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anonymous
 4 years ago
The fluorescent light tubes made by the company Welllit have lifetimes which are normally distributed with mean 2010 hours and standard deviation 20 hours. The company descides to promote its sales of the tubes by guaranteeing a minimum life of the tubes, replacing free of charge any tubes that fail to meet this minimum life. If the company wishes to have to replace free only 3% of the tubes sold, find the guaranteed minimum it must set.
A purchaser buys four of these fluorescent light tubes. What is the probability that one of the four fails before the guaranteed minimum lifetime?
anonymous
 4 years ago
The fluorescent light tubes made by the company Welllit have lifetimes which are normally distributed with mean 2010 hours and standard deviation 20 hours. The company descides to promote its sales of the tubes by guaranteeing a minimum life of the tubes, replacing free of charge any tubes that fail to meet this minimum life. If the company wishes to have to replace free only 3% of the tubes sold, find the guaranteed minimum it must set. A purchaser buys four of these fluorescent light tubes. What is the probability that one of the four fails before the guaranteed minimum lifetime?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Let the minimum guaranteed lifetime be G. Now, standardizing the variable L ~ N(2010,400), let us define a new variable such that  \[(L  2010) / 20 = Z\] ........ (1) Where Z ~ N(0,1), now the corresponding minimum guaranteed value of z be g. \[P (z < g) = 0.03\] hence, \[P (z > g) = 0.03\] hence, \[\phi (g) = 1  0.03 = 0.97\] (Please let me know if you need the explanation) Now, from the Normal distribution table  \[\phi (1.88) = 0.96995\] and \[\phi (1.89) = 0.97062\] Using the method of linear interpolation we have g = 1.88075 Hence, replacing L with G and z with g (=1.88705) in equation (1) we get  \[G \approx 1973 hrs\] This is the desired answer for the first part. For the second part if only one tube does not cover the min guaranteed lifetime is 0.03 and all other will have a probability of surviving that hence they all will individually have probabilities of 0.97. Hence for a particular tube to miss the guaranteed lifetime is given by \[P (X = 1) = 0.03 * (.0.97)^{3}\] Any of the 4 tubes can miss the min lifetime hence, we actually should multiply the above probability by \[\left(\begin{matrix}4 \\ 1\end{matrix}\right)\] Hence the required probability is  \[= 4 * 0.03 * 0.97^{3}\] = 0.1095 = 10.95%

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks :)... I just don't understand one part: Now, from the Normal distribution table  ϕ(1.88)=0.96995 and ϕ(1.89)=0.97062 Using the method of linear interpolation we have g = 1.88075 Hence, replacing L with G and z with g (=1.88705) in equation (1) we get  G≈1973hrs

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, we need to have a value of z such that \[\phi(z) = 0.97\] Assuming in the small interval of z = 1.88 and z = 1.89 the Normal PDF (probability Density Function) curve is linear then we have  \[[\phi(g)  \Phi(1.88)] / [\Phi(1.89)  \Phi(1.88)] = (0.97  0.96995) / (0.97062  0.96995)\] solving above equation we will get g =  1.88075
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