Question

Latent learning is a type of learning which is not apparent in the learner's be- haviour at the time of learning, but which manifests later when a suitable mo- tivation and circumstances appear. In his experiments, Tolman placed hungry rats in a maze with no reward for finding their way through it. He also studied a comparison group that was rewarded with food at the end of the maze. Suppose
Latent learning is a type of learning which is not apparent in the learner’s be- haviour at the time of learning, but which manifests later when a suitable mo- tivation and circumstances appear. In his experiments, Tolman placed hungry rats in a maze with no reward for finding their way through it. He also studied a comparison group that was rewarded with food at the end of the maze. Suppose that the number of days that a rat spends in a maze is a random variable with probability mass function : f(x)=kx+1, x=0,1,2,3,4. 25 1. Find the value of k that makes f a probability mass function. Graph f. 2. What is the expected number of days a rat will spend in the maze? 3. Find and graph the cumulative distribution of the number of days a rat will spend in the maze. Conditional on the fact that a rat spends at most 3 days in the maze, what is the probability that a rat spends two days in? 4. Suppose that a rat is to receive from this researcher a food that costs 10 dollars for each of the first two days spent in the maze and 5 dollars for each day after the first two days, what is the expected expenditure of this researcher?

Answer 1

Page do Son ter Number or a maze days that a rat spends 1) & fcal) Probability mass Reenchon (PMF) of 3 is F(0) kati x=0,1,2,3,4 25 1 poperty of PMI E fmi) 11 s (kt K*0+1 25 + KAI 25 + 2k + 25 + 311 25 4 +1 28 +4** ) 1+ k +1 +2K+ + 3let! +48 +! 25 25-5 10 Ik =2 11 is i The PmF Teal CP 21 23+ 25 2-0, 1.23.4. 2) Expected number OF spend the maze days afat 111 is given as Ea) Elifrar) 0 * 1 f 25 * 3 25 3*7 25 + 2 * + 25 + 9 + 9 25

Dete te 2.8 ~ 3 25 E (0) 3 days The crepected number of days a rod. 10911 Spend in the maze 2 0 91 frag 3 2 4 125 3/25 15/25 (1/259/25 Now, cumulative distribution function of a Y given as 91 0 2. 3 4 F(X)/5 4/259/25 16/25 25/25 We have to find the probability that arat Spends two days to maze YH Spends at 3 days in mare P [ 9 = 2/2 539 Pon=2, ac 3 Prac3) pourded most ie P (932) Plac) 5/25 16/25 PL=2]M43) = 0.3125

Dale 4) Espeoled spend Mure of the researcher BP 10* (FCO) + C) + 5* (FC2J + FCB) + F(4) 10* f 44751 + 5* (25) 40f 105 25 5.8 . The expected eapenditure or rats 5.8 dollars. of the researcher