An insurance policy pays 100 per day for up to three days of hospitalization and 50 per day for each day of hospitalization thereafter. The number of days of hospitalization, [math]X[/math], is a discrete random variable with probability function

[[math]] \operatorname{P}[X=k] = \begin{cases} \frac{6-k}{15}, \, k =1,2,3,4,5 \\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

Determine the expected payment for hospitalization under this policy.

• 123
• 210
• 220
• 270
• 367

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

Solution: C

If [math]k[/math] is the number of days of hospitalization, then the insurance payment [math]g(k)[/math] is

[[math]] g(k) = \begin{cases} 100k, \quad k =1,2,3 \\ 300 + 50(k-3), \quad k =4,5. \end{cases} [[/math]]

Thus, the expected payment is

[[math]] \begin{align*} \sum_{k=1}^5 g(k)p_k &= 100 p_1 + 200 p_2 + 300 p_3 + 350 p_4 + 400 p_5 \\ &= \frac{1}{15}(100 × 5 + 200 × 4 + 300 × 3 + 350 × 2 + 400 × 1) \\ &= 220. \end{align*} [[/math]]

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