exercise:8b126d52c7: Difference between revisions
(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> Jones puts in two new lightbulbs: a 60 watt bulb and a 100 watt bulb. It is claimed that the lifetime of the 60 watt bulb has an exponential density with average lifetime 200 hours (<math>\lambda = 1/200</math>). The 100 watt bulb also has an ex...") |
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Jones puts in two new lightbulbs: a 60 watt bulb and a 100 watt bulb. It is claimed that the lifetime of the 60 watt bulb has an exponential | |||
density with average lifetime 200 hours (<math>\lambda = 1/200</math>). The 100 watt bulb also has an exponential density but with average lifetime of only 100 hours (<math>\lambda = 1/100</math>). Jones wonders what is the probability that the 100 watt bulb will outlast the 60 watt bulb. | |||
100 watt bulb. It is claimed that the lifetime of the 60 watt bulb has an exponential | |||
density with average lifetime 200 hours (<math>\lambda = 1/200</math>). The 100 watt | |||
bulb also has an exponential density but with average lifetime of only 100 hours | |||
(<math>\lambda = 1/100</math>). Jones wonders what is the probability that the 100 watt bulb will | |||
outlast the 60 watt bulb. | |||
If <math>X</math> and <math>Y</math> are two independent random variables with exponential densities <math>f(x) = \lambda e^{-\lambda x}</math> and <math>g(x) = \mu e^{-\mu x}</math>, respectively, then the | |||
If <math>X</math> and <math>Y</math> are two independent random variables with exponential densities | |||
<math>f(x) = \lambda e^{-\lambda x}</math> and <math>g(x) = \mu e^{-\mu x}</math>, respectively, then the | |||
probability that | probability that | ||
<math>X</math> is less than <math>Y</math> is given by | <math>X</math> is less than <math>Y</math> is given by | ||
<math display="block"> | |||
P(X < Y) = \int_0^\infty f(x)(1 - G(x))\,dx, | |||
</math> | </math> | ||
<math | where <math>G(x)</math> is the cumulative distribution function for <math>g(x)</math>. Explain why this is the | ||
case. Use this to show that | case. Use this to show that | ||
< | <math> | ||
P(X < Y) = \frac \lambda{\lambda + \mu} | P(X < Y) = \frac \lambda{\lambda + \mu} | ||
<math> | </math> and to answer Jones's question. | ||
Latest revision as of 02:15, 14 June 2024
Jones puts in two new lightbulbs: a 60 watt bulb and a 100 watt bulb. It is claimed that the lifetime of the 60 watt bulb has an exponential density with average lifetime 200 hours ([math]\lambda = 1/200[/math]). The 100 watt bulb also has an exponential density but with average lifetime of only 100 hours ([math]\lambda = 1/100[/math]). Jones wonders what is the probability that the 100 watt bulb will outlast the 60 watt bulb.
If [math]X[/math] and [math]Y[/math] are two independent random variables with exponential densities [math]f(x) = \lambda e^{-\lambda x}[/math] and [math]g(x) = \mu e^{-\mu x}[/math], respectively, then the probability that [math]X[/math] is less than [math]Y[/math] is given by
where [math]G(x)[/math] is the cumulative distribution function for [math]g(x)[/math]. Explain why this is the case. Use this to show that [math] P(X \lt Y) = \frac \lambda{\lambda + \mu} [/math] and to answer Jones's question.