Welcome to a lesson
on the Poisson distribution.
There are two main characteristics of a Poisson experiment.
Number one, the Poisson probability distribution
gives the probability of a number of events occurring
in a fixed interval of time or space
if these events happen with a known average rate
and independently of the time since the last event.
For example, a book editor might be interested in the number
of words spelled incorrectly in a particular book.
It might be that, on average, there are five words
spelled incorrectly in 100 pages.
The interval is the 100 pages.
Number two, the Poisson distribution may be used
to approximate the binomial
if the probability of success is small, such as 0.01 or 1%,
and the number of trials as large, such as 1000,
where N is the number of trials
and P is the probability of a success.
The random variable X is equal to the number of occurrences
in the interval of interest.
We use the notation shown here for the Poisson distribution.
We read this as X is a random variable
with a Poisson distribution.
The parameter Mu or Lambda is the mean
for the interval of interest.
As an example, a company receives about 16 telephone calls
between eight AM and 10 AM.
We want to find probabilities over the time interval
of 15 minutes.
We let X equal the number of calls received in 15 minutes.
Notice the interval of interest is 15 minutes,
or one fourth of an hour.
So X takes on the values
of zero one, two, three, four, and so on.
If the company receives, on average, 16 calls in two hours,
and there are eight 15-minute intervals in two hours,
then Mu or Lambda is equal to 16 divided by eight,
or one eighth times 16, which is equal to two,
which we indicate using this notation here,
shown at the bottom.
Now let's find some probabilities.
First, what is the probability the company receives
exactly three calls in the next 15 minutes?
If we want to find the probability, X is equal to three.
To find the probabilities, we will be using
the Poisson PDF and the Poisson CDF functions
on the TI-84.
To find the probability X equals three,
we use the Poisson PDF function, enter the mean, comma,
the value of X shown here.
So going to the calculator, we press second VAR
for the distribution menu, arrow up to Poisson PDF,
which is option C.
Be careful not to select CDF,
which we'll use in the next two examples.
Press enter, Lambda or Mu is two, enter,
X is three, enter, press enter on paste,
and enter again.
To four decimal places, we have approximately 0.1804.
Next, what is the probability the company receives three
or less calls in the next 15 minutes, which is a probability
that X is less than or equal to three.
To find this probability,
we use the Poisson CDF function on the TI-84.
CDF stands for cumulative distribution function,
and the Poisson CDF function accumulates
the probabilities from the left.
So to find the probability X is less than or equal three,
we need to use the Poisson CDF function.
So going back to the calculator,
again, we press second VAR for the distribution menu,
arrow up to Poisson CDF, which is option D, press enter,
enter Mu or Lambda, which is two, X, which is three,
enter on paste, enter again.
To four decimal places, we have approximately 0.8571.
And for third probability, we went to find that probability
the company receives more than three calls
in the next 15 minutes,
which is the probability that X is greater than three.
And again, because the Poisson CDF function accumulates
the probabilities from the left,
starting at X equals zero, to find the probability X
is greater than three, we need to find one minus
the Poisson CDF function value
when the mean is two and X is three.
We enter one minus, second VARS, arrow up to Poisson CDF,
enter, two, enter, three, enter, enter, enter.
And we have approximately 0.1429.
Notice how, if we were to add the probabilities
from the second and third questions, we get one, or 100%.
Let's take a look at one more example.
The text message users receive or send an average
of 41.5 text messages per day.
How many text messages does a text message user
receive or send per hour?
Notice how the time interval is one hour,
and we're given information about the number
of text messages per day.
So to find Mu, or the mean, we'll have to take 41.5
and divide by the number of hours in one day,
which is 24.
Going to the calculator.
Mu is approximately 1.7292.
Now, if we use this approximation, we will have more
of an error than if we just use the quotient.
So I'm going to go ahead and use the quotient
to find these probabilities.
So the first question is, what is the probability
that a text message user receives or sends
two messages per hour,
which is a probability X is equal to two.
So because we have X equals two,
we will use the Poisson PDF function,
not the Poisson CDF function.
PDF stands for probability density function.
We will enter Poisson PDF.
And then the mean is a fraction of 41.5 divided by 24,
comma, the X value of two.
So going back to the calculator.
Again, second VARS for distribution,
arrow up to Poisson PDF, which is C,
press enter, enter the mean, which is 41.5 divided by 24,
enter, X is equal to two, enter,
enter on paste, enter again.
To four decimal places, we have approximately 0.2653.
And then, for the last example, what is the probability
that a text message user receives or sends
more than two messages per hour,
which is a probability X is greater than two.
So because the Poisson CDF function gives us a probability
when X is less than or equal to a particular value,
to find the probability X is greater than two,
we'll have to take one and then subtract
the function value of Poisson CDF,
or again, the mean is 41.5 divided by 24,
and X is equal to two.
We enter a one minus second VARS, Poisson CDF,
which is option D, enter, Mu, again, is 41.5 divided by 24,
X is equal to two, enter, enter, enter.
And we have approximately 0.2505.
I hope you found this helpful.