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.

Step 1: Determine the Proportionality Constant

The lifetime of a machine part has a continuous distribution on the interval [math](0, 40)[/math]. The probability density function (PDF) is given to be proportional to [math](10+x)^{-2}[/math], so we can write it in the form:

[[math]]f(x) = C(10+x)^{-2} \quad \text{for } 0 \lt x \lt 40[[/math]]

and [math]f(x) = 0[/math] otherwise. To find the proportionality constant [math]C[/math], we use the normalization condition that the total probability over the entire support must be equal to 1:

[[math]]\int_{0}^{40} f(x) dx = 1[[/math]]

Substitute the expression for [math]f(x)[/math] and perform the integration:

[[math]]\int_{0}^{40} C(10+x)^{-2} dx = 1[[/math]]

[[math]]C \left[ -(10+x)^{-1} \right]_{0}^{40} = 1[[/math]]

[[math]]C \left[ -(10+40)^{-1} - (-(10+0)^{-1}) \right] = 1[[/math]]

[[math]]C \left[ -\frac{1}{50} + \frac{1}{10} \right] = 1[[/math]]

To simplify the expression in the brackets, find a common denominator:

[[math]]C \left[ \frac{-1 + 5}{50} \right] = 1[[/math]]

[[math]]C \left[ \frac{4}{50} \right] = 1[[/math]]

[[math]]C \frac{2}{25} = 1[[/math]]

Now, solve for [math]C[/math]:

[[math]]C = \frac{25}{2} = 12.5[[/math]]

Step 2: Calculate the Probability [math]P(X \lt 6)[/math]

With the proportionality constant [math]C = 12.5[/math], the probability density function is [math]f(x) = 12.5(10+x)^{-2}[/math]. We need to calculate the probability that the lifetime of the machine part is less than 6. This is found by integrating the PDF from 0 to 6:

[[math]]P(X \lt 6) = \int_{0}^{6} f(x) dx = \int_{0}^{6} 12.5(10+x)^{-2} dx[[/math]]

Perform the integration:

[[math]]P(X \lt 6) = 12.5 \left[ -(10+x)^{-1} \right]_{0}^{6}[[/math]]

[[math]]P(X \lt 6) = 12.5 \left[ -(10+6)^{-1} - (-(10+0)^{-1}) \right][[/math]]

[[math]]P(X \lt 6) = 12.5 \left[ -\frac{1}{16} + \frac{1}{10} \right][[/math]]

To simplify the expression in the brackets, find a common denominator (e.g., 80):

[[math]]P(X \lt 6) = 12.5 \left[ \frac{-5}{80} + \frac{8}{80} \right][[/math]]

[[math]]P(X \lt 6) = 12.5 \left[ \frac{3}{80} \right][[/math]]

Substitute [math]12.5 = \frac{25}{2}[/math]:

[[math]]P(X \lt 6) = \frac{25}{2} \times \frac{3}{80} = \frac{75}{160}[[/math]]

Simplify the fraction by dividing the numerator and denominator by 5:

[[math]]P(X \lt 6) = \frac{15}{32}[[/math]]

As a decimal, this is:

[[math]]P(X \lt 6) \approx 0.46875[[/math]]

Rounding to two decimal places, the probability is approximately [math]0.47[/math].

Key Insights

• For any probability density function (PDF), the total probability over its entire support must integrate to 1. This normalization condition is crucial for determining proportionality constants.
• To calculate the probability of a continuous random variable falling within a specific range, integrate its PDF over that given interval.
• Problems involving proportionality often require an initial step to determine the exact functional form of the PDF, typically by finding a constant of proportionality.
• Familiarity with basic integration techniques, especially for power functions, is essential for solving problems in continuous probability distributions.

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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.