Binomial DistributionDate: 8/25/96 at 20:1:58 From: Anonymous Subject: Binomial distribution The mean life of a certain television tube is 1200 hours with a standard deviation of 200 hours. What is the 0.8 quantile, ie. the length of life that we would expect 80 percent of the tubes to reach? The factory operates three machines. The probability that they will be in use on any one day are 0.8,0.7 and 0.6 respectively. What is the probability that at least one will be in use on a particular day? Date: 8/26/96 at 15:14:22 From: Doctor Robert Subject: Re: Binomial distribution If you use the normal distribution table, you find that 80 percent of the values of a distribution lie below .84 standard deviations above the mean. P(z<=.84)=.7995 which is approximately .8 or 80 percent. Therefore you would expect 80 percent of the lightbulbs to have a life of 1200 + .84(200) = 1368 hr. The probability that none of the machines is running is (1-.8)(1-.7)(1-.6) = .2(.3)(.4) = .024. Now, the negation of "none of the machines os running" is "at least one of the machines is running". Therefore, the probability that at least one of the machines is running is 1 -.024 = .976.. -Doctor Robert, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/