For a special whole life insurance policy issued on (40), you are given:

(i) Death benefits are payable at the end of the year of death

(ii) The amount of benefit is 2 if death occurs within the first 20 years and is 1 thereafter

(iii) [math]Z[/math] is the present value random variable for the payments under this insurance (iv) [math]\quad i=0.03[/math]

(v)

(vi) [math]\quad E\left[Z^{2}\right]=0.24954[/math]

Calculate the standard deviation of [math]Z[/math].

• 0.27
• 0.32
• 0.37
• 0.42
• 0.47

For a special 2-year term insurance policy on [math](x)[/math], you are given:

(i) Death benefits are payable at the end of the half-year of death

(ii) The amount of the death benefit is 300,000 for the first half-year and increases by 30,000 per half-year thereafter

(iii) [math]\quad q_{x}=0.16[/math] and [math]q_{x+1}=0.23[/math]

(iv) [math]\quad i^{(2)}=0.18[/math]

(v) Deaths are assumed to follow a constant force of mortality between integral ages

(vi) [math]Z[/math] is the present value random variable for this insurance

Calculate [math]\operatorname{Pr}(Z\gt277,000)[/math].

• 0.08
• 0.11
• 0.14
• 0.18
• 0.21

You are given:

(i) [math]\quad q_{60}=0.01[/math]

(ii) Using [math]i=0.05, A_{60: 31}=0.86545[/math]

Using [math]i=0.045[/math] calculate [math]A_{60: 3}[/math].

• 0.866
• 0.870
• 0.874
• 0.878
• 0.882

For a special increasing whole life insurance on (40), payable at the moment of death, you are given:

(i) The death benefit at time [math]t[/math] is [math]b_{t}=1+0.2 t, \quad t \geq 0[/math]

(ii) The interest discount factor at time [math]t[/math] is [math]v(t)=(1+0.2 t)^{-2}, \quad t \geq 0[/math]

(iii) [math]\quad{ }_{t} p_{40} \mu_{40+t}= \begin{cases}0.025, & 0 \leq t\lt40 \\ 0, & \text { otherwise }\end{cases}[/math]

(iv) [math]Z[/math] is the present value random variable for this insurance

Calculate [math]\operatorname{Var}(Z)[/math].

• 0.036
• 0.038
• 0.040
• 0.042
• 0.044

For a 30 -year term life insurance of 100,000 on (45), you are given:

(i) The death benefit is payable at the moment of death

(ii) Mortality follows the Standard Ultimate Life Table

(iii) [math]\delta=0.05[/math]

(iv) Deaths are uniformly distributed over each year of age

Calculate the [math]95^{\text {th }}[/math] percentile of the present value of benefits random variable for this insurance.

• 30,200
• 31,200
• 35,200
• 36,200
• 37,200

For a 3-year term insurance of 1000 on (70), you are given:

(i) [math]\quad q_{70+k}^{\text {SULT }}[/math] is the mortality rate from the Standard Ultimate Life Table, for [math]k=0,1,2[/math]

(ii) [math]\quad q_{70+k}[/math] is the mortality rate used to price this insurance, for [math]k=0,1,2[/math]

(iii) [math]\quad q_{70+k}=(0.95)^{k} q_{70+k}^{S U L T}[/math], for [math]k=0,1,2[/math]

(iv) [math]i=0.05[/math]

Calculate the single net premium.

• 29.05
• 29.85
• 30.65
• 31.45
• 32.25

For a 25 -year pure endowment of 1 on [math](x)[/math], you are given:

(i) [math]\quad Z[/math] is the present value random variable at issue of the benefit payment

(ii) [math]\operatorname{Var}(Z)=0.10 E[Z][/math]

(iii) [math]{ }_{25} p_{x}=0.57[/math]

Calculate the annual effective interest rate.

• 5.8%
• 6.0%
• 6.2%
• 6.4%
• 6.6%

For a whole life insurance of 1000 on (50), you are given:

(i) The death benefit is payable at the end of the year of death

(ii) Mortality follows the Standard Ultimate Life Table

(iii) [math]i=0.04[/math] in the first year, and [math]i=0.05[/math] in subsequent years

Calculate the actuarial present value of this insurance.

• 187
• 189
• 191
• 193
• 195

You are given:

(i) [math]\quad A_{35: 15}=0.39[/math]

(ii) [math]\quad A_{35: 15}^{1} 0.25[/math]

(iii) [math]\quad A_{35}=0.32[/math]

Calculate [math]A_{50}[/math].

• 0.35
• 0.40
• 0.45
• 0.50
• 0.55

The present value random variable for an insurance policy on [math](x)[/math] is expressed as:

Determine which of the following is a correct expression for [math]E[Z][/math].

• [math]{ }_{10 \mid} \bar{A}_{x}+{ }_{20} \bar{A}_{x}-{ }_{30 \mid} \bar{A}_{x}[/math]
• [math]\bar{A}_{x}+{ }_{20} E_{x} \bar{A}_{x+20}-2{ }_{30} E_{x} \bar{A}_{x+30}[/math]
• [math]{ }_{10} E_{x} \bar{A}_{x}+{ }_{20} E_{x} \bar{A}_{x+20}-2{ }_{30} E_{x} \bar{A}_{x+30}[/math]
• [math]{ }_{10} E_{x} \bar{A}_{x+10}+{ }_{20} E_{x} \bar{A}_{x+20}-2{ }_{30} E_{x} \bar{A}_{x+30}[/math]
• [math]{ }_{10} E_{x}\left[\bar{A}_{x+10}+{ }_{10} E_{x+10} \bar{A}_{x+20}-{ }_{10} E_{x+20} \bar{A}_{x+30}\right][/math]