Sam keeps a record of his work outs over several years. X is the number of times he goes to the gym in a week. He calculates relative frequencies to create the empirical probability distribution below. X 0 1 2 3 4 P(X) 0.1 0.15 0.40 0.25 0.10 For example, for 10% of the weeks he did not go to the gym. For 15% of the weeks he went to the gym once. What is the mean of X (the expected number of times he visits the gym per week in the long run)?

Answer:2.1Step-by-step explanation:We know that for a discrete  random variable X, the expected value of X is given by E(X) = [tex]\sum x_{i}f_{X}(x_{i})[/tex], where the values [tex]x_{i}[/tex] are the different values that can be taked by X, and [tex]f_{X}[/tex] is the probability mass function of X. In other words, if the discrete random variable X can take n different values [tex]x_{1}, x_{2}, \cdots, x_{n}[/tex] with probabilities [tex]p_{1}, p_{2}, \cdots, p_{n}[/tex] respectively, then, the expected value of X is given by E(X) = [tex]x_{1}p_{1} + x_{2}p_{2} + \cdots + x_{n}p_{n}[/tex]So, in this case we have that the expected number of times that Sam visits the gym per week in the long run is given byE(X) = (0)(0.1) + (1)(0.15) + (2)(0.4) + (3)(0.25) + (4)(0.1) = 2.1