An insurance company has found that 1% of all applicants for life insurance have diabetes.

Calculate the probability that five or fewer of 200 randomly selected applicants have diabetes.

• 0.85
• 0.88
• 0.91
• 0.95
• 0.98

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: E

The number of applicants with diabetes has a binomial distribution and thus

[[math]] \begin{align*} \operatorname{P}(X \leq 5 ) &= \sum_{x=0}^5 \binom{200}{x} (0.01)^x(0.99)^{200-x} \\ &= 0.134 + 0.271 + 0.272 + 0.181 + 0.090 + 0.036 \\ &= 0.984. \end{align*} [[/math]]

A faster solution is to use the Poisson distribution with λ − 200(0.01) =2 as an approximation. Then,

[[math]] \begin{align*} \operatorname{P}(X \leq 5) = \sum_{x=0}^5 \frac{e^{-2}2^x}{x!} &= e^{-2} \left ( \frac{1}{1} + \frac{2}{1} + \frac{4}{2} + \frac{8}{6} + \frac{16}{24} + \frac{32}{120}\right) \\ &= 0.983. \end{align*} [[/math]]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.