17. Suppose an actuary for an i knows that the life expecta popl ary ife Pan of
ID: 3354537 • Letter: 1
Question
17. Suppose an actuary for an i knows that the life expecta popl ary ife Pan of people cu r. rently 30 years of age is 74 years and 77 (s = 8) for women. Asumi distribution is normal in each case following questions (round s decimals, percents to two digits, and the z scores to two otoages to whole years) a. What percentage of current 30 ear-old aditional b. What percentage of these men and women C. The longest living 10% will live to at least men and women will live to the tr retirement age of 65 years? 70 years? will die by age 75 years? 80 years? what age? The longest 20%?Explanation / Answer
A)
FEMALE
i) P(X>65)
1-P(X<65)
1-P(Z<(65-77)/8)
1-P(Z<-1.5)
=1-NORMSDIST(-1.5)
0.9332
ii)P(X>70)
1-P(X<70)
1-P(Z<(70-77)/8)
1-P(Z<-.875)
=1-NORMSDIST(-0.875)
0.8092
MALE
iii) P(X>65)
1-P(X<65)
1-P(Z<(65-74)/7)
1-P(Z<-1.28571)
=1-NORMSDIST(-1.28571)
0.9007
iv) P(X>70)
1-P(X<70)
1-P(Z<(70-74)/7)
1-P(Z<-.57143)
=1-NORMSDIST(-.57143)
0.7161
b)
FEMALE
i) P(X<65)
P(Z<(65-77)/8)
P(Z<-1.5)
=NORMSDIST(-1.5)
0.0668
ii) P(X<70)
P(Z<(70-77)/8)
P(Z<-.875)
=NORMSDIST(-0.875)
0.1908
MALE
iii) P(X<65)
P(Z<(65-74)/7)
P(Z<-1.28571)
=NORMSDIST(-1.28571)
0.0993
iv) P(X<70)
P(Z<(70-74)/7)
P(Z<-.57143)
=NORMSDIST(-.57143)
0.2839
c)
longest living 10%
i) female
P(Z>z) 10%
p(z<z)= 0.9
z= =NORMSINV(0.9)
z= 1.281551566
(X-mean)/sd= 1.281551566
X= =1.281552*8+77
X= 87.252416
X= 87
ii) male
P(Z>z) 10%
p(z<z)= 0.9
z= =NORMSINV(0.9)
z= 1.281551566
(X-mean)/sd= 1.281551566
X= =1.281552*7+74
X= 82.970864
X= 83
longest living 20%
iii) female
P(Z>z) 20%
p(z<z)= 0.8
z= =NORMSINV(0.8)
z= 0.841621234
(X-mean)/sd= 0.841621234
X= =.841621234*8+77
X= 83.73296987
X= 84
iv) male
P(Z>z) 20%
p(z<z)= 0.8
z= =NORMSINV(0.8)
z= 0.841621234
(X-mean)/sd= 0.841621234
X= =.841621234*7+74
X= 79.89134864
X= 80
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