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p review solutions 3rd ed fall 2007

# p review solutions 3rd ed fall 2007 - Probability Third...

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Sheet1 Page 1 Probability, Third Edition By David J Carr & Michael A Gauger Published by BPP Professional Education Solutions to review questions Solution 1 Answer: A This topic is covered in Chapter 7. Claim amounts follow a lognormal distribution: We need to calculate the parameter values: . . +( )=102,881 Equation 2 make solving the equations easier. Squaring Equation 1 and substituting into Equation 2: 2 102,881 2 .102,881 . (exps ) -= .s=ln +1 =0.99841 ~log N(,s2) .ln X ~ N(h,s2) X h exp .h+s.=245 Equation 1 exp(2 hs2 ) (exp s2) -1 These formulas are given in Section 7.4. Here, we have pulled out a factor of exp(2 h+s) from the variance expression to

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Sheet1 Page 2 245 .245 . We require Pr( X >300) : Pr( X >300) =Pr(ln X >ln 300) .ln X -5.0021 ln 300 -5.0021 . > . . . 0.99841 0.99841 . Err:509 Probability Solutions to review questions Solution 2 .. Answer: C This topic is covered in Chapter 1. .. The definition of exhaustive events is in Section 1.2. Since they are exhaustive events, we have: Pr( ) Pr( ) 1A A B'+ n Err:520 But: Pr( ) Pr( ) Pr( ) 0.65 A B B A B' n Err:520 - n Err:520 So: Pr( ) 0.35A = Alternatively, we could show this situation on a Venn diagram and calculate the probability from there: A B S 0.65 0.15 0.2 h=5.0021
Sheet1 Page 3 .. .. .. .. Solutions to review questions Probability Solution 3 Answer: A This topic is covered in Chapter 9. We are using the method of distribution functions here. Examples of this method can be found in Section 9.2. Note that Y =1.07 X . Consider () = Y <y): Fy Pr( . y . () =Pr( <y) =Pr(1.07 X <y) =Pr X Fy Y .< . . We can calculate this by integrating the probability density function over the appropriate range: y y 32 .32( x +4) . .y . 16 1.07 dx =. . h BPP Professional Education

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Sheet1 Page 4 -16 +4 11 Err:510 .. .(x +4) 3 .-2 . (y 4 . + 1.07) 0 .. Hence: Fy() 1 Err:520 (y +4.28) 2 We need to differentiate this to get the probability density function: 2 18.3184 36.6368 fy() F () == (y +4.28) 3 (y +4.28) 3 Solution 4 Answer: D This topic is covered in Chapter 2. If the order of allocation matters we need to use permutations. This can be found in Section 2.2. We need to allocate the difficult case first. It can be allocated to any of the three experts. There are three ways of doing this. Now we have three claims to be allocated to the remaining nine people. This can be done in the following
Sheet1 Page 5 number of ways: 9! P ==504 6! The total number of ways that the claims can be allocated is: 3 504 =1, 512 Probability Solutions to review questions Solution 5 Answer: B This topic is covered in Chapters 4 and 6. The probability density function given in the question is that of the Weibull distribution, which can be found in Section 6.6. in Section 4.4. We need ( ) Fx : x x 8 t . t . -x h BPP Professional Education You donht actually need to know the distribution in order to answer the question. The definition of the interquartile range is

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Sheet1 Page 6 () = te dt =. Fx - e . 1 e Err:520 . 32 .8 32 . 0 .. The lower quartile, L , is such that: LL -- 256 256 0.25 1 e .e =0.75 .L =1.71157 Err:520 The upper quartile, U , is such that: UU -- 256 256 0.75 1 e .e =0.25 .U =2.08335 Err:520
Sheet1 Page 7 So the interquartile range is 2.08335 -1.71157 =0.3718 .

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p review solutions 3rd ed fall 2007 - Probability Third...

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