soklan macam senang tapi....

L is the loss at issue random variable for a fully discrete whole life insurance of 1 on (x). the annual premium charged for this insurance is 0.044. you are given:

(i) Ax = 0.40

(ii) a double dot x =10

(iii) Var (L) = 0.12

an insurer has a portfolio of 100 such insurances on 100 independent lives. 80 of these insurances have death benefit of 4 and 20 have death benefits of 1. assume that the total loss for this portfolio is distributed normally. calculate the probability that the present value of the gain for this portfolio is greater than 22.

okay..

aku taw ape yg dia nak..

dia nak P(X > 22) 

bila dia ckp assume normally distributed..

so P(X > 22) = P(Z > (22-mean)/standard deviation )

so sebelum dapat probability ikut normal graph tu..

aku kene cari mean ngan variance la dulu..

kan..??

tapi..

mcm mane nk cari mean..??

E(S) = 80x4(__??__) + 20x1(__??__)

cmne pulak nk cari variance..??

Var(S) = (__??__)Var(L) + (__??__)

argh..!!

aku xtaw ape yg patut ade dlm blank tu..!!

hiyargh..!!!

jom bunuh diri same2 nak..??

liana..!!

chill la babe..

okay..

dia  ade kasik Ax ngan a double dot x kan..?

hurm..

daripade 2 bende tu..

ko boleh carik ape..??

yer..

carik la d..

d = (1-Ax)/ a double dot x

tekan calculator..

ko akan dapat d= 0.06

belek la example ke ape..

ko akan jumpe ni..

E(L) = Ax + P a double dot x

so kalau untuk benefits of 1..

kacang je la nk cari E(L)..

sebab dia dah bagi sume..

E(L) = 0.4-0.044(10) = -0.04

kalau benefits of 4..??

kali 4 je ke..??

hurm..

bukan.

kene cari P dia sendiri sebab dia x bagi P ntuk 4..

pastu guna formula yg sama..

E(L) = 4(0.4)- P(10) = __

pastu Var (L) = (1+P/d)^2 * (2kecik Ax - Ax ^2)

bla bla bla..

then aku dapat Var (S) = 11312.704

E(S) = 95.2

masukkan value..

P(Z > (22-95.2)/ sqrt(11312.704))

= P(Z > -0.6822)

= 1 - 0.2547

= 0.7453 #

tapi..

jawapan dia 0.25

grrrrr...

jom bunuh orang yg buat soklan ni..!!