c) The waiting time (in minutes) at a dental surgery is represented by a random variable $X$ which has a probability density function:
\[
f(x)=\left\{\begin{array}{cl}
0, & \text { if } x< 0 \\
\frac{x}{40000}\left(400-x^{2}\right) \quad \text { if } 0 \leq x \leq 20 \\
0, & \text { if } x> 20
\end{array}\right.
\]
The surgery conducted a customer satisfaction survey that involved rating the surgery from 1 star (extremely dissatisfied) to 5 stars (extremely satisfied). The results of the survey were normally distributed, with a mean of 3 stars and a standard deviation of 0.5 stars. $8 \%$ of patients with a waiting time greater than 18 minutes gave a rating of 4 or more stars. Use the empirical rule when required to answer the following questions:
i) Determine the most common waiting time. Give your answer to 2 decimal places.
ii) Find $P(X> 18)$. Give your answer as an exact fraction.
2
1
iii) Hence, find the probability that a patient waited for more than 18 minutes given that they gave 4 or more stars on the survey. Give your answer to 4 decimal places.

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 ：\(\boxed{\text{iii) The probability that a patient waited for more than 18 minutes given that they gave 4 or more stars on the survey is 0.2216}}\)