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Question 103

Interest rate: 6% per year, compounded annually.

Annuity payment: $1,000 payable at the beginning of each month for life, where the annuity is payable at least until an amount equal to the gross single premium has been paid.

Policy load: 6% of the gross premium.

Selected values:

= 3.10 = 2.85

What is the gross single premium for a policy issued at age 65?

(A) Less than $146,000

(B) $146,000 but less than $149,000

(D) $152,000 but less than $155,000

(E) $155,000 or more

Question 104

Assume a uniform distribution of decrement over each interval [x, x+1].

_.3p_35.7 = 0.990

m_36.8 = 0.03

l₃₇ = 100,000

What is l₃₅?

(A) Less than $102,000

(B) $102,000 but less than $105,000

(D) $108,000 but less than $111,000

(E) $111,000 or more

Question 105

m_x is constant for x>0.

_.6q₃₅= 0.018

_.4q_36.6= 0.016

_.8p_37.2 = 0.991

l₃₈ = 100,000

What is l₃₅?

(A) Less than $102,000

(B) $102,000 but less than $105,000

(D) $108,000 but less than $111,000

(E) $111,000 or more

Question 106

Assume a Balducci distribution of decrement over each interval [x, x+1].

m_40.1 = 0.015

_.7p₄₁ = 0.988

l₄₂ = 100,000

What is l₄₀?

(A) Less than $102,000

(B) $102,000 but less than $105,000

(D) $108,000 but less than $111,000

(E) $111,000 or more

Question 107

The following facts are known concerning a retirement plan:

(i) The plan effective date is 1/1/1996.

(ii) There were 5 participants on 1/1/1996, and no new participants have entered the plan since that date.

(iii) 3 of the 5 participants have died, on 1/1/1997, 1/1/1999, and 1/1/2002.

(iv) The plan is subject to a uniform survival distribution over the interval [0,w].

(v) The observation period of the plan has been from inception through 1/1/2003.

What is the maximum likelihood estimate w?

(A) Less than 8.0

(B) 8.0 but less than 9.5

(D) 11.0 but less than 12.5

(E) 12.5 or more

Question 108

Selected values from a select and ultimate table with a two-year select period:

l₂₅ = 100,000 q_[20] = 0.10 ₄p_[21] = 0.82 ₂p_[20]+1 = 0.9 ´ ₂p_[21]

What is l_[20]?

(A) Less than 115,000

(B) 115,000 but less than 125,000

(D) 135,000 but less than 145,000

(E) 145,000 or more

Question 109

s(x) = , 0 £ x £ c

l₀ = 100,000

l₃₆ = 50,000

A = the probability that a life age 20 will die after age 50.

B = the probability that a life age 30 will die between age 55 and age 65.

What is the absolute value of A - B?

(A) Less than 0.30

(B) 0.30 but less than 0.35

(D) 0.40 but less than 0.45

(E) 0.45 or more

Question 110

Terms of a life annuity: $1,000 payable each year on October 1.

Selected values:

i = .07

q_x = 0.05, for x < 100

q_x = 1.00, for x = 100

Deaths are assumed to occur uniformly.

Date of birth of Smith: 1/1/1938

What is the present value of the annuity to Smith as of 1/1/2003?

(A) Less than $8,000

(B) $8,000 but less than $8,050

(D) $8,100 but less than $8,150

(E) $8,150 or more

Solution To Question 103

i⁽¹²⁾/12 = 1.06^1/12 – 1 = .004868 Þ d⁽¹²⁾/12 = = = .004844

Þ d⁽¹²⁾ = .004844 ´ 12 = .058128

Let G be the gross premium. Since the load is 6% of the gross premium, the net premium is equal to 94% (100% - 6%) of the gross premium. The net premium is equal to the present value of the future benefit payments. The equation of value (per dollar of monthly benefit) is:

.94G = + Þ + - .94G = 0

Since values for the deferred annuities are given for deferrals of 12 and 13 years, it is reasonable to assume that it will take between 12 and 13 years for the total annuity value paid to be equal to G.

Substituting G = 12 into the equation of value:

+ - (.94 ´ 12) = + 3.10 – 11.28 = 8.6538 + 3.10 – 11.28 = .4738

Substituting G = 13 into the equation of value:

+ - (.94 ´ 13) = + 2.85 – 12.22 = 9.1378 + 2.85 – 12.22 = -.2322

Using linear interpolation,

G = 12 + = 12.6711

The total gross single premium for the policy is:

$1,000 ´ 12 ´ 12.6711 = 152,053

Solution To Question 104

The following formulas based upon the uniform distribution of decrement assumption can be found in the chart on page 66 of “Survival Models and Their Estimation” by Dick London:

_1-sp_x+s = and m_x+s =

Using the provided data,

_.3p_35.7 = = .99 Þ p₃₅ = .99 ´ (1 - .7q₃₅)

Þ 1 - q₃₅ = .99 - .693q₃₅

Þ q₃₅ = .032573

m_36.8 = = .03 Þ q₃₆ = .03 ´ (1 - .8q₃₆)

Þ q₃₆ = .03 - .024q₃₆

Þ q₃₆ = .029297

l₃₇ = l₃₅ ´ p₃₅ ´ p₃₆ = l₃₅ ´ (1 - q₃₅) ´ (1 - q₃₆)

= l₃₅ ´ (1 - .032573) ´ (1 - .029297) = 100,000 Þ l₃₅ = 106,487

Solution To Question 105

The following formulas based upon a constant force of mortality can be found in the chart on page 66 of “Survival Models and Their Estimation” by Dick London:

