Continuous_solu -...

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Continuous Distribution Problems for Reliability in Engineering – Solutions. 1. The lifetime of all Everglo bulbs is normally distributed with mean 400 hours and standard deviation 50  hours.   a.  What percentage of the bulbs will burn out in less than 360 hours? b. After how many hours will 80% of the bulbs burn out? c. Let X = lifetimes.  Given X ~ N(400,50) a. We want P(X < 360) =NORMDIST(360,400,50,1)=0.2119 Alternatively by hand and tables we need to convert X to standard Z using Z = (X –  μ )/ σ : P(X < 360) = P[Z < (360 – 400)/50] = P[Z < -0.80] = 0.2119 obtained directly from looking up in Normal  tables. b. We want k such that P(X < k ) = 0.80; use NORMINV(0.8,400,50) = 442.1 hours Alternatively first compute a z value to satisfy P(Z < z 1  ) = 0.80.  Looking in the body of normal tables  for 0.8000 we get the closest number to be 0.7995 yielding z 1  = 0.74.  Now form an equation  connecting z 1  and k then solve for k:  z 1  = 0.74 = (k – 400)/50.  We have k – 400 = 0.74*50 = 37 giving k  = 400+37 = 437 which is within rounding of 442.1 2. High-Tech Inc., produces an electronic component, GX-7, that has an average life span of 4500 hours.  The lifespan is normally distributed with a standard deviation of 500 hours. The company is considering  a 3800 hour warranty on the GX-7.  If this warranty policy is adopted, what proportion of GX-7  components should High-Tech expect to replace under warranty? Let X = lifetimes.  Given X ~ N(4500,500) and k = 3800 a. We want P(X < 3800) =NORMDIST(3800,4500,500,1)=0.081.  Thus High-Tech can expect to replace  about 8% of the GX-7 components. 3.
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This note was uploaded on 09/22/2011 for the course MECH 2110 taught by Professor Clark,b during the Spring '08 term at Auburn University.

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Continuous_solu -...

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