Question

A company that sells annuities (a form of insurance or investment entitling the investor to a series of annual sums) must base the annual payout on the probability distributions of the length of life of the participants in the plan. Suppose the probability distribution of the lifetime of the participants is approximately a normal distribution with a mean of 68 year and a standard deviation of 4 years.  

1. The company knows that approximately:

1. 2 of every 3 participants lifespan is between 64 and 72 years
2. 19 of every 20 participants lifespan is between 60 and 76 years
3. Both of the above statement is true
4. We can’t say without taking a sample first

2. The median lifespan of the participant’s in the program is:

1. 68 years
2. 64 years
3. 72 years
4. We can’t say without taking a sample first

3. Approximately the % of participants who will live beyond 76 years is:

1. 5%
2. 2.5%
3. We can’t say without a sample first

4. Approximately, the %of participants who will NOT live up to 60 years is

1. 5%
2. 2.5%
3. We can’t say without a sample first

5. The Z-score of a participant’s life span was Z= -1.33. This implies:

1. The participant’s life span was 1.33 standard deviation below the average lifespan of participant’s in the plan
2. He/she lived up to 62.68 years
3. Both of the above

6. The Z-score of a participant’s lifespan was Z = -1.33. Approximately what % of participants die before this age

1. 10%
2. 90%
3. 81%
4. 9%

Answer 1

1) Option - b) 19 of every 20 participants lifespan is between 64 and 72 years.

2) Option - a) 68 years

3) P(X > 76)

= P((X - )/ > (76 - )/)

= P(Z > (76 - 68)/4)

= P(Z > 2)

= 1 - P(Z < 2)

= 1 - 0.9772

= 0.0228

Option - b) 2.5%

4) P(X < 60)

= P((X - )/ < (60 - )/)

= P(Z < (60 - 68)/4)

= P(Z < -2)

= 0.0228

Option - b) 2.5%

5) Option - c) Both of the above

6) P(Z < -1.33)

= 0.0918 = 9.18%

Option - d) 9%