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(ISOM111)[2007](f)midterm~2546^_10367.pdf
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Solution (2007)
1. (a) The score distribution of Group I has longer tail to the left (skew to the left).
(b) (i) The mean of Group II is lower because of one outlier (there is one student scored 0, probably due to cheating or absence from the exam). If exclude the outlier Group II has mean 78.63, which is higher than that of Group I.
(ii) The upper quartile 88 (Group II) > 84.5 (Group I)
(iii) The median 80 > 76
(iv) Except for the outlier no one scored below 60 in Group II, while there were three students scored below 60 in Group I.
2. (a) The prior probabilities in the problem is the probabilities that an email message is junk (0.3) or not (0.7) before filtered.
(b)
0.30.9 + 0.7 0.05 = 0.305

(c)
0.3 0.9 / 0.305 = 0.885

(d)
(b) is the empirical probability, the probability of an event that can be observed; and (c) is the posterior probability, the conditional probability of the true state of the email given the observable event.


3. (a) They are subjective probabilities because they represent personal degrees of belief that the project will be successful.
(b)
2 0.7 . 0.5 + 2.5 0.5 . 0.7 = 1.45

(c)
(1.8 . 0.55)2 0.5 + (.0.7 . 0.55)2 0.5=1.5625


4. (a) No, the probability of winning is not the same for all matches.
3 21
(b)
C2 (0.7) (0.3) = 0.441

(c)
E(X ) = np = 5 0.65 = 3.25 ,2 = npq = 5 0.65 0.35 = 1.138


32 5
(d) CC / C = 0.6 .
21 3
(e) P(W = 2) = P(H = 2,C = 0) + P(H = 1,C = 1) + P(H = 0,C = 2) = 0.01643
5. (a) P{X< 69} = P{Z < (69 . 65) / 3} = P(Z < 1.33) = 0.5 + 0.4082 = 0.9082
(b)
P(Y < h) = 0.85 . z . score = 1.04 , h = 63 +1.04 2.5 = 65.6

(c)
P(60.9 < Y < 65.1) = P(.0.84 < Z < 0.84) = 0.599 , = 100 0.599 = 59.9


=

npq = 4.9 , P(56.5 B 63.5) = P(.0.69 < Z < 0.73) = 0.522
(d)
E(X . Y ) = 65 . 63 = 2, var(X. Y ) = var(X ) + var(Y ) = 32 + (2.5)2 = 15.25

(e)
P(X . Y > 0) = P(Z >.0.51) = 0.695