Download Below you will find a set of notes outlining Binomial Distributions

Below you will find a set of notes outlining Geometric Distributions. Please read through
these and fill in the blanks, where appropriate. You will know that you need to complete the
blanks because you will see a pencil (). Once you are done with that task, you will
complete the accompanying review exercises on both binomial and geometric distributions.
Feel free to highlight or add information on these note sheets, as they are yours to work
with.
Geometric Distributions:
Suppose we have an experiment where we repeat binomial trials until we get our first success,
then we stop. Let n be the number of the trial on which we get our first success. In this context, n
is not a fixed number. In fact, n could be any of the numbers.
In real life, there are many situations where we keep trying until we achieve success. This is true
in areas such as military science, real estate, general marketing, medical science, and technology.
Suppose the random variable X = the number of trials until success is achieved.
Then X is a geometric random variable if:
1. There are only two outcomes: success or failure.
2. The variable of interest is the number of trials required to obtain the first success.
3. The n observations are independent.
4. The probability of success p is the same for each observation.
Because n is not fixed, there could be an infinite number of X-values.
The probability histogram for a geometric distribution is always skewed to the right. This is
because the probability of getting your first success decreases with each trial.
Your TI-83 calculator knows the geometric distribution.
The probability distribution function (or p.d.f.) for a geometric distribution allows you to find the
probability of an individual outcome. The format is geometpdf(p, x).
The cumulative distribution function (or c.d.f.) for a geometric distribution is the cumulative
density function, meaning it finds the sum (or running total) of the probability of several outcomes.
The format, is geometcdf(p, x). This calculates the probability of finding the first success on or
before the xth trial.
Binomial or Geometric???
If you flip two coins ten times and record the number of times you get two heads, you are doing a
binomial experiment. The binomial distribution summarizes probabilities for how often you’d get
2 heads out of 10 tosses if you repeated the experiment several times. Since there is a given set n
value (it is finite), this represents a binomial experiment.
If you were interested in how many times you had to toss the coins before getting 2 heads for the
first time, you’d be dealing with a geometric experiment. The geometric distribution summarizes
the probabilities for getting 2 heads after tossing the coins once, twice, and so on.
Example:
Suppose each child born to Jay and Kay has probability 0.25 of having blood type O. What is the
probability that their first success of having a child with type O will be on child 1, child 2, child 3,
child 4, and child 5?
Let X = the number of trials until success is achieved (child with type O blood = success)
1. There are only two outcomes: success (type O blood) or failure (not type O blood).
2. The variable of interest is the number of trials required to obtain the first success (there is
not a fixed number of observations, since we don’t know when our first success will
happen).
3. Each of the 5 observations is independent, since one child’s blood type will not influence the
next child’s blood type.
4. The probability of success (type O blood) is the same for each of the 5 observations.
So X is a geometric random variable.
The probability that X is equal to n is given by the following
formula:
P( X  n)  (1  p)n1 p
The probability that X is greater than n is given by the
following formula:
P( X
n)  (1  p)n
X = n = the number of trials until success is achieved
q = (1 - p) = the number of failures
p = the probability for success
*Notice that often the x is interchanged with an n. This is simply a variable substitution!
(If you think about it, the substitution makes sense.)
Explain why the variable substitution of X = n makes sense when given a geometric probability
model. Write your answer in the space below.
Another Example:
An automobile assembly plant produces sheet metal door panels. Each panel moves on an
assembly line. As the panel passes a robot, a mechanical arm will perform spot welding at
different locations. Each location has a magnetic dot painted where the weld is to be made. The
robot is programmed to locate the magnetic dot and perform the weld. However, experience
shows that on each trial the robot is only 85% successful at locating the dot. If it cannot locate the
magnetic dot, it is programmed to try again. The robot will keep trying until it finds the dot (and
does the weld) or the door panel passes out of the robot’s reach.
a) What is the probability that the robot’s first success will be on attempts n = 1, 2, or 3?
Use the formula: P(n)  (1  p)n1 p
(n = 1): (1  0.85)11 (0.85)  (0.15)0 (0.85)  0.85
This means the probability for success
on the first trial is 0.85, or 85%.

