Assume that a new light bulb will burn out after \(t\) hours, where \(t\) is chosen from \([0, \infty)\) with an exponential density \(f(t) = \lambda e^{-\lambda t}\) In this context, \(\lambda\) is often called the failure rate of the bulb.
\(f(t) = \lambda e^{-\lambda t}\)
We can use the Fundamental Theorem of Calculus with this equation.
\(F(x) = \int^{x}_{a}f(t)dt\)
We can also keep in mind that \(\int e^{\lambda t}dt\) can also be written as \(\frac{1}{\lambda} e^{\lambda t} + C\)
Keeping this in mind, we can work on solving this problem.
\(\int^{T}_{0}\frac{1}{100}e^{-\frac{1}{100}t}\)
This can also be written as….
\(\frac{1}{100}|\frac{e^{-\frac{1}{100}t}}{-\frac{1}{100}}|^{T}_{0}\)
The \(\frac{1}{100}\) and \(-\frac{1}{100}\) solve to a \(-1\), so…
\(|-e^{-\frac{1}{100}t}|^{T}_{0}\)
This solves to be…
\(|-e^{-\frac{1}{100}T}|\) and \(|-e^{-\frac{1}{100}0}|\)
To be perfectly honest, we do not care about the case for 0, just for T, as outlined in the problem.
\(|-e^{-\frac{1}{100}T}|\)
\(e^{-\frac{1}{100}T}\)
Thus, we can come to the conclusion that the probability the bulb will not burn out before \(T\) hours is \(e^{-\frac{1}{100}T}\).
To start with, we have the equation \(\frac{1}{2} = e^{-\frac{1}{100}T}\).
From our knowledge of natural logarithms, we know that if two different functions are equal to each other, then their natural logarithms are equal to each other as well.
\(ln(\frac{1}{2}) = ln(e^{-\frac{1}{100}T})\)
Using a rule of natural logarithms, we know that \(log_{a}(\frac{1}{x}) = -log(x)\). We can rewrite our equation once more with this knowledge:
\(-ln(2) = ln(e^{-\frac{1}{100}T})\)
Using yet another rule of natural logarithms, we know that \(log_{a}(x^{b}) = b\times log_{a}(x)\). Once more, our equation gets rewritten…
\(-ln(2) = (-\frac{1}{100}T)ln(e)\)
The natural logarithm of e is 1, so…
\(-ln(2) = (-\frac{1}{100}T)\times1\)
\(-ln(2) = -\frac{1}{100}T\)
Now it’s a matter of solving the equation.
\(-\frac{1}{100}T = -ln(2)\)
\(100\times(-\frac{1}{100}T) = 100\times(-ln(2))\)
\(-T = -100ln(2)\)
\(\frac{-T}{-1} = \frac{-100ln(2)}{-1}\)
\(T = 100ln(2)\)
\(T = 100\times0.6931472\)
\(T = 69.31472\)
So, when a light bulb has a reliability of \(\frac{1}{2}\), you can rest assured that its T value is 69.31472.