Question

A city has just added 100 new female recruits to its police force. The city will provide a pension to each new hire who remains with the force until retirement. In addition, if the new hire is married at the time of her retirement, a second pension will be provided for her husband. A consulting actuary makes the following assumptions: (i) Each new recruit has a 0.4 probability of remaining with the police force until retirement. (ii) Given that a new recruit reaches retirement with the police force, the probability that she is not married at the time of retirement is 0.25. (iii) The events of different new hires reaching retirement and the events of different new hires being married at retirement are all mutually independent events. Calculate the probability that the city will provide at most 90 pensions to the 100 new hires and their husbands. (A) 0.60 (B) 0.67 (C) 0.75 (D) 0.93 (E) 0.99

Answer 1

E) 0.99

Explanation:

100 recruits x 0.4 chance of retiring as police officer = 40 officers

probability of being married at time of retirement = (1 - 0.25) x 40 = 30 officers

each new recruit will result in either 0, 1 or 2 new pensions

• 0 pensions when the recruit leaves the police force (0.6 prob.)
• 1 pension when the recruit stays until retirement but doesn't marry (0.1 prob.)
• 2 pensions when the recruit stays until retirement and marries (0.3 prob.)

mean = µ = E(Xi) = (0 x 0.6) + (1 x 0.1) + (2 x 0.3) = 0.7

σ² = (0² x 0.6) + (1² x 0.1) + (2² x 0.3) - µ² = 0 + 0.1 + 1.2 - 0.49 = 0.81

in order for the total number of pensions (X) that the city has to provide:

the normal distribution of the pension funds = 100 new recruits x 0.7 = 70 pension funds

the standard deviation = σ = √100 x √σ² = √100 x √0.81 = 10 x 0.9 = 9

P(X ≤ 90) = P [(X - 70)/9] ≤ [(90 - 70)/9] = P [(X - 70)/9] ≤ 2.22

z value for 2.22 = 0.9868 ≈ 0.99