Answer:
(a) The probability that the lifetime is at most 2000 h is 0.8647.
(b) The probability that the lifetime is at most 2000 h is 0.8647.
(c) The probability that the lifetime is between 500 h and 2000 h is 0.4712.
(d) The variance of the lifetime of a particular component is 10⁻⁶.
Step-by-step explanation:
Let <em>X </em>= lifetime of a particular component
The random variable ![X\sim Exp(\lambda = \frac{1}{1000} )](https://tex.z-dn.net/?f=X%5Csim%20Exp%28%5Clambda%20%3D%20%5Cfrac%7B1%7D%7B1000%7D%20%29)
The probability distribution function of an exponential distribution is:
![f(x)=\left \{ {{\lambda e^{-\lambda x}; x>0\atop {0};\ otherwise} \right.](https://tex.z-dn.net/?f=f%28x%29%3D%5Cleft%20%5C%7B%20%7B%7B%5Clambda%20e%5E%7B-%5Clambda%20x%7D%3B%20x%3E0%5Catop%20%7B0%7D%3B%5C%20otherwise%7D%20%5Cright.)
(a)
Compute the probability that the lifetime is at most 2000 h as follows:
![P(X\leq 2000)=\int\limits^{2000}_{0} {\lambda e^{-\lambda x}} \, dx \\=\lambda[\frac{e^{-\lambda x}}{-\lambda} ]^{2000}_{0} \\=[-e^{\frac{-2000}{1000}}+e^{\frac{-0}{1000} } }]\\=1-0.1353\\=0.8647](https://tex.z-dn.net/?f=P%28X%5Cleq%202000%29%3D%5Cint%5Climits%5E%7B2000%7D_%7B0%7D%20%7B%5Clambda%20e%5E%7B-%5Clambda%20x%7D%7D%20%5C%2C%20dx%20%5C%5C%3D%5Clambda%5B%5Cfrac%7Be%5E%7B-%5Clambda%20x%7D%7D%7B-%5Clambda%7D%20%5D%5E%7B2000%7D_%7B0%7D%20%5C%5C%3D%5B-e%5E%7B%5Cfrac%7B-2000%7D%7B1000%7D%7D%2Be%5E%7B%5Cfrac%7B-0%7D%7B1000%7D%20%7D%20%7D%5D%5C%5C%3D1-0.1353%5C%5C%3D0.8647)
Thus, the probability that the lifetime is at most 2000 h is 0.8647.
(b)
Compute the probability that the lifetime is more than 1000 h as follows:
![P(X>1000)=\int\limits^{\infty}_{1000} {\lambda e^{-\lambda x}} \, dx \\=\lambda[\frac{e^{-\lambda x}}{-\lambda} ]^{\infty}_{1000} \\=[-e^{\frac{-\infty}{1000}}+e^{\frac{-1000}{1000} } }]\\=0+0.3679\\=0.3679](https://tex.z-dn.net/?f=P%28X%3E1000%29%3D%5Cint%5Climits%5E%7B%5Cinfty%7D_%7B1000%7D%20%7B%5Clambda%20e%5E%7B-%5Clambda%20x%7D%7D%20%5C%2C%20dx%20%5C%5C%3D%5Clambda%5B%5Cfrac%7Be%5E%7B-%5Clambda%20x%7D%7D%7B-%5Clambda%7D%20%5D%5E%7B%5Cinfty%7D_%7B1000%7D%20%5C%5C%3D%5B-e%5E%7B%5Cfrac%7B-%5Cinfty%7D%7B1000%7D%7D%2Be%5E%7B%5Cfrac%7B-1000%7D%7B1000%7D%20%7D%20%7D%5D%5C%5C%3D0%2B0.3679%5C%5C%3D0.3679)
Thus, the probability that the lifetime is more than 1000 h is 0.3679.
(c)
Compute the probability that the lifetime is between 500 h and 2000 h as follows:
![P(500](https://tex.z-dn.net/?f=P%28500%3CX%3C2000%29%3D%5Cint%5Climits%5E%7B2000%7D_%7B500%7D%20%7B%5Clambda%20e%5E%7B-%5Clambda%20x%7D%7D%20%5C%2C%20dx%20%5C%5C%3D%5Clambda%5B%5Cfrac%7Be%5E%7B-%5Clambda%20x%7D%7D%7B-%5Clambda%7D%20%5D%5E%7B2000%7D_%7B500%7D%20%5C%5C%3D%5B-e%5E%7B%5Cfrac%7B-2000%7D%7B1000%7D%7D%2Be%5E%7B%5Cfrac%7B-500%7D%7B1000%7D%20%7D%20%7D%5D%5C%5C%3D-0.1353%2B0.6065%5C%5C%3D0.4712)
Thus, the probability that the lifetime is between 500 h and 2000 h is 0.4712.
(d)
The variance of an exponential distribution is, ![Var(X)=\frac{1}{\lambda^{2}}](https://tex.z-dn.net/?f=Var%28X%29%3D%5Cfrac%7B1%7D%7B%5Clambda%5E%7B2%7D%7D)
The variance of the lifetime of a particular component is:
![Var(X)=\frac{1}{\lambda^{2}}=(\frac{1}{\lambda})^{2}=(\frac{1}{1000} )^{2}=10^{-6}](https://tex.z-dn.net/?f=Var%28X%29%3D%5Cfrac%7B1%7D%7B%5Clambda%5E%7B2%7D%7D%3D%28%5Cfrac%7B1%7D%7B%5Clambda%7D%29%5E%7B2%7D%3D%28%5Cfrac%7B1%7D%7B1000%7D%20%29%5E%7B2%7D%3D10%5E%7B-6%7D)
Thus, the variance of the lifetime of a particular component is 10⁻⁶.