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Suzie has a tetrahedral die with numbered faces that she uses to grade your exam

ID: 3642579 • Letter: S

Question

Suzie has a tetrahedral die with numbered faces that she uses to grade your exam She tosses the die twice. The first toss is the first digit of your score, and the second toss is the second digit of your score. Your score is then doubled. The exam is graded on a 90/80/70/60 scale, so even though the highest possible score on the exam is an 88, a student who receives an 88 will still not get an A. To make matters worse, the die is loaded so that p(1) = 4, p(2) = .35. and p(3) =2. Complete the following table. Find the probability that a randomly chosen student will pass the exam Find the mean score on the exam.

Explanation / Answer

a) First toss is biased but second toss is fair.

P(score) = P(1st digit biased) x P(2nd digit biased)

P(1)= 0.4, P(2)= 0.35, P(3) = 0.2, P(4) = 0.05

Score      Digits On Die         Probability                        

22                1,1              P(1)xP(1) = .4 x .4 = .16           

24                 1,2                        0.14                        

26                1,3                        0.08                                  

28                1,4                        0.02                                  

42                  2,1                      .35 x0.4 = 0.14             

44                 2,2                       0.1225                            

46                 2,3                       0.07                            

48                  2,4                       0.0175                            

62                  3,1                        0.08

64                  3,2                        0.07

66                  3,3                        0.04

68                  3,4                        0.01

82                  4,1                       0.02

84                  4,2                       0.0175

86                  4,3                       0.01

88                  4,4                      0.0025

So you can fill out the table with score and its probability.

b) Probability that a student will pass

= P(X>=60(If that's the passing marks)) = P(62) +P(64) +P(66)....+P(88) = 0.3075

I think the passing marks are 60. If it is different, then also you can calculate the probabilty. It's just the sum.

c) mean score = score x P(score)     for score = 22,24,...88

= .16 x 22 + .14 x 24 + .08x26 +.02 x28 +...... .0025 x 88

= 41.8

If you have any problem in understanding ant step then ask me.