Question

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.

Answer 1

Answer:

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

$W = \frac{X-\mu}{\sigma} = \frac{X-42.91}{6.3171}$

The cummulative distribution function of W will be denoted $\phi$ . The values of $\phi$ 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).

$P(24.52 < X < 61.3) = P(\frac{24.52-42.91}{6.3171} < \frac{X-42.91}{6.3171} < \frac{61.3-42.91}{6.3171} ) =\\P(-2.91 < W < 2.911) = \phi(2.91) - \phi(-2.91) = 2 \phi(2.91) - 1 = 2*0.9982-1 = 0.9964$

Note that, since the density function of a standard normal random variable si symmetric, then $\phi(-2.91) = 1- \phi(2.91)$ .

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

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