Asked by homework 4 years ago

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.

The lifetime

homework

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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Asked: 4 years ago