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Uniform distribution is a type of probability distribution where all outcomes are equally likely. An easy way to imagine it is if you are rolling a fair six-sided die. Each side has an equal chance of landing face up, which is 1/6. With uniform distribution, the probability of any interval is proportional to its length.
In the context of our exercise, the waiting time for the bus is uniformly distributed between 0 and 10 minutes. This means all minutes within this time frame have an equal probability of being the waiting time. This is described by the probability density function (pdf), which, for a uniform distribution on the interval \(a, b\), is determined as:

• \( f(x) = \frac{1}{b - a} \text{ for } a \leq x \leq b \)
• \( f(x) = 0 \text{ otherwise} \)

This means for our specific problem, the pdf is \( f(x) = \frac{1}{10} \text{ for } 0 \leq x \leq 10 \) because our interval from 0 to 10 has a length of 10.

The expected value, or mean, provides the average or central value of a random variable. It's the long-run average value that you would get if you repeated an experiment a lot of times.
For a discrete random variable, the expected value is the sum of all possible values, each multiplied by the probability of its occurrence. In a continuous setting, like our uniform distribution, it's calculated via integration using the probability density function (pdf).
In our exercise, we find the expected value of the largest waiting time using its pdf \(f_Y(y) = \frac{5y^4}{10^5}\). The integration formula for expected value is \( E[Y] = \int_{a}^{b} y \cdot f_Y(y) \, dy \). Performing this calculation gives us an expected value of 8 minutes, indicating that on average, the largest wait time in a week is 8 minutes. The same method is used to determine the expected value of the median and the smallest waiting time, revealing more insights into how bus wait times vary throughout the week.

A probability density function (pdf) describes the likelihood of a random variable to take on a particular value. For continuous random variables, it helps to determine how probabilities are distributed across different values.
In the context of our exercise, the pdf is derived for several variables: each of the five individual waiting times, the largest and smallest waiting times, and the median wait time.
For the waiting times \(X_1, X_2, \, ..., \, X_5\), each is uniformly distributed as \( f(x) = \frac{1}{10}\). For the largest time \(Y = \max(X_1, X_2, \, ..., \, X_5)\), the unique pdf is found via: \[ f_Y(y) = \frac{d}{dy} \left(\left( \frac{y}{10} \right)^5 \right) = \frac{5y^4}{10^5} \]This pdf allows us to analyze just how the largest times can vary.
Understanding pdfs enables us to compute important statistical measures like expected value and variance, offering a complete picture of probability behavior within samples.

Standard deviation is a measure of the amount of variation or dispersion in a set of values. It tells us how much the values in a data set differ from the mean of the data set.
In our uniform distribution exercise, we calculated the standard deviation of the largest waiting time to grasp its variability. The standard deviation is derived from the variance. Variance itself is the expected value of the squared deviation from the mean. For the largest time's variance, the formula is:

• \( \text{Var}(Y) = E[Y^2] - (E[Y])^2 \)

Where \ E[Y^2] \ can be found through integration. After plugging in our values, the variance is found to be 4, and thus, the standard deviation is \( \sqrt{4} = 2 \).
Standard deviation provides intuitive understanding of spread - for instance, in our case, it tells us how much the largest waiting times flux around their average of 8 minutes.