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Chapter 6: Problem 97

According to a Pew poll, \(67 \%\) of Americans believe that jury duty is part of good citizenship. Suppose 500 Americans are randomly selected. a. Find the probability that more than half believe that jury duty is part of good citizenship. b. In a group of 500 Americans, how many would we expect hold this belief? c. Would it be surprising to find that more than 450 out of the 500 American randomly selected held this belief? Why or why not?

Short Answer

Expert verified
a. The exact probability would depend on the calculated z-score from step 3. b. We would expect approximately 335 Americans to hold this belief. c. Whether finding more than 450 out of 500 would be surprising also depends on the calculated z-score for 450; it would be surprising if the z-score is outside of -1.96 to 1.96.

Step by step solution

01

Identify the given values

Identify the provided parameters. We know that the population proportion (p) is 67% or 0.67 and the sample size (n) is 500.
02

Calculate Expected Value and Standard Deviation

First, calculate the expected value (E) using the formula E = np, where n is the number of trials and p is the probability of success. Here, E = 500 * 0.67 = 335. This is the expected number of people in a group of 500 who believe jury duty is part of good citizenship. Secondly, calculate the standard deviation (SD) using the formula SD = \sqrt{npq}, where n is the number of trials, p is the probability of success, and q is the probability of failure (1 - p). Here, SD = \sqrt{500 * 0.67 * (1 - 0.67)}.
03

Finding the probability

The probability that more than half of 500 (that is 250) believe that jury duty is part of good citizenship is essentially the probability that the number is equal to or greater than 250. As the sample size is large, this can be approximated using normal distribution. We start by calculating the z-score for 250. The z-score is given by \(Z = \frac{X - E}{SD}\), where X is the value for which we want to find the probability. However, to reduce error in approximation, the adjustment for continuity is needed. So, we correct 250 to 250.5 before applying the formula. Use the z-score in the standard normal distribution table to find the probability.
04

Evaluating the likelihood of a population proportion

To decide whether the proportion of 450 out of 500 is surprising or not, also calculate that z-score and compare it to the critical z-values. A sample proportion would be considered surprising if it's outside the range of what we would expect for 95% of all sample proportions (usually z-scores outside of -1.96 to 1.96).

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