Revision as of 02:13, 9 June 2024 by Bot (Created page with "<div class="d-none"><math> \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}</math></div> Assume that a new light bulb will burn out after <math>t</math> hours, where <math>t</math> is chosen from <math>[0,\infty)</math> with an exponential density <math display="block"> f(t) = \lambda e^{-\lambda t}\ . </math> In this context, <math>...")
BBy Bot
Jun 09'24
Exercise
[math]
\newcommand{\NA}{{\rm NA}}
\newcommand{\mat}[1]{{\bf#1}}
\newcommand{\exref}[1]{\ref{##1}}
\newcommand{\secstoprocess}{\all}
\newcommand{\NA}{{\rm NA}}
\newcommand{\mathds}{\mathbb}[/math]
Assume that a new light bulb will burn out after [math]t[/math]
hours, where [math]t[/math] is chosen from [math][0,\infty)[/math] with an exponential density
[[math]]
f(t) = \lambda e^{-\lambda t}\ .
[[/math]]
In this context, [math]\lambda[/math] is often called the failure rate of the bulb.
- Assume that [math]\lambda = 0.01[/math], and find the probability that the bulb will not burn out before [math]T[/math] hours. This probability is often called the reliability of the bulb.
- For what [math]T[/math] is the reliability of the bulb [math] = 1/2[/math]?