Question 7.11: The lifetime X of a device is assumed to follow an exponenti......
The lifetime X of a device is assumed to follow an exponential distribution with parameter 𝜆 > 0.
(i) Calculate the probability that the lifelength of the device
(a) exceeds its mean value;
(b) is at least twice as much as its mean value.
(ii) Find the length of time t_0, such that the probability the device works at time t_0 is 95%.
Let F denote the distribution function of the lifetime X. Then by (7.16)
F(t)= \begin{cases} 0, \qquad \qquad t \lt 0, \\ 1-\mathrm{e}^{-\lambda t}, \quad t \geq 0. \end{cases} (7.16)
F(t)=1-\mathrm{e}^{-\lambda t} t for any t ≥ 0, while F(t) = 0 for t < 0. Further, the expected lifetime of the device is \lambda ^{-1}.
(i) (a) The required probability is P(X\gt \lambda^{-1}), which is given by
P(X\gt \lambda^{-1})=1-F(\lambda^{-1})=\mathrm{e}^{-\lambda\cdot \lambda^{-1}}=\mathrm{e}^{-1}\cong0.3679\cong37%.
(b) We now seek P(X\gt 2\lambda^{-1}), which by an appeal to (7.16) again, equals
P(X\gt 2\lambda^{-1})=1-F(2\lambda^{-1})=\mathrm{e}^{-\lambda \cdot {2}\lambda^{-1}}=\mathrm{e}^{-2}\cong0.1353,or about 13.5%. Observe that both probabilities for this part are independent of the parameter 𝜆.
(ii) Compared to the previous part, we see here that the problem is in the opposite direction, since we seek the value t_0 such that
P(X\geq t_{0})=0.95.Using once more the particular form of the distribution function for X, this becomes 1-\mathrm{e}^{-\lambda t_{0}}=0.05, which in turn yields
\mathrm{e}^{-\lambda t_{0}}=0.95.Taking logarithms on both sides, we get -\lambda t_{0}=\ln(0.95), so that we obtain finally
t_{0}=-{\frac{1}{\lambda}}\ln(0.95).In the terminology used in Example 7.4, this is the 95%-quantile of the ε(λ) distribution.