document.write( "Question 223170: A light bulb manufacturer claims his light bulbs will last 500 hours on the average. The lifetime of a light bulb is assumed to follow an exponential distribution.
\n" ); document.write( "a. What is the probability that the light bulb will have to replaced within 500 hours?
\n" ); document.write( "b. What is the probability that the light bulb will last more than 1000 hours?
\n" ); document.write( "c. What is the probability that the light bulb will last between 200 and 800 hours?
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Algebra.Com's Answer #167041 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
A light bulb manufacturer claims his light bulbs will last 500 hours on the average. The lifetime of a light bulb is assumed to follow an exponential distribution.
\n" ); document.write( "a. What is the probability that the light bulb will have to replaced within 500 hours?
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\n" ); document.write( "lamda = 1/500
\n" ); document.write( "Ans: P(x<=500) = 1 - e^(-lamda*x) = 1-e^[(-1/500)*500] = 1-e^-1 = 0.6321\r
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\n" ); document.write( "b. What is the probability that the light bulb will last more than 1000 hours?
\n" ); document.write( "Ans:P(x>=1000) = 1 - P(x<=1000) = 1 -[1-e^(-1/500*1000)
\n" ); document.write( "= 1 -[1-e^(-2)] = e^(-2) = 0.1353
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\n" ); document.write( "\n" ); document.write( "c. What is the probability that the light bulb will last between 200 and 800 hours?
\n" ); document.write( "Answer will be P(x<=800)-P(x<=200)
\n" ); document.write( "Remember P(x<=k) = 1-e^(-lambda*k)
\n" ); document.write( "I'll leave that to you.
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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