Question #115400
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. Calculate the probability that the lifetime of the machine part is less than 10 years. Hint: Show that f(t) is legitimate and find the proportionality constant.
1
Expert's answer
2020-05-12T18:37:54-0400

Let k is a constant, thenf(t)=k[(t+10)2]=k(t+8),0<t<40k040(t+8)dt=1k[t22+8t]040=1k[4022+8(40)]=1k(1120)=1k=11120P(t<10)=11120010(t+8)dt=11120[t22+8t]010=11120[1022+8(10)]=11120×130=13112\text {Let k is a constant, then}\\ f(t)=k[(t+10)-2]=k(t+8), 0<t<40\\ k\int _0^{40}(t+8)dt=1\\ k\left[\frac{t^2}{2}+8t\right]^{40}_0=1\\ k[\frac{40^2}{2}+8(40)]=1\\ k(1120)=1\\ \therefore k=\frac{1}{1120}\\ P(t<10)=\frac{1}{1120}\int _0^{10}(t+8)dt\\ =\frac{1}{1120}\left[\frac{t^2}{2}+8t\right]^{10}_0\\ =\frac{1}{1120}[\frac{10^2}{2}+8(10)]\\ =\frac{1}{1120}\times 130\\ =\frac{13}{112}




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