Answer:
E) 0.99
Step-by-step 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
