Question

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 1

2.1

Explanation:

We know that for a discrete random variable X, the expected value of X is given by E(X) =
`\sum x_(i)f_(X)(x_(i))`, where the values
`x_(i)` are the different values that can be taked by X, and
`f_(X)` is the probability mass function of X. In other words, if the discrete random variable X can take n different values
`x_(1), x_(2), \cdots, x_(n)` with probabilities
`p_(1), p_(2), \cdots, p_(n)` respectively, then, the expected value of X is given by E(X) =
`x_(1)p_(1) + x_(2)p_(2) + \cdots + x_(n)p_(n)`

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 by

E(X) = (0)(0.1) + (1)(0.15) + (2)(0.4) + (3)(0.25) + (4)(0.1) = 2.1