**Check that for uniform distribution [a,b] mean is (a+b)/2
(22) If the student takes third test her expected score is 50.
If the student takes 2nd test she scores >50 (mean=75) with prob=1/2 and less than 50 with prob=1/2.
If she scores >50 she stops otherwise opt for third test.
So, expected score before she takes second test will be 0.5*75 + 0.5*50 = 62.5.
Now, the student will take second test only if she scores <62.5 (prob=5/8) in the first and stops if she scores >62.5 (mean = 81.25; prob = 3/8).
So, expected score before first test = 3/8*81.25 + 5/8*62.5 ~ 70.
(23) In this case she can take third test if her score in second <40. But expected score remains 50.
If the student takes second test she scores > 40 (mean=70 prob =3/5) and <40(prob=2/5)
Expected score before second test is 3/5*70 + 2/5*50 = 62.
So she will take second test if her score on first < 62.
Let x axis(0-20) represent hours and y-axis utility. Take 0-end of x-axis as student 1(S1) origin and 20-end as S2 origin.
Now plot utility of S1 and S2. Note that if S1 is assigned < 2hr or >12 hrs. U1 and U2 both increase (<2) or decrease (>12).
U1 and U2 can be raised by given more work to S1 (<2 hr case) or less work (>12 hr). So these allocations are not PO.
If 2<=S1<=12. One increases and other decreases. So these will be PO because if you change the allocation then at least one agent will be worse off.