Re: [商管] [統計]-東吳98-財務工程與精算數學-貝ꐠ…

看板Grad-ProbAsk作者 (IceSnowSmart)時間16年前 (2009/11/11 22:31), 編輯推噓2(205)
留言7則, 2人參與, 最新討論串2/2 (看更多)
: 1. : There are two types of driver. Good driver makes 75% of the population and : in one year have zero claims with probability 0.7, one claim with probability : 0.2, and two claims with probability 0.1. Bad drivers makes up 25% of the : population and have zero, one, or two claims with probability 0.5, 0.3, 0.2, : respectively. For a particular policyholder, suppose we have observed zero : claim in year 1 and one claim in year 2. : 1) What is the probability that the policyholder is a good driver? Pr(所求)=0.75*0.7*0.2/[0.75*0.7*0.2+0.25*0.5*0.3] =14/19 : 2) What is the probability that the policyholder makes no claims next year? Pr(所求)=14/19*0.7+5/19*0.5=123/190 : 3) What is the expected number of claims next year? What is the Variance? Pr(0 claim)=123/190 Pr(1 claim)=14/19*0.2+5/19*0.3=43/190 Pr(2 claim)=14/19*0.1+5/19*0.2=24/190=12/95 E(X)=[43/190]*1+[12/95]*2=91/190 Var(X)=123/190*[91/190]^2+43/190*[99/190]^2+12/95*[289/190]^2=0.5021 訂正了變異數 第一次算沒算到0 claim的 -- 參考答案 歡迎討論 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 59.120.53.85

11/12 00:30, , 1F
第二小題的作法是假設good driver 但題目好像沒指定~
11/12 00:30, 1F

11/12 00:34, , 2F
是我想錯囉 SORRY~~
11/12 00:34, 2F

11/12 00:38, , 3F
變異數 43/190+4*24/190-(91/190)^2=0.5021
11/12 00:38, 3F

11/12 00:42, , 4F
這一題好酷 因為加了條件 讓原本好司機跟壞司機機率變了
11/12 00:42, 4F

11/12 09:03, , 5F
但是題目中的 policyholder 是只 zero claim in one year,
11/12 09:03, 5F

11/12 09:05, , 6F
one claim in year 2, 但題目中的 0.2 不是指 one year
11/12 09:05, 6F

11/12 09:06, , 7F
所以是直接用0.2就可以嗎?
11/12 09:06, 7F
※ 編輯: alice90426 來自: 140.114.32.87 (11/12 13:11)
文章代碼(AID): #1A-igote (Grad-ProbAsk)
文章代碼(AID): #1A-igote (Grad-ProbAsk)