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PPT – Chapter 5 Discrete Probability Distributions PowerPoint presentation | free to view - id: 4883c-NDFmM

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Chapter 5 Discrete Probability Distributions

- Mrs. Mandy Wimpey
- Palmetto High School

Real-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?

Introduction 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.

Constructing 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.

Constructing 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.

Requirements for a Probability Distribution

- The total (sum) of the probabilities must equal

1.0. - Each individual probability must be between 0 and

1, inclusively.

Mean, 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.

Steps 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.

Using 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.

Examples

- 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.

The 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

More 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.

If 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.

- 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.

The 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.

Notation 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

The Binomial Probability Formula

Using the TI-84 will help simplify this formula.

You will use the MATH menu. The formula will

then look like

Examples

- 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

- 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

Mean and Standard Deviation for a Binomial

Distribution

- Mean
- Standard Deviation

Example

- 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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