# The lifetime of a machine part has a continuous distribution on the interval (0, 60) with probability density function f

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The lifetime of a machine part has a continuous distribution on the interval (0, 60) with probability density function f, where f(x) is proportional to (10 + x) -2. Calculate the probability that the lifetime of the machine part is less than 21.

homework

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First we define our probability density function, which is on the interval (0,60) and proportional to $(10+x)^{-2}$. It also must be equal to 1 since the sum of probabilities over a probability space is always equal to 1. And so we have,

$$\int_0^{60} k(10+x)^{-2}dx=1$$

$$\rightarrow k \int_0^{60} (10+x)^{-2}dx=1$$

$$\rightarrow k (-(10+x)^{-1}) \big|_0^{60}=1$$

$$\rightarrow -k\bigg(\frac{1}{10+60}-\frac{1}{10+0}\bigg)=1$$

$$\rightarrow \frac{3k}{35}=1 \rightarrow \boxed{k=\frac{35}{3}}$$

now we have the proportional constant k=35/3, we can find the probability that the lifetime of the machine part is less than 21. This means integrating from 0 to 21 over the function $\frac{35}{3}(10+x)^{-2}$,

$$\int_0^{21} \frac{35}{3}(10+x)^{-2}dx$$

$$\rightarrow \frac{35}{3}(-(10+x)^{-1})\big|_0^{21}$$

$$\rightarrow -\frac{35}{3}\bigg(\frac{1}{10+21}-\frac{1}{10+0}\bigg)$$

$$\approx 0.79$$

Hence the probability that the lifetime of the machine part is less than 21 is 0.79

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