\[ f(t)=.01e^{-0.1t} \]
Where t is measured in hours.
Answer
The failure rate can be found with the following equation: \[ -\frac{R'(t)}{R(t)} \] The ratio of realiability prime over realiability https://en.wikipedia.org/wiki/Failure_rate
F(t) represents the realiability function. This can derive the reliability function from the given density function as per http://www.engineeredsoftware.com/nasa/reliabil.htm
We can derive the reliability as follows: \[ R(t)=1-\int _{ 0 }^{ t }{ f(t)dt } \]
First compute the integration by hand before finding R(t)
\[ \int _{ 0 }^{ t }{ .01e^{-0.1t}dt } \]
Consider the technique here https://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/expondirectory/Exponentials.html to compute integral of exponents We perform u-substitution after pulling out the 0.01
\[ 0.01\int _{ 0 }^{ t }{ e^{-0.1t}dt }\\ u=-0.01t\\ du=-0.01dt\\ -\frac{du}{0.01}=dt \]
Convert upper and lower limits of integration using our u to x map
\[ upper=-0.01t\\ lower=0 \]
Although, the upper limit is negative. Mapping a value in the domain of t to a new value in the domain of u. Integrate as is in order to avoid confusion. Replace our new dt in for u
\[ -\frac{0.01}{0.01}\int _{ 0 }^{ -0.01t }{ e^{u}du }\\ -(e^{-0.01t}-1)\\ 1+e^{-0.01t} \]
Applying fundamental theorem of calculus part 2 and consolidate the negative on the outside of the integral.
Computing R(t) as follows: \[ R(t)=1-(1+e^{-0.01t})\\ R(t)=e^{-0.01t} \]
Finding the failure involves computing derivative of R(t) Using the method as defined under http://tutorial.math.lamar.edu/Classes/CalcI/DiffExpLogFcns.aspx, computing the failure rate and the exponentials cancel out.
\[ -\frac{R'(t)}{R(t)}=-\frac{-0.01e^{-0.01t}}{e^{0.01t}}=0.01 \]
Answer Using derived reliability formula and plug in 20 to replace t
\[ R(t)=e^{-0.01t}=e^{-0.01(20)} \]
This comes out to be roughly .8
Answer We want to find R(40) because we need to compute the probability that the bulb will last an additional 20 hours given that the bulb lasts first 20 hours. You can think of it as R(20+20). We divide this quantitiy by R(20) since it is our benchmark. https://en.wikipedia.org/wiki/Conditional_probability
\[ P=\frac{R(40)}{R(20)}=\frac{e^{-0.01(40)}}{e^{-0.01(20)}} \]
Above results to roughly .8
Answer In order to compute this probability we need to consider the numerator as the difference between the failure functions F(41)-F(40). The denominator will be the reliability evaluated at 40. Our probability is as follows : We are only interested in what happens when t=41. Since F(t) is the result of the integral we computed area. It is the area under the curve from 0 to T. We are simply subtracting the area under the curve associated with values [0,40] leaving only the line at F(41)
\[ P=\frac{F(41)-F(40)}{R(40)}\\ =\frac{(e^{-0.01t}-1)-(e^{-0.01t}-1)}{e^{-0.01(40)}}=0.01 \]