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 that jackson goes to the gym is 2.05. the probability that he goes 0 days is 0.1, the probability that he goes 2 days is , and the probability that he goes 3 days is .

For a probability distribution the expected value is the summation of product of probabilities with their respective data values. Let x be the probability that Jackson goes gym for 2 days and y be the probability that he goes gym for 3 days.

For the given case we have following values and their probabilities:

0 : 0.1

2 : x

3 : y

So the expected value will be = 0(0.1) + 2(x) + 3(y)

Expected value is given to be 2.05. So we can write the equation as:

2x + 3y = 2.05 (Equation 1)

Also for a probability distribution, the sum of probabilities must always equal to 1. So we can set up the second equation as:

0.1 + x + y = 1

x + y = 0.9 (Equation 2)

From Equation 2 we can write the value of x to be x = 0.9 - y. Using this value in equation 1, we get:

2(0.9 - y) + 3y = 2.05

1.8 - 2y + 3y = 2.05

1.8 + y = 2.05

y = 0.25

Using the value of y in equation 2 we get value of x to be 0.65

Therefore we can conclude that:

The probability that Jackson goes to gym for 2 days is 0.65 and the probability that he goes to gym for 3 days is 0.25

He goes to the gym 0, 2, or 3 days per week.  

Let's assume P(0 days)=p, P(2 days)=q, and P(3 days) = r.

So, p+q+r = 1

E(X)= E(x=0)*P(0 days) + E(x=2)*P(2 days) + E(x=3)*P(3 days)

E(X) = 0*p + 2*q + 3*r = 2q + 3r

Given is the value of P(0 days) = p = 0.1

Given is the expected value of number of days is E(X) = 2.05

Therefore, we have 0.1 + q + r = 1 and 2q + 3r = 2.05

Or, q + r = 0.9 and 2q + 3r = 2.05

Solving the two equations :-

(2q+3r)-2*(q+r) = 2.05-2(0.9)

r = 0.25

q = 0.65

Therefore, P(2 days) = 0.65 and P(3 days) = 0.25

one that is bigger than 3 1/2

found it xd it ok yt