This problem deals with the exponential distribution, which is commonly used to model waiting times. Given the average waiting time of 35 minutes, we can use the exponential distribution's properties to answer each part.

(a) Standard Deviation

For an exponential distribution with mean $\mu$, both the mean and the standard deviation $SD(X)$ are equal to $\mu$. Given: $\mu = 35 \text{ minutes}$ So: $SD(X) = 35 \text{ minutes}$

(b) Probability of Waiting Between 27 and 38 Minutes

The probability density function for an exponential distribution with mean $\mu$ is: $f(x) = \frac{1}{\mu} e^{-x / \mu}$ However, the cumulative distribution function (CDF) is more useful here: $P(X \leq x) = 1 - e^{-x / \mu}$

To find $P(27 \leq X \leq 38)$, we calculate: $P(27 \leq X \leq 38) = P(X \leq 38) - P(X \leq 27)$ where: $P(X \leq 38) = 1 - e^{-38 / 35}$ $P(X \leq 27) = 1 - e^{-27 / 35}$ Substitute the values and calculate to four decimal places.

(c) Conditional Probability of Total Waiting Time

This part asks for the probability of waiting at most 46 minutes, given that you've already waited 30 minutes. For an exponential distribution, the memoryless property applies: $P(X > t + s | X > s) = P(X > t)$ So: $P(\text{wait in total at most 46 minutes}) = P(X \leq 46 | X \geq 30) = \frac{P(X \leq 46) - P(X \leq 30)}{1 - P(X \leq 30)}$

(d) 50th Percentile (Median) Waiting Time

For an exponential distribution, the median is given by: $\text{Median} = \mu \ln(2)$ Substitute $\mu = 35$ and calculate