Apackage 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.

0.0115

Step-by-step explanation:

The binomial probability formula :-

, 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 :-

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

step-by-step explanation:

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

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