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

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e. Yes

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Step 1 ：a. $X \sim N (6.26, {2.35}^{2})$

Step 2 ：b. $\bar{x} \sim N (6.26, \frac{{2.35}^{2}}{47})$

Step 3 ：c. $P (6.6758 \leq X \leq 7.0972) = P (\frac{6.6758 - 6.26}{2.35} \leq Z \leq \frac{7.0972 - 6.26}{2.35}) \approx P (0.1765 \leq Z \leq 0.3565) = 0.0383$

Step 4 ：d. $P (6.6758 \leq \bar{x} \leq 7.0972) = P (\frac{6.6758 - 6.26}{\frac{2.35}{\sqrt{47}}} \leq Z \leq \frac{7.0972 - 6.26}{\frac{2.35}{\sqrt{47}}}) \approx P (2.3340 \leq Z \leq 4.6937) = 0.0098$

Step 5 ：e. Yes

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