Answer to Question #119288 in Calculus for Nharnha

Question #119288
The lifetime of a machine is continuous on the interval (0, 40) with probability density
function f, where f(t) is proportional to (t+ 10)−2
, and t is the lifetime in years. Find
the proportionality constant that makes the f(t) a probability function.
1
Expert's answer
2020-06-01T16:11:49-0400

By the definition, f(t)=(t+10)2f(t)=(t+10)-2 is the probability density function if



+f(t)dt=1040(t+102)dt==040(t+8)dt=C(t22+8t)040=C(16002+320)=1C1120=1C=11120\int\limits_{-\infty}^{+\infty}f(t)dt=1\to \int\limits_0^{40}\left(t+10-2\right)dt=\\[0.3cm] =\int\limits_0^{40}\left(t+8\right)dt=C\cdot\left.\left(\frac{t^2}{2}+8t\right)\right|_0^{40}=\\[0.3cm] C\cdot\left(\frac{1600}{2}+320\right)=1\to C\cdot1120=1\longrightarrow\\[0.3cm] \boxed{C=\frac{1}{1120}}

Comment: Perhaps the probability distribution function was incorrectly indicated and actually meant f(t)=1/(t+10)2f(t)=1/(t+10)^2 , then



+f(t)dt=1040(1(t+10)2)dt=C(1t+10)040=C(150110)=1C1550=1C=504=252=12.5C=252=12.5\int\limits_{-\infty}^{+\infty}f(t)dt=1\to \int\limits_0^{40}\left(\frac{1}{(t+10)^2}\right)dt=\\[0.3cm] -C\cdot\left.\left(\frac{1}{t+10}\right)\right|_0^{40}=-C\cdot\left(\frac{1}{50}-\frac{1}{10}\right)=1\\[0.3cm] -C\cdot\frac{1-5}{50}=1\longrightarrow C=\frac{50}{4}=\frac{25}{2}=12.5\\[0.3cm] \boxed{C=\frac{25}{2}=12.5}


ANSWER



If   f(t)=(t+10)2C=11120If   f(t)=1(t+10)2C=252=12.5\text{If}\,\,\,f(t)=(t+10)-2\longrightarrow C=\frac{1}{1120}\\[0.3cm] \text{If}\,\,\,f(t)=\frac{1}{(t+10)^2}\longrightarrow C=\frac{25}{2}=12.5\\[0.3cm]


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