The lifetime of a machine part has a continuous distribution on the interval (0, 40) with probability density function [math]f(x)[/math], where [math]f(x)[/math] is proportional to [math](10+x)^{-2}[/math] on the interval.

Calculate the probability that the lifetime of the machine part is less than 6.

• 0.04
• 0.15
• 0.47
• 0.53
• 0.94

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

Solution: C

We know the density has the form [math]C (10 + x )^{-2}[/math] for [math]0 \lt x \lt 40 [/math] (equals zero otherwise). First, determine the proportionality constant [math]C[/math] from the condition [math]\int_{0}^{40} f(x) dx = 1 [/math]:

[[math]] 1 = \int_{0}^{40} C(10+x)^{-2} dx = -C(10+x)^{-1} \Big |_0^{40} = \frac{C}{10} - \frac{C}{50} = \frac{2}{25}C [[/math]]

so [math]C = 25/2 [/math], or 12.5. Then, calculate the probability over the interval (0, 6):

[[math]] 12.5 \int_0^6 (10+x)^{-2} dx = -(10 + x)^{-1} \Big |_0^6 = (\frac{1}{10} - \frac{1}{16})(12.5) = 0.47. [[/math]]

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