Jackson goes to the gym 0, 2, or 3 days per week, depending on work demands. The expected value of the number of days per week t

Jackson attends the gym either 0, 2, or 3 days weekly. Let’s denote P(0 days)=p, P(2 days)=q, and P(3 days)=r. Hence, the equation p+q+r equals 1. The expectation formula gives us E(X) = E(x=0)*P(0 days) + E(x=2)*P(2 days) + E(x=3)*P(3 days) or E(X) = 0*p + 2*q + 3*r = 2q + 3r. We're given P(0 days) = p = 0.1 and E(X) = 2.05, establishing the equations 0.1 + q + r = 1 and 2q + 3r = 2.05. These can be rearranged to q + r = 0.9 and 2q + 3r = 2.05. By solving these equations: (2q + 3r) - 2*(q + r) = 2.05 - 2(0.9) simplifies to r = 0.25, leading to q = 0.65. Thus, we find that P(2 days) = 0.65 and P(3 days) = 0.25.

In a probability distribution, the expected value is computed by summing the products of probabilities and their corresponding data values. Denote x as the probability of Jackson attending the gym for 2 days and y as the probability for 3 days. The probabilities and their values are as follows: 0 days: 0.1, 2 days: x, and 3 days: y. Therefore, the expected value can be expressed as: 0(0.1) + 2(x) + 3(y). We're given that the expected value is 2.05, leading to the equation 2x + 3y = 2.05 (Equation 1). Additionally, the total probability must equal 1, allowing us to set the second equation as: 0.1 + x + y = 1, which simplifies to x + y = 0.9 (Equation 2). By rearranging Equation 2, we substitute x as 0.9 - y into Equation 1, yielding: 2(0.9 - y) + 3y = 2.05, simplifying to 1.8 - 2y + 3y = 2.05, hence 1.8 + y = 2.05, inferring y = 0.25. Substituting this back into Equation 2 leads to x = 0.65. Finally, we conclude that the probability of Jackson going to the gym for 2 days is 0.65, while for 3 days is 0.25.