91Ó°ÊÓ

We are given the probability mass function \(f(x) = \frac{1}{4}\), for \(x = 0, 1, 2, 3\). This means that the probability for each value of x is \(\frac{1}{4}\).

The cumulative distribution function, F(t), is the sum of probabilities for values of x less than or equal to t. Mathematically, it can be expressed as: \[F(t) = P(X \leq t) = \sum_{x=0}^{t}f(x)\] For each possible value of t, substitute the value of f(x) from the given PMF, and calculate the sum of probabilities: 1. If \(t < 0\), \(F(t) = 0\) (as there are no values of x less than 0) 2. If \(0 \leq t < 1\), \(F(t) = f(0) = \frac{1}{4}\) 3. If \(1 \leq t < 2\), \(F(t) = f(0) + f(1) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\) 4. If \(2 \leq t < 3\), \(F(t) = f(0) + f(1) + f(2) = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}\) 5. If \(t \geq 3\), \(F(t) = f(0) + f(1) + f(2) + f(3) = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = 1\)

To sketch the graph of F(t), plot the values of F(t) we calculated in Step 2, and connect the points with horizontal lines. Note that this graph will look like a staircase, with steps at t = 0, 1, 2, and 3: 1. For \(t < 0\), F(t) = 0 2. For \(0 \leq t < 1\), F(t) = \(\frac{1}{4}\) 3. For \(1 \leq t < 2\), F(t) = \(\frac{1}{2}\) 4. For \(2 \leq t < 3\), F(t) = \(\frac{3}{4}\) 5. For \(t \geq 3\), F(t) = 1 Remember to also label the axes and the points on the graph. The x-axis should be labeled t, and the y-axis should be labeled F(t). The points on the graph should be labeled with their corresponding values of F(t). For example, the point where t = 1 should be labeled with F(t) = \(\frac{1}{2}\).