91Ó°ÊÓ

In a uniform distribution, the cumulative distribution function (CDF) helps you understand the probability of a random variable being less than or equal to a certain value. For this exercise, we consider the time between a visitor's arrival and the 8:30 A.M. start time as a random variable \( X \). Since the arrival time is uniformly distributed between 8:30 A.M. and 10:00 A.M, this period covers 90 minutes, marking 8:30 A.M. as 0 minutes and 10:00 A.M. as 90 minutes. The range of possible arrival times is hence \([0, 90]\) minutes.
The CDF for a uniform distribution over this interval can be expressed using the formula:\[F(x) = \frac{x-a}{b-a}, \quad \text{for} \; a \leq x \leq b\]Here, \( a = 0 \) and \( b = 90 \). Substituting these into our formula, we get:\[F(x) = \frac{x}{90}, \quad 0 \leq x \leq 90\] This function tells us the probability that a visitor arrives by minute \( x \). For instance, \( F(45) = \frac{45}{90} = 0.5 \), meaning there's a 50% chance the visitor arrives by 9:15 A.M.

The mean and variance offer insight into the center and spread of the distribution of arrival times. In a uniform distribution, the mean or average, symbolized as \( \mu \), is the midpoint of the interval. It is calculated using:\[\mu = \frac{a+b}{2}\]For our uniform distribution interval \([0, 90]\), the mean is \[\mu = \frac{0 + 90}{2} = 45 \text{ minutes}\]This implies that on average, a visitor arrives 45 minutes after 8:30 A.M., or at 9:15 A.M.
Variance, denoted \( \sigma^2 \), measures the spread of the distribution. For a uniform distribution, it is given by:\[\sigma^2 = \frac{(b-a)^2}{12}\]Thus, for our interval, the variance is:\[\sigma^2 = \frac{(90-0)^2}{12} = 675\]A larger variance would indicate a wider spread in arrival times. Here, a variance of 675 provides a sense of how dispersed the arrival times are from the mean.

Calculating probabilities for events within a uniform distribution involves using the CDF to determine the likelihood of various outcomes. Let's consider two scenarios: a visitor waiting less than 10 minutes for a show, and more than 20 minutes.

If a visitor waits less than 10 minutes, this means they must arrive at least 10 minutes before a show starts. We are interested in the probability for a show starting at 9:00, 9:30, and 10:00. From our CDF: \[F(40) = \frac{40}{90} = \frac{4}{9} \approx 0.444\]This means there is a 44.4% chance that the visitor will wait less than 10 minutes if they arrive before 9:40 A.M. for the 9:30 A.M. show.

To find the probability that a visitor waits more than 20 minutes, they need to arrive more than 20 minutes before the show. For the 9:00 A.M. show:\[P(X > 10) = 1 - F(10) = 1 - \frac{10}{90} = \frac{8}{9} \approx 0.889\]This indicates that there's an 88.9% probability of a visitor arriving such that they wait more than 20 minutes till the show begins. Calculating these probabilities helps in assessing visitor behaviors and managing show admissions efficiently.