Question

The lifetime X of an electronic component has an exponential distribution such that \({\bf{P}}\left( {{\bf{X}}\, \le {\bf{1000}}} \right){\bf{ = 0}}{\bf{.75}}\). What is the expected lifetime of the component?

Short answer

\(\beta = 721.35\)

Step-by-step solution

Given information

The random variable X denotes an electronic component’s lifetime, which is an exponential distribution.

It is also given\(P\left( {X\, \le 1000} \right) = 0.75\).

Calculating the prerequisite value

We have to find\(\beta \)since \(E\left( X \right) = \beta \).

Therefore,

\(\begin{array}{l}f\left( {x|\beta } \right) = \frac{1}{\beta }{e^{ - \frac{x}{\beta }}},\,0 < x < \infty \\\end{array}\)

For any,

\(\begin{aligned}{c}\alpha > 0,\\P\left[ {x \le a} \right) &= F\left( a \right)\\ &= \int\limits_0^\alpha {\frac{1}{\beta }{e^{ - \frac{y}{\beta }}}dy} \\ &= 1 - {e^{ - \frac{\alpha }{\beta }}}\end{aligned}\)

Calculating the required probability

\(\begin{array}{l}So,\,\,P\left\{ {X \le 1000} \right\}\\ \Rightarrow 1 - {e^{ - \frac{{1000}}{\beta }}} = 0.75\\ \Rightarrow \beta = 721.35\end{array}\)

Hence the required value is\({\bf{\beta = 721}}{\bf{.35}}\)