To find the probability that the sample proportion of first-time customers is greater than 0.4, we will use the normal approximation to the binomial distribution, since we have a sample size of 79, which is sufficiently large for the Central Limit Theorem to apply.
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Identify parameters:
- Population proportion (\(p\)) = 0.28
- Sample size (\(n\)) = 79
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Calculate the mean (\(\mu\)) and standard deviation (\(\sigma\)) of the sample proportion: \[ \mu = p = 0.28 \] \[ \sigma = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.28(1 - 0.28)}{79}} = \sqrt{\frac{0.28 \times 0.72}{79}} \approx \sqrt{\frac{0.2016}{79}} \approx \sqrt{0.002548} \approx 0.0505 \]
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Calculate the z-score for a sample proportion (\(\hat{p}\)) of 0.4: \[ z = \frac{\hat{p} - \mu}{\sigma} = \frac{0.4 - 0.28}{0.0505} \approx \frac{0.12}{0.0505} \approx 2.3762 \]
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Find the probability that the z-score is greater than 2.3762. We will look up this z-score in the standard normal distribution table or use a calculator to find the right tail probability: \[ P(Z > 2.3762) \approx 1 - P(Z \leq 2.3762) \] Using standard normal distribution tables or calculators, we find: \[ P(Z \leq 2.3762) \approx 0.9915 \] Therefore, \[ P(Z > 2.3762) \approx 1 - 0.9915 = 0.0085 \]
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Final result: The probability that the sample proportion is greater than 0.4 is approximately: \[ \boxed{0.0085} \]
Hence, the answer to the problem is \(0.0085\) rounded to four decimal places.
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