Title: Chapter 5 Discrete Probability Distributions
1Chapter 5 Discrete Probability Distributions
- Mrs. Mandy Wimpey
- Palmetto High School
2Real-World Application
- DIAGNOSTIC TESTING
- When a disease is rare, health care workers
will pool the blood samples of many people and
test the combination of blood. This saves time
and money. - If the pooled sample results in a negative test,
no further testing is needed (no one has the
disease). - If the pooled sample results in a positive test,
further testing will be conducted on an
individual basis (since at least one person has
the disease). - Lets suppose the probability of having a disease
is 0.05 and a pooled sample of 15 individuals is
tested. What is the probability that no further
testing will be needed?
3Introduction and Terms
- Random Variable a variable whose values are
determined by chance - Discrete Random Variable a variable whose
values are found by counting - Discrete Probability Distribution the values a
random variable can assume and their
corresponding probabilities.
4Constructing a Discrete Probability Distribution
Construct a probability distribution for rolling
one die. List all possible outcomes on the
first row. These are all your x-values. List
each x-values chance of occurring on the second
row. These are the corresponding
probabilities for each x-value.
5Constructing a Discrete Probability Distribution
- A coin is tossed 3 times. Construct a
probability distribution for the number of heads
you get in 3 tosses.
Recall the sample space for a coin tossed 3
times. Do a tree diagram if you have forgotten.
6Requirements for a Probability Distribution
- The total (sum) of the probabilities must equal
1.0. - Each individual probability must be between 0 and
1, inclusively.
7Mean, Standard Deviation, and Expected Value
- The mean (average) of a probability distribution
is calculated as - The standard deviation of a probability
distribution is calculated as - The expected value of a probability distribution
is the mean of the probability distribution.
8Steps in calculating the Mean and Standard
Deviation
- The Mean
- Multiply each x by its corresponding P(x).
- Add them up.
- This is the mean or expected value for the
distribution. - The Standard Deviation
- Square each x value.
- Multiply each squared x by its corresponding
P(x) value. - Add them up.
- Subtract the mean from the sum you found in step
3. - Take the square root of the difference you found
in step 4.
9Using the TI-84 to calculate the Mean Standard
Deviation for a Probability Distribution
- Enter your x values into L1.
- Enter your P(x) values into L2.
- Press STAT
- Go over to CALC
- Select 1-VAR STATS
- Type L1, L2 to tell the calculator to use both
lists at one time - Press ENTER.
10Examples
- Example 1
- The probability that 0, 1, 2, 3, or 4 people will
be placed on hold when they call a radio talk
show is shown in the distribution. - Find the expected number of people on hold and
the standard deviation. - Should the radio station get an additional phone
line? Explain.
11The expected number of people on hold is the mean
number of people on hold. This is calculated by
The radio station can expect an average of 1.6
people to be on hold at any given time.
The standard deviation of the number of people on
hold is found as follows
12More Expected Value
- Example One thousand tickets are sold at 1
each for a color TV valued at 350. What is the
expected value of the gain if a person purchases
one ticket? - The problem must be set up using a table as
follows in the next slide.
13If you win, you have gained only 349 since you
paid a dollar to enter the contest. Since there
are 1000 tickets and 1 winning ticket, you have a
1/1000 chance of winning. If you lose, you lost
1, so its a negative amount. Since there is
only 1 winning ticket, there are 999 that are
non-winning tickets.
14- To calculate the Expected Value, you are
calculating the Mean. Multiply each x by its
corresponding P(x) value. Then, add. - E(x) (349)(1/1000) (-1)(999/1000)
- -0.65
- One individual person either wins a 350 TV or
loses 1. So the -0.65 concerns the average
loss over a period of time.
15The Binomial Probability Distribution
- Requirements for a Binomial Distribution
- Each trial can have only 2 outcomes. We call
these successes and failures. - There is a fixed number of trials. (The
experiment doesnt go on forever.) - Each trial is independent.
- The probability of a success remains constant for
each trial.
16Notation for Binomial Distributions
- P(S) probability of a success
- P(F) probability of a failure
- p numerical probability of success
- q numerical probability of failure
- P(S) p and P(F) q
- n total number of trials
- x number of successes in n trials
- q 1- p
17The Binomial Probability Formula
Using the TI-84 will help simplify this formula.
You will use the MATH menu. The formula will
then look like
18Examples
- Example 1
- A survey found that one out of five Americans say
he or she has visited a doctor in any given
month. If 10 people are selected at random, find
the probability that exactly 3 will have visited
a doctor last month. - In this case, n 10 x 3 p 1/5 and q 4/5.
- So, the formula should look like this
19- Example 2
- A survey found that 30 of teenage consumers
receive their spending money from part-time jobs.
If 5 teenagers are randomly selected, find the
probability that at least 3 of them have
part-time jobs. - For this problem, n 5 x 3, 4, 5 p 0.30
and q 0.70 - Putting it all together in the formula, we get
20Mean and Standard Deviation for a Binomial
Distribution
21Example
- Example 1
- A coin is tossed 4 times. Find the mean and
standard deviation of the number of heads
obtained. - Mean
- Standard Deviation
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