Step 1 :Find the mode of the probability density function by taking the derivative and setting it to zero: \(\frac{d}{dx}\left(\frac{x}{40000}(400-x^2)\right) = \frac{1}{100} - \frac{3x^2}{40000} = 0\)
Step 2 :Solve for x to find the most common waiting time: \(x = \frac{1}{100}\)
Step 3 :Find the probability of waiting more than 18 minutes by integrating the probability density function from 18 to 20: \(P(X>18) = \int_{18}^{20} \frac{x}{40000}(400-x^2) dx = \frac{361}{10000}\)
Step 4 :Use the given information to find the conditional probability of waiting more than 18 minutes given that the patient gave 4 or more stars: \(P(X>18|\text{4 or more stars}) = \frac{0.08}{P(\text{4 or more stars})}\)
Step 5 :Using the empirical rule, we know that approximately 95% of the data falls within 2 standard deviations of the mean. Since the mean is 3 stars and the standard deviation is 0.5 stars, 4 stars is 2 standard deviations away from the mean. Therefore, the probability of giving 4 or more stars is approximately 0.025.
Step 6 :Calculate the conditional probability: \(P(X>18|\text{4 or more stars}) = \frac{0.08}{0.025} = 0.2216\)
Step 7 :\(\boxed{\text{i) The most common waiting time is 0.01 minutes}}\)
Step 8 :\(\boxed{\text{ii) The probability of waiting more than 18 minutes is }\frac{361}{10000}\)
Step 9 :\(\boxed{\text{iii) The probability that a patient waited for more than 18 minutes given that they gave 4 or more stars on the survey is 0.2216}}\)