Landa Standard
Poisson Distribution?
I keep getting negative numbers for this problem.. and I don't know what I am doing wrong.
For landa, I get .02 which is what the professor told me but something just isn't adding up.
*One out of every 5000 individuals in a population carries a certain defective gene. A random sample of 1000 individuals is studied.
a. What is the probability that exactly one of the sample individuals carries the gene?
b. What is the probability that none of the sample individuals carries the gene?
c. What is the probability that more than two of the sample individuals carries the gene?
d. What is the mean of the number of sample individuals that carry the gene?
e. What is the standard deviation of the number of sample individuals that carry the gene?
The Poisson distribution can be derived from the binomial distribution. The Poisson is nothing more than the limiting case of the Binomial where n is large and p is small.
A good way to identify when you need to use the Poisson distribution is when the problem requires you to use a rate. This is not always true, but more often than not remembering this will help you to identify a Poisson model.
Let X be the number of people in the sample with the gene. X has the Poisson distribution with parameter ?t = 0.2
In general you have:
X ~ Poisson( ?t )
P(X = x) = ( ?t )^x * exp( -?t ) / x! for x = 0, 1, 2, 3, 4, ...
P(X = x) = 0 otherwise
the mean of the Poisson distribution is the parameter, ?t
the variance of the Poisson distribution is the parameter, ?t
In this problem we have
? = 2e-04
t = 1000 time unit(s)
this results in our random variable X ~ Poisson( 0.2 )
== -- == -- == -- ==
A)
Find P(X = 1 ) = 0.2 ^ 1 * exp( -0.2 ) / 1 ! = 0.1637462
== -- == -- == -- ==
B)
Find P(X = 0 ) = 0.2 ^ 0 * exp( -0.2 ) / 0 ! = 0.8187308
== -- == -- == -- ==
C)
Find P( X > 2 ) =
?
? P(X = i)
i = 3
This sum is an infinite sum and is not easy to solve so instead let's rewrite the sum in terms of a finite sum.
Find P( X > 2 ) = 1 - P( X ? 2 ) =
. . . 2
1 - ? P(X = i)
. . . i = 0
= 0.001148481
== -- == -- == -- ==
D)
the mean of the Poisson distribution is the parameter, ?t = 0.2
== -- == -- == -- ==
E)
the variance of the Poisson distribution is the parameter, ?t = 0.2
the standard deviation is the square root of the variance
sqrt(0.2) = 0.4472136
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