More Than 5