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# 08ffx - Marks Question 1 x 0 1 2 3 4 5 Mean = P(x(x-mean)^2...

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Marks Question 1 x P(x) xP(x) (x-mean)^2 * P(x) 0 0.1 0 0.73 1 0.11 0.11 0.32 2 0.2 0.4 0.1 3 0.23 0.69 0.02 4 0.3 1.2 0.51 5 0.06 0.3 0.32 Mean = 2.7 Var = 1.99 6 (a) Mean 2.7 Var 1.99 SD 1.41 6 (b) Mean 5.4 Var 3.98 SD 1.19 8 © Mean 5.4 Corr = 0.52 Var 6.05 SD 2.46 Marks Question 2 6 (a) Poisson x x! lambda P(x) 0 1 8.15 0 1 1 8.15 0 2 2 8.15 0.01 P(<3) = 0.01 2 (b) Last week was exceptional compared to the recent year, since the proba 6 © HyperG N= 35 S= 10 n= 3 x= 1 P(x=1) = CSx * C(N-S)(n-x) / CNn 0.46 6 (d) Binomial p= 0.3

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n 10 x 2 P(x=2) = Cnx p^x q^(n-x) 0.24 Assumption: Nova Scotia is typical of Canada with respect to these accid Max_MarksQu.# Sub N.B. This Excel File should be printed out in the Landscape or Qu.#3 Question: 3 a. P(P) = 0.150 P(A|P) = 0.100 P(R) = 0.850 P(A|R) = 0.025 P(A) = P(P)*P(A|P)+P(R)*P(A|R) P(A) = 0.04 0.96 3 b. Although this is an application of Bayes' Theorem, it can be d Simple Conditional Probability with the answer found in part 'a P(P|A) = {P(P) * P(A|P)}/ P(A) P(P|A) = 0.41 2 c. Similarly, P(R|A) = {P(R) * P(A|R)}/ P(A) 0.59 Actually, P(R|A) = 1- P(P|A) Hence, 0.59 3 d. 1 - (1 - 0.1) 0.271 2 e. (1-0.025)^3 0.93 2 f. 0.07 0.271/0.073 3.71 Premium = P(A C ) = P(R|A) = P(R|A) = P{[(A|P) 3 ]>= 1} = 1 - {P(A|P) 3 = 0} = P{[(A|P) 3 ]>= 1} = P{[(A|R) 3 ] = 0} = {1 - P(A|R)} 3 = P{[(A|R) 3 ] = 0} = P{[(A|R) 3
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08ffx - Marks Question 1 x 0 1 2 3 4 5 Mean = P(x(x-mean)^2...

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