Question

Suppose that the lifetime of a particular component has an exponential distribution with mean value 1000h.

Answer 1

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 X = lifetime of a particular component

The random variable $X\sim Exp(\lambda = \frac{1}{1000} )$

The probability distribution function of an exponential distribution is:

$f(x)=\left \{ {{\lambda e^{-\lambda x}; x>0\atop {0};\ otherwise} \right.$

(a)

Compute the probability that the lifetime is at most 2000 h as follows:

$P(X\leq 2000)=\int\limits^{2000}_{0} {\lambda e^{-\lambda x}} \, dx \\=\lambda[\frac{e^{-\lambda x}}{-\lambda} ]^{2000}_{0} \\=[-e^{\frac{-2000}{1000}}+e^{\frac{-0}{1000} } }]\\=1-0.1353\\=0.8647$

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:

$P(X>1000)=\int\limits^{\infty}_{1000} {\lambda e^{-\lambda x}} \, dx \\=\lambda[\frac{e^{-\lambda x}}{-\lambda} ]^{\infty}_{1000} \\=[-e^{\frac{-\infty}{1000}}+e^{\frac{-1000}{1000} } }]\\=0+0.3679\\=0.3679$

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:

$P(500$

Thus, the probability that the lifetime is between 500 h and 2000 h is 0.4712.

(d)

The variance of an exponential distribution is, $Var(X)=\frac{1}{\lambda^{2}}$

The variance of the lifetime of a particular component is:

$Var(X)=\frac{1}{\lambda^{2}}=(\frac{1}{\lambda})^{2}=(\frac{1}{1000} )^{2}=10^{-6}$

Thus, the variance of the lifetime of a particular component is 10⁻⁶.