A statistician calculates that 7% of Americans own a Rolls Royce. If the statistician is right, what is the probability that the proportion of Rolls Royce owners in a sample of 613 Americans would differ from the population proportion by greater than 3%? Round your answer to four decimal places.

The probability that the proportion of Rolls Royce owners in a sample of 613 Americans would differ from the population proportion by greater than 3% is 0.036.

Step-by-step explanation:

The mean of the people owning a Roll Royce in the sample is 0.07*613 = 42.91, the standard deviation is √(42.91*(1-0.07)) = 6.3171. Since we are working with a binomial distribution with high parameter n, then we can approximate it by a Normal random variable instead, lets denote it X. In order to make  computations easier, we will take W, the standarization of X, which is given by the formula

The cummulative distribution function of W will be denoted . The values of are well known are they can be found in the attached file. We will calculate the probability that X is between 613*0.04 = 24.52 and 613*0.010 = 61.3, and then we will obtain the probability we want by substracting this result from 1 (because we are calculating the probability of the complementary event).

Note that, since the density function of a standard normal random variable si symmetric, then .

Thus, the probability that the proportion will differ in more than 3% is 1-0.9964 = 0.036.