This problem describes a uniform probability distribution for a random variable $X$, which represents the output of a random-number generator that generates numbers between 0 and 1.

(a) Identify the graph of the uniform density function.

Since the random variable $X$ follows a uniform probability distribution between 0 and 1, the probability density function (PDF) for this uniform distribution is constant over the interval [0, 1].

The uniform distribution's PDF for $X$ is given by:

\begin{cases} 1, & 0 \leq x \leq 1 \\ 0, & \text{otherwise} \end{cases}$$ Graphically, this is a horizontal line at height 1 from $$x = 0$$ to $$x = 1$$, forming a rectangle over the interval [0, 1]. ### (b) What is the probability of generating a number between 0.53 and 0.95? In a uniform distribution, the probability is simply the length of the interval over the total range. So the probability of generating a number between 0.53 and 0.95 is the length of this interval: $$P(0.53 \leq X \leq 0.95) = \text{length of the interval} = 0.95 - 0.53 = 0.42$$ Thus, the probability of generating a number between 0.53 and 0.95 is 0.42. ### (c) What is the probability of generating a number greater than 0.83? Similarly, the probability of generating a number greater than 0.83 is the length of the interval from 0.83 to 1: $$P(X > 0.83) = 1 - 0.83 = 0.17$$ Thus, the probability of generating a number greater than 0.83 is 0.17. --- Do you need any further explanation or details? Here are 5 related questions: 1. What is the probability of generating a number less than 0.25? 2. How do we calculate the expected value for a uniform distribution? 3. What is the cumulative distribution function (CDF) for a uniform distribution? 4. How does the shape of the density function change if the uniform distribution is on an interval [a, b] instead of [0, 1]? 5. What is the variance of the uniform distribution on [0, 1]? **Tip**: In a uniform distribution, probabilities are simply the relative lengths of intervals within the total range.