A telephone exchange operator assumes that 8% of the phone calls are wrong numbers. If the operator is right, what is the probability that the proportion of wrong numbers in a sample of 421 phone calls would differ from the population proportion by greater than 3%? Round your answer to four decimal places.

The probability that the sample proportion differ from the population proportion by greater than 3% is 0.0241.

Step-by-step explanation:

Let X = number of phone calls that are wrong numbers.

The proportion of phone calls that are wrong numbers is, p = 0.08.

A sample of n = 421 phone calls is selected to determine the proportion of wrong numbers in this sample.

The random variable X follows a Binomial distribution with parameters n and p.

The probability mass function of a Binomial distribution is:

Now, for the sample proportion to differ from the population proportion by 3% the value of the sample proportion should be:

So when the sample proportion is less than 5% or greater than 11% the difference between the sample proportion and population proportion will be greater than 3%.

        Compute the value of P (X ≤ 21) as follows:

        Compute the value of P (X ≥ 47) as follows:

Then the probability that the sample proportion differ from the population proportion by greater than 3% is:

Thus, the probability that the sample proportion differ from the population proportion by greater than 3% is 0.0241.

b. would be the answer : )

step-by-step explanation:

step 1: add -16x to both sides.

16x−4y+−16x=64+−16x  

−4y=−16x+64

step 2: divide both sides by -4.

−4y /-4 = -16x + 64/-4

answer:   y=4x−16

hope i'm right