Brian Lovins <[EMAIL PROTECTED]> wrote in sci.stat.edu:
>You have a sample of 475 adults. The following is a distribution:
>
>1 prior-25 adults
>2 prior-30 adults
>3 priors-35 adults
>4 priors-40 adults
>5 priors-45 adults
>6 priors-50 ad.
>7 p.-55 ad.
>8p- 60 ad
>9p-65 ad
>10 p- 65 ad
>
>Suppose you want to select four adults at random from this population. What
>is the probability of getting the following in 4 selections?
>
>At least two adults with 5 priors, with replacement.
First question: with _exactly_ five priors, or with five or more?
Either way, figure the relative frequency of the class you're
interested in (which is the number of adults in that class, divided
by sample size). That is the probability of getting the desired
class in any one selection. Let's call that number p.
>From that point, you have a probability distribution where each
selection either will or will not get you an adult from the desired
class. Since you are doing this with replacement, p does not change
from trial to trial. So your question boils down to the probability
of 2, 3, or 4 successes out of 4 trials, where the probability of
success on any one trial is p.
You should have learned a particular type of probability
distribution that is useful in this situation.
For future homework questions, please don't just post the question
but show us specifically what you have tried to solve it. Or else
just give us your teacher's e-mail address so we can cut out the
middleman and mail solutions to him or her directly.
--
Stan Brown, Oak Road Systems, Cortland County, New York, USA
http://OakRoadSystems.com
"Honesty always gives you the advantage of surprise."
-- /Yes, Prime Minister/
.
.
=================================================================
Instructions for joining and leaving this list, remarks about the
problem of INAPPROPRIATE MESSAGES, and archives are available at:
. http://jse.stat.ncsu.edu/ .
=================================================================
