Mathematics: Suppose that the lifetime of a particular component has an exponential distribution with mean value 1000h. (a) What is the probability that the lifetime is at most 2000h? (b) What is the probability that the lifetime is more than 1000h? (c) What is the probability that the lifetime is between 500h and 2000h? (d) Compute the variance of the distribution?

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Suppose that the lifetime of a particular component

Ответ:

eweqwee3147

18.01.2022

Mathematics

(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.

(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:

$f(x)=\left \{ {{\lambda e^{-\lambda x}; x0\atop {0};\ otherwise} \right.$

(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:

$P(X1000)=\int\limits^{\infty}_{1000} {\lambda e^{-\lambda x}} \, dx \\=\lambda[\frac{e^{-\lambda x}}{-\lambda} ]^{\infty}_{1000} \\=[-e^{\frac{-\infty}{1000}}+e^{\frac{-1000}{1000} } }]\\=0+0.3679\\=0.3679$