Questions: In an article in Advertising Age, Nancy Giges studies global spending patterns. Giges presents data concerning the percentage of adults in various countries who have purchased various consumer items (such as soft drinks, athletic footwear, blue jeans, beer, and so on) in the past three months. (Round answer to 4 decimal places.) (a) Suppose we wish to justify the claim that fewer than 50 percent of adults in Germany have purchased blue jeans in the past three months. The survey reported by Giges found that 44 percent of the respondents in Germany had purchased blue jeans in the past three months. Note: The actual figure in the survey is different; the figure has been changed here for instructional purposes. Assume that a random sample of 405 German adults was employed, and let p be the proportion of all German adults who have purchased blue jeans in the past three months. If, for the sake of argument, we assume that p=0.5, use the normal approximation to the binomial distribution to calculate the probability that 44 percent or fewer of 405 randomly selected German adults would have purchased blue jeans in the past three months. Note: Because 44 percent of 405 is 178, you should calculate the probability that 178 or fewer of 405 randomly selected German adults would have purchased blue jeans in the past three months.

In an article in Advertising Age, Nancy Giges studies global spending patterns. Giges presents data concerning the percentage of adults in various countries who have purchased various consumer items (such as soft drinks, athletic footwear, blue jeans, beer, and so on) in the past three months. (Round answer to 4 decimal places.) (a) Suppose we wish to justify the claim that fewer than 50 percent of adults in Germany have purchased blue jeans in the past three months. The survey reported by Giges found that 44 percent of the respondents in Germany had purchased blue jeans in the past three months. Note: The actual figure in the survey is different; the figure has been changed here for instructional purposes.

Assume that a random sample of 405 German adults was employed, and let p be the proportion of all German adults who have purchased blue jeans in the past three months. If, for the sake of argument, we assume that p=0.5, use the normal approximation to the binomial distribution to calculate the probability that 44 percent or fewer of 405 randomly selected German adults would have purchased blue jeans in the past three months. Note: Because 44 percent of 405 is 178, you should calculate the probability that 178 or fewer of 405 randomly selected German adults would have purchased blue jeans in the past three months.
Transcript text: In an article in Advertising Age, Nancy Giges studies global spending patterns. Giges presents data concerning the percentage of adults in various countries who have purchased various consumer items (such as soft drinks, athletic footwear, blue jeans, beer, and so on) in the past three months. (Round answer to 4 decimal places.) (a) Suppose we wish to justify the claim that fewer than 50 percent of adults in Germany have purchased blue jeans in the past three months. The survey reported by Giges found that 44 percent of the respondents in Germany had purchased blue jeans in the past three months. Note: The actual figure in the survey is different; the figure has been changed here for instructional purposes. Assume that a random sample of 405 German adults was employed, and let $p$ be the proportion of all German adults who have purchased blue jeans in the past three months. If, for the sake of argument, we assume that $p=.5$, use the normal approximation to the binomial distribution to calculate the probability that 44 percent or fewer of 405 randomly selected German adults would have purchased blue jeans in the past three months. Note: Because 44 percent of 405 is 178 , you should calculate the probability that 178 or fewer of 405 randomly selected German adults would have purchased blue jeans in the past three months. $\square$ $P$
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Solution

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Solution Steps

Step 1: Calculate Mean and Standard Deviation

Given the sample size \( n = 405 \) and the assumed population proportion \( p = 0.5 \), we can calculate the mean \( \mu \) and standard deviation \( \sigma \) of the binomial distribution as follows:

\[ \mu = n \cdot p = 405 \cdot 0.5 = 202.5 \]

\[ \sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{405 \cdot 0.5 \cdot 0.5} \approx 10.0623 \]

Step 2: Calculate Probability Using Normal Approximation

To find the probability that 178 or fewer of the 405 randomly selected German adults purchased blue jeans, we need to calculate the Z-score for \( x = 178 \):

\[ Z = \frac{x - \mu}{\sigma} = \frac{178 - 202.5}{10.0623} \approx -2.4348 \]

Using the standard normal distribution, we find:

\[ P(X \leq 178) = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(-2.4348) - \Phi(-\infty) \approx 0.0074 \]

Final Answer

The probability that 178 or fewer have purchased blue jeans is approximately \( P \approx 0.0074 \).

Thus, the final answer is:

\[ \boxed{P = 0.0074} \]

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