A manufacturer of window frames knows from long experience that 5 percent of the 		
production will have some type of minor defect that will require an
adjustment. What is
the probability that in a sample of 20 window frames:		
		
  a.) None will need adjustments?		
		
		
  b.) At least one will need adjustment?		
		
		
  c.) More than two will need adjustment?

Hi!!


Here we must use the Poisson distribution:
"The Poisson distribution is most commonly used to model the number of
random occurrences of some phenomenon in a specified unit of space or
time. For example.
·The number of phone calls received by a telephone operator in a 10-minute period.
·The number of flaws in a bolt of fabric.
·The number of typos per page made by a secretary. "
"The Poisson Distribution":
http://www.stat.tamu.edu/stat30x/notes/node70.html


The only parameter in this distribution is lambda, the rate at which
the events happens. In this case, lambda = 5% of 20 = 1 .

a.) None will need adjustments?

lambda = 1.0
n = 0

Just plug the values in the formula showed in the link above and use a
calculator, you will find that:
P(X = 0) = 0.36787945

Note you can use the following online calculator (note that lambda is m).
http://www.changbioscience.com/stat/prob.html


b.) At least one will need adjustment?

P(X >= 1) = 1 - P(X < 1) =
          = 1 - P(X = 0) =
          = 1 - 0.36787945 =
          = 0.63212055


c.) More than two will need adjustment?

P(X > 2) = 1 - P(X =< 2) =
         = 1 - [P(X = 0) + P(X = 1) + P(X = 2)] =
         = 1 - [0.36787945 + 0.36787945 + 0.18393973] =
         = 0.08030137


I hope this helps you. Please do not hesitate to request for a
clarification if you find something unclear.

Regards.
livioflores-ga

Thanks again!!

Livioflores consistently demonstrates a state of confusion as to what
Poisson distribution is. Cf.<a
href="http://answers.google.com/answers/threadview?id=558519">http://answers.google.com/answers/threadview?id=558519</a>

Again, the researcher tries to use Poisson where the variable has
Binomial distribution.

In the problem posed by ml123 the number of the windows with defect is
a Binomial with p=0.05 and N=20. So for X>2 it'll yield
P{X>2}=0.07548366, which is quite different from the answer given
above.

To better understand the fallacy, try to use the method suggested by
Livioflores in computiing the probability of that there will be MORE
than 20 windows with defects. This would give you a positive (albeit
very small) probability, in spite of that there are only 20 windows
total.

-- RK