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The mean of a continuous uniform distribution is like its balancing point. Imagine a seesaw, where the mean is the point where it perfectly balances. In a uniform distribution over the interval \[a, b\], the mean, denoted as \( \mu \), is found as the average of the endpoints of the interval.
For the formula, we use: \[ \mu = \frac{a + b}{2} \]
This formula simply adds the starting point \(a\) and the ending point \(b\) and then divides by 2. It's like finding the midpoint on a line.

• In our problem, with \(a = 1.5\) and \(b = 5.5\), the mean is calculated as \( \mu = \frac{1.5 + 5.5}{2} = 3.5 \).

This tells us that the average value of the random variable \((X)\) is 3.5, located precisely in the middle of the interval.

The variance in a uniform distribution is a measure of how spread out the values are around the mean. Imagine each number in the interval taking its turn jumping around the mean. Variance tells us how energetic these jumps are.
The formula for the variance \( \sigma^2 \) with a uniform distribution is: \[ \sigma^2 = \frac{(b - a)^2}{12} \]
This formula cleverly uses 12 in the denominator, which normalizes the square of the interval's length and accounts for the uniform spread of values.

• For our example where \(a = 1.5\) and \(b = 5.5\), it calculates to: \( \sigma^2 = \frac{(5.5 - 1.5)^2}{12} = \frac{16}{12} = \frac{4}{3} \).

A smaller variance implies numbers are closely packed around the mean, while a larger variance indicates they are more dispersed. Here, the variance of \( \frac{4}{3} \) conveys a moderate spread around the mean of 3.5.

The cumulative distribution function (CDF) is like a step-by-step guide on the probability landscape. It tells you the probability that the random variable \(X\) will assume a value less than or equal to \(x\).
For a continuous uniform distribution between \(a\) and \(b\), the CDF is: \[ F(x) = \begin{cases} 0 & \text{for } x < a, \ \frac{x - a}{b - a} & \text{for } a \leq x \leq b, \ 1 & \text{for } x > b. \end{cases} \]
This function gradually moves from 0 to 1 as \(x\) progresses from \(a\) to \(b\).

• In our scenario, \(F(x)\ = \ \frac{x - 1.5}{4}\ ext{ for values of } x\ ext{ between 1.5 and 5.5}\).

Simply put, the CDF grows as \(x\) travels across the interval, providing a cumulative perspective of the probability.

The probability density function (PDF) represents how likely the random variable is to take on a particular value. Think of it as a guide showing where the variable often visits in its entire journey.
In a continuous uniform distribution, the PDF is constant, meaning every point within the interval \(a\) to \(b\) is equally likely. The formula is: \[ f(x) = \frac{1}{b-a} \]
This equation gives a straight line on a graph, showing no preference for any interval values.

• For the interval \( [1.5, 5.5] \), the PDF becomes \( f(x) = \frac{1}{4} \).

This means every value has the same likelihood of occurring. Essentially, the PDF provides a map indicating that any point is just as probable as another within our defined range.