A telephone exchange operator assumes that 8% of the phone calls are wrong numbers.
If the operator is correct, what is the probability that the proportion of wrong numbers in a sample of 494 phone calls would differ from the population proportion by more than 3%? Round your answer to four decimal places.

The probability that the phone calls would differ from the population proportion by more than 3% is 0How to determine the probability value?From the question, the given parameters about the distribution areProportion, p = 8%Sample size, n = 494The actual proportion, x = greater than 3%The mean is calculated asMean = npSo, we haveMean = 8% * 494 = 39.52The standard deviation is calculated asStandard deviation = √np(1 - p)So, we haveStandard deviation = √494 * 8% (1 - 8%) = 0.77The z-score is calculated using the following formula[tex]z = \frac{\bar p - p}{\sqrt{p(1 - p)}/n}[/tex]Substitute the given parameters in the above equation[tex]z = \frac{8\% - 3\%}{\sqrt{8\% * (1 - 8\%)}/494}[/tex]Evaluatez = 90.9The probability is then calculated as:P(x > 3%) = P(z > 90.9)From the z table of probabilities, we have;P(x > 3%) = 0Hence, the probability is 0Read more about probability at:brainly.com/question/25870256#SPJ1