Re: DSE 2016 doubt

Posted by Abhitesh on Jun 18, 2017; 4:40pm
URL: http://discussion-forum.276.s1.nabble.com/DSE-2016-doubt-tp7605789p7605812.html

**Check that for uniform distribution [a,b] mean is (a+b)/2

Now,
(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.