_sq_x = 1 – (1 – q_x)^s, _1-sq_x+s = 1 – (1 – q_x)^1-s and _1-sp_x+s = (p_x)^1-s

Using the provided data,

_.6q₃₅ = 1 – (1 – q₃₅)^.6 = .018 Þ .982 = (1 - q₃₅)^.6

Þ .970180 = 1 - q₃₅

Þ q₃₅ = .029820

_.4q_36.6 = 1 – (1 – q₃₆)^.4 = .016 Þ .984 = (1 - q₃₆)^.4

Þ .960479 = 1 - q₃₆

Þ q₃₆ = .039521

_.8p_37.2 = (p₃₇)^.8 = .991 Þ p₃₇ = .988763

l₃₈ = l₃₅ ´ p₃₅ ´ p₃₆ ´ p₃₇ = l₃₅ ´ (1 - q₃₅) ´ (1 - q₃₆) ´ p₃₇

= l₃₅ ´ (1 - .029820) ´ (1 - .039521) ´ .988763 = 100,000

Þ l₃₅ = 108,534

Solution To Question 106

The following formulas based upon the Balducci distribution of decrement assumption can be found in the chart on page 66 of “Survival Models and Their Estimation” by Dick London:

_sp_x = and m_x+s =

Using the provided data,

m_40.1 = = .015 Þ q₄₀ = .015 ´ (1 - .9q₄₀)

Þ q₄₀ = .015 - .0135q₄₀

Þ q₄₀ = .014800

_.7p₄₁ = = .988 Þ 1 - q₄₁ = .988 ´ (1 - .3q₄₁)

Þ 1 - q₄₁ = .988 - .2964q₄₁

Þ q₄₁ = .017055

l₄₂ = l₄₀ ´ p₄₀ ´ p₄₁ = l₄₀ ´ (1 - q₄₀) ´ (1 - q₄₁)

= l₄₀ ´ (1 - .014800) ´ (1 - .017055) = 100,000 Þ l₄₀ = 103,263

Solution To Question 107

The total number of years in the observation period (from 1/1/1996 through 1/1/2003) is 7 years. There were 3 deaths during the observation period. Therefore, the average number of years between deaths is:

7/3 = 2a years

There are 2 participants still alive on 1/1/2003. It is expected that there will be 2a years between each death for the period beginning on 1/1/2003. Therefore, the maximum likelihood estimate w is:

w = 7 + (2 ´ 2a) = 11.6667

Solution To Question 108

₄p_[21] = l₂₅/l_[21] = 100,000/ l_[21] = 0.82 Þ l_[21] = 121,951

₂p_[20]+1 = .9 ´ ₂p_[21] Þ l₂₃/l_[20]+1 = .9 ´ l₂₃/l_[21]

Þ 1/l_[20]+1 = .9 ´ 1/121,951

Þ l_[20]+1 = 135,501

p_[20] = l_[20]+1/l_[20] Þ 0.90 = 135,501/ l_[20] Þ l_[20] = 150,557

Solution To Question 109

s(36) = = Þ c = 108

Recall that s(n) represents the probability of a newborn surviving n years (_np₀).

The probability that a life age 20 will die after age 50 is simply the probability that the life age 20 survives at least 30 years. Therefore,

A = ₃₀p₂₀ = = = = = .5339

The probability that a life age 30 will die between age 55 and 65 is the difference between the probabilities that the life age 30 survives 25 years and that the life survives 35 years. Therefore,

B = ₂₅p₃₀ - ₃₅p₃₀ =

= = .1355

A – B = .5339 - .1355 = .3984

Solution To Question 110

Recall that under the uniform distribution of death assumption,

_sp_x = (1 – s ´ q_x), for s<1.

Also recall the summation formula for a geometric series:

1 + x + x² + … + x^n-1 =

Since q_x = .05 for all ages under 100, p_x = .05 for all ages under 100. In addition, since q₁₀₀ = 1.0, there are no lives surviving at age 101. Therefore, the last age at which an annuity payment could possibly be made is age 100.

The present value of the annuity is:

1,000v^.75 _.75p₆₅ + 1,000v^1.75 _1.75p₆₅ + … + 1,000v^34.75 _34.75p₆₅ + 1,000v^35.75 _35.75p₆₅

= 1,000v^.75 ´ [_.75p₆₅ + (v)(p₆₅)(_.75p₆₆) + … + (v³⁴)( ₃₄p₆₅)(_.75p₉₉) + (v³⁵)( ₃₅p₆₅)(_.75p₁₀₀)]

= 1,000v^.75 ´ [(1 - .75q₆₅) + (v)(p₆₅)(1 - .75q₆₆) + …

+ (v³⁴)( ₃₄p₆₅)(1 - .75q₉₉) + (v³⁵)( ₃₅p₆₅)(1 - .75q₁₀₀)]

= 1,000v^.75 ´ [(.9625) + (v)(.95)(.9625) + … + (v³⁴)(.95³⁴)(.9625) + (v³⁵)(.95³⁵)(.25)]

= (1,000v^.75(.9625) ´ [1 + .95v + (.95v)² + … + (.95v)³⁴])

+ 1,000(v^35.75)(.95³⁵)(.25)

= (1,000v^.75(.9625) ´ ) + 1,000(v^35.75)(.95³⁵)(.25)

= 8,031 + 4

= 8,035