The distribution of the size of claims paid under an insurance policy has probability density function

[[math]] f(x) = \begin{cases} cx^a, \, 0 \lt x \lt 5 \\ 0, \, \textrm{otherwise} \end{cases} [[/math]]

Where [math]a \gt 0[/math] and [math]c \gt 0 [/math]. For a randomly selected claim, the probability that the size of the claim is less than 3.75 is 0.4871.

Calculate the probability that the size of a randomly selected claim is greater than 4.

• 0.404
• 0.428
• 0.500
• 0.572
• 0.596

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

Solution: B

Because the density function must integrate to 1,

[[math]] 1 = \int_0^5 cx^a dx = c \frac{5^{a+1}}{a+1} \Rightarrow \frac{a+1}{5^{a+1}}. [[/math]]

From the given probability,

[[math]] \begin{align*} 0.4871 &= \int_{0}^{3.75} cx^a dx = c \frac{3.75^{a+1}}{a+1} = \frac{a+1}{5^{a+1}}\frac{3.75^{a+1}}{a+1} = \left( \frac{3.75}{5}\right)^{a+1} \\ \ln(0.4871) &= -0.71929 = (a + 1) \ln(3.75 / 5) = -0.28768(a+1) \\ a &= (−0.71929) / (−0.28768) − 1 =1.5. \end{align*} [[/math]]

The probability of a claim exceeding 4 is,

[[math]] \int_4^5 cx^a dx = c \frac{5^{a+1} - 4^{a+1}}{a+1} = \frac{a+1}{5^{a+1}} \frac{5^{a+1} - 4^{a+1} }{a+1} = 1- \left(\frac{4}{5}\right)^{1.5 + 1} = 0.42757. [[/math]]

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