Binomial Distribution

Date: 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/