Problem
The average amount of money spent for lunch per person in the college cafeteria is \( \$ 6.26 \) and the standard deviation is \( \$ 2.35 \). Suppose that 47 randomly selected lunch patrons are observed. Assume the distribution of money spent is normal, and round all answers to 4 decimal places where possible. a. What is the distribution of \( X \) ? \( X \sim N(6.26 \) 2.35 b. What is the distribution of \( \bar{x} ? \bar{x} \sim N( \) c. For a single randomly selected lunch patron, find the probability that this patron's lunch cost is between \( \$ 6.6758 \) and \( \$ 7.0972 \). d. For the group of 47 patrons, find the probability that the average lunch cost is between \( \$ 6.6758 \) and \( \$ 7.0972 \). e. For part d), is the assumption that the distribution is normal necessary? O Yes No
Solution
Step 1 ：a. \( X \sim N(6.26, 2.35^2) \)
Step 2 ：b. \( \bar{x} \sim N\left(6.26, \frac{2.35^2}{47}\right) \)
Step 3 ：c. \( P(6.6758 \le X \le 7.0972) = P\left(\frac{6.6758 - 6.26}{2.35}\le Z \le \frac{7.0972 - 6.26}{2.35}\right) \approx P(0.1765 \le Z \le 0.3565) = 0.0383 \)
Step 4 ：d. \( P(6.6758 \le \bar{x} \le 7.0972) = P\left(\frac{6.6758 - 6.26}{\frac{2.35}{\sqrt{47}}}\le Z \le \frac{7.0972 - 6.26}{\frac{2.35}{\sqrt{47}}}\right) \approx P(2.3340 \le Z \le 4.6937) = 0.0098 \)
Step 5 ：e. Yes
