Question

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? Round to four decimal places.

Answer 1

Answer: 0.0115

Step-by-step explanation:

The binomial probability formula :-

$P(X)=^nC_x\ p^x(1-p)^{n-x}$, here n is the number total of trials , p is the probability of getting success in each trial and P(x) is the probability of getting success in x trial.

Given : A package delivery service claims that no more than 5 percent of all packages arrive at the address late.

We assume that the conditions for the binomial​ hold with parameters :-

n=10 ; p=0.05

Let x be the random variable that represents the number of packages arrive at the address late.

Now, the probability that more than 2 packages will be delivered​ late :-

$P(x>2)=1-P(x\leq2)\\\\=1-(P(0)+P(1)+P(2))\\\\=1-(^{10}C_0(0.05)^{0}(1-0.05)^{10}+^{10}C_{1}(0.05)^1(1-0.05)^9+^{10}C_{2}(0.05)^2(1-0.05)^8)\\\\=1-((0.95)^{10}+(10)(0.05)(0.95)^9+45(0.05)^2(0.95)^8}\\\\=1-0.988496442621=0.011503557379\approx0.0115$

Hence, the probability that more than 2 packages will be delivered​ late = 0.9885