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Lightbulb Probability Density Function Problem

oglemetender

New member
Joined
Feb 15, 2011
Messages
1
I have been trying to solve a problem about the probability that 3 lightbulbs will last for 1500 hours of operation if the probability density function of a lightbulb is f(x) = 6[.25-(x-1.5)^2] when 1 <= x <= 2 and f(x) = 0 otherwise. x is measured in multiples of 1,000 hours. I tried doing [the integral from 1 to 1.5 of f(x)] CUBED, but I got an answer greater than 1. Does anyone know what I am doing wrong?
 

tkhunny

Moderator
Staff member
Joined
Apr 12, 2005
Messages
9,708
Always keep track of yourself.

\(\displaystyle \int_{1}^{2}6\cdot\left[\frac{1}{4}-\left(x-\frac{3}{2}\right)^{2}\right]\;dx\;=\;1\)

This is a proper pdf

When you apply the exponent, 3, you get...

\(\displaystyle \int_{1}^{2}\left{6\cdot\left[\frac{1}{4}-\left(x-\frac{3}{2}\right)^{2}\right]\right}^{3}\;dx\;=\;\frac{54}{35}\;>\;1\)

Clearly, this is no good.

Try this:

\(\displaystyle \int_{1.5}^{2}6\cdot\left[\frac{1}{4}-\left(x-\frac{3}{2}\right)^{2}\right]\;dx\;=\;\frac{1}{2}\)

After that, it's a Binomial problem with p = 1/2 and n = 3. Let's see what you get.
 
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