(n = 2):
This means the probability for success
on the ________ trial is _____, or
_____.

(n = 3):
This means the probability for success
on the ________ trial is _____, or
_____.
b) The assembly line moves so fast that the robot only has a maximum of three chances before
the door panel is out of reach. What is the probability that the robot will be successful before the
door panel is out of reach? (hint: this is an example of the cumulative geometric distribution
function)
 P(n  1 or 2 or 3)  _____________________________
This means that the weld should be correctly located about _______________ of the time.
c) What is the probability that the robot will not be able to locate the correct spot within three tries?
If 10,000 panels are made, what is the expected number of defectives? Comment on the meaning
of this answer in the context of “forecasting failures” and the “limits of design”.
Answer:
Although this formula can be used quite easily once you understand the model and know
the variables, your graphing calculator will do the work for you. Here’s how:
1. Push the “Dist” button;
2. Choose “Geometpdf” (it is choice D);
3. Enter in “Geometpdf(p, x) or “Geometcdf(p,x)” depending on the type of problem
you are working on;
4. Push “Enter”.
Your calculator will now give you the probability to each value x.
Let’s look back at our example from before about the robots. Complete the following using your
calculator.
The following table shows the probability distribution function (p.d.f.) for the geometric random
variable, X. Complete the table by following the process.
Part A:
X
P(X)
0
1
2
3
4
5
P(X = n = 0):
By calculator: Geometpdf (0.85, 0) = _______________
P(X = n = 1)
By calculator: Geometpdf (
P(X = n = 2)
By calculator: Geometpdf (
P(X = n = 3)
By calculator: Geometpdf (
P(X = n = 4)
By calculator: Geometpdf (
P(X = n = 5)
By calculator: Geometpdf (
Part B:
Does your answer for P(X = n = 0) make sense? Why or why not. Explain in detail using complete
sentences (I am looking for 2-3 well-written sentences here!)
Part C:
Make a probability histogram for the geometric probability distribution function (p.d.f.).
Part D:
Is this distribution skewed right? Why or why not. Again, write a detailed response.
The following table shows the cumulative distribution function (c.d.f.) for the geometric random
variable, X. Complete the table by following the process.

X
0
1
2
3
4
5
P(X)
P( X  0) P( X  1) P( X  2) P( X  3) P( X  4) P( X  5)
F(X)
0
0.85
Geometcdf ( 0.85, 0) = 0
Geometcdf ( 0.85, 1) = 0.85
Geometcdf ( 0.85, 2) =
Geometcdf ( 0.85, 3) =
Geometcdf ( 0.85, 4) =
Geometcdf ( 0.85, 5) =
The easiest way to complete the cumulative distribution function is by using your graphing
calculator. You will use the “Geometcdf” choice on your calculator.
In the space below, construct a histogram of the cumulative distribution function (c.d.f.).
You will also need to know these formulas!
If X is a binomial random variable with parameters n and p, then the mean (expected value) of X
is:
x  1
p
Let’s look at another example of a cumulative geometric distribution.
Example:
There are 10 different prizes in boxes of Frosted Flakeys. Prize #4 is the most valuable to
collectors. What is the probability that you will get Prize #4 (considered the “success” in this case)
without having to buy more than four boxes of Frosted Flakeys? (Assume all prizes are equally
likely in each box.)
What we are trying to find is: P( X  4)  G (2)  G (2)  G (3)  G (4)
Geometcdf(0.01, 4) = ________________
Why did I use the c.d.f. instead of the p.d.f.?
Explain below.
Given the situation above, suppose you are unlucky and don’t get your prize in one of the first four
boxes. On average, how long will you have to wait to get Prize #4?
Average Wait Time = Expected Value =
Answer:
What does this answer mean?
x  1
p