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Drupady [299]
3 years ago
10

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.
Mathematics
1 answer:
aalyn [17]3 years ago
6 0

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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Answer:

A sample of 1032 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is:

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How large a sample is needed to estimate the true proportion of defective components to within 2.5 percentage points with 99% confidence?

A sample of n is needed.

n is found when M = 0.025. So

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0.025 = 2.575\sqrt{\frac{0.0667*0.9333}{n}}

0.02\sqrt{n} = 2.575\sqrt{0.0667*0.9333}

\sqrt{n} = \frac{2.575\sqrt{0.0667*0.9333}}{0.02}

(\sqrt{n})^2 = (\frac{2.575\sqrt{0.0667*0.9333}}{0.02})^2

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Rounding up

A sample of 1032 is needed.

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