1. Derive the mean, variance, mg and cumulative distribution function for the uniform.

The uniform distribution is one of the simplest types of probability distributions. In the case of a continuous uniform distribution over the interval \([a, b]\), each outcome in this interval is equally likely. Let's derive the mean, variance, moment-generating function (mgf), and cumulative distribution function (CDF) for the continuous uniform distribution.

### 1. **Probability Density Function (PDF)**

The probability density function for a uniform distribution on the interval \([a, b]\) is given by:

\[
f(x) = \begin{cases} 
\frac{1}{b-a} & \text{if } a \leq x \leq b \\
0 & \text{otherwise} 
\end{cases}
\]

### 2. **Mean (Expected Value)**
The mean \( \mu \) of a uniform distribution is calculated as:

\[
\mu = \mathbb{E}[X] = \int_{a}^{b} x f(x) \, dx
\]

Substitute \( f(x) = \frac{1}{b-a} \):

\[
\mu = \int_{a}^{b} x \cdot \frac{1}{b-a} \, dx
\]

The integral becomes:

\[
\mu = \frac{1}{b-a} \int_{a}^{b} x \, dx
\]

We integrate \( x \) with respect to \( x \):

\[
\mu = \frac{1}{b-a} \left[ \frac{x^2}{2} \right]_{a}^{b} = \frac{1}{b-a} \left(\frac{b^2}{2} - \frac{a^2}{2}\right)
\]

Simplifying:

\[
\mu = \frac{1}{b-a} \cdot \frac{b^2 - a^2}{2} = \frac{1}{b-a} \cdot \frac{(b-a)(b+a)}{2} = \frac{b+a}{2}
\]

So, the mean of the uniform distribution is:

\[
\mu = \frac{a+b}{2}
\]

### 3. **

Explore how to derive the mean, variance, moment-generating function (MGF), and cumulative distribution function (CDF) for the uniform distribution on an interval [a, b]. This mathematical problem delves into probability theory and is suitable for advanced undergraduate students and beyond.

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