Question: 1.\tAn online retailer claims that 90% of all orders are shipped within 12 hours of being received. A consumer group placed 121 orders of different sizes and at different times of day; 102 orders were shipped within 12 hours. \na.\tIf the online retailer\u2019s claim is correct, what is the probability that 102 or fewer out of 121 orders are shipped within 12 hours?
1. **State the problem:** We want to find the probability that 102 or fewer orders out of 121 are shipped within 12 hours, assuming the true shipping rate is 90% (0.9).\n\n2. **Identify the distribution:** This is a binomial problem where \(n=121\) and \(p=0.9\). The number of successes \(X\) (orders shipped within 12 hours) follows \(X \sim \text{Binomial}(n=121, p=0.9)\).\n\n3. **Use normal approximation:** Since \(n\) is large, we approximate the binomial distribution with a normal distribution \(N(\mu, \sigma^2)\) where \(\mu = np = 121 \times 0.9 = 108.9\) and \(\sigma = \sqrt{np(1-p)} = \sqrt{121 \times 0.9 \times 0.1} = \sqrt{10.89} \approx 3.3\).\n\n4. **Apply continuity correction:** We want \(P(X \leq 102)\), so use \(P(X \leq 102.5)\) for the normal approximation.\n\n5. **Calculate the z-score:**\n$$z = \frac{102.5 - 108.9}{3.3} = \frac{-6.4}{3.3} \approx -1.94$$\n\n6. **Find the probability:** Using standard normal tables or a calculator, \(P(Z \leq -1.94) \approx 0.0262\).\n\n7. **Interpretation:** The probability that 102 or fewer orders are shipped within 12 hours, assuming the claim is true, is approximately 2.62%. This is quite low, suggesting the observed data is somewhat unlikely if the claim is correct.