A restaurant server believes the distribution of their tips has a model that is slightly skewed to the left, with a mean of \( \$ 10.90 \) and a standard deviation of \( \$ 4.50 \). They usually wait on about 30 parties over a weekend of work. a) Estimate the probability that they will earn at least \( \$ 350 \) in tips. b) How much do they earn on the best \( 5 \% \) of such weekends? a) P(tips from 30 parties \( >\$ 350)=\square \) (Round to four decimal places as needed.) b) The total amount that they earn on the best \( 5 \% \) of such weekends is at least \( \$ \square \) (Round to two decimal places as needed.)

To tackle this problem, we need to first find out how the tips from 30 parties can be modeled. Assuming tips are approximately normally distributed due to the Central Limit Theorem (CLT), we can calculate the mean and standard deviation of the total tips over the 30 parties. a) The mean total tips from 30 parties can be: \[ \mu = 30 \times 10.90 = 327 \] The standard deviation for the total tips can be calculated as: \[ \sigma = \sqrt{30} \times 4.50 \approx 24.5 \] Next, we need to find \( P(X > 350) \), where \( X \) is the total tips from 30 parties. We can standardize this using the z-score formula: \[ z = \frac{X - \mu}{\sigma} = \frac{350 - 327}{24.5} \approx 0.94 \] Now, using z-tables or calculators, we find \( P(Z > 0.94) \). This gives us: \[ P(Z > 0.94) = 1 - P(Z < 0.94) \approx 1 - 0.8264 = 0.1736 \] Thus, the probability that the server will earn at least $350 in tips is: \[ P(tips from 30 parties >\$ 350) = 0.1736 \] b) To find how much they earn on the best 5% of weekends, we need to determine the 95th percentile (since we are looking for the top 5% which equates to 100% - 5% = 95%) for a normal distribution. From z-tables, the z-score that corresponds to the 95th percentile is approximately 1.645. Now we convert back to the original scale using the following formula: \[ X = \mu + z\sigma = 327 + (1.645 \times 24.5) \approx 327 + 40.2 \approx 367.2 \] Thus, the total amount that they earn on the best 5% of weekends is at least: \[ \$367.20 \] In conclusion: a) \( P(tips from 30 parties >\$ 350) = 0.1736 \) b) The total amount that they earn on the best 5% of such weekends is at least \( \$367.20 \)