problems problems 4.17, 4.36, 4.40, (try: 4.43). 4. random variables a random variable is a way of...

Problems

Problems 4.17, 4.36, 4.40, (TRY: 4.43)

4. Random Variables

A random variable is a way of recording a quantitative variable of a random experiment.

4. Random Variables


This variable has a distribution, mean and standard deviation, so we can discuss outliers using the same procedures as back in Chapter 2.

4. Random Variables


This variable has a distribution, mean (expected value) and standard deviation, so we can discuss outliers using the same procedures as back in Chapter 2.

This includes percentiles, Chebyshev’s Rule and the Empirical Rule!

4. Random Variables

…outliers using the same procedures as back in Chapter 2.

This includes percentiles, Chebyshev’s Rule and the Empirical Rule!

The difference in this Chapter is we talk about the probabilities of what is to occur and in Chapter 2 we talked about the frequency of what did occur.

4. Random Variables

The difference in this Chapter is we talk about the probabilities of what is to occur and in Chapter 2 we talked about the frequency of what did occur.

In Chapter 2 we are talking about the sample and in Chapter 4 we are talking about the population.

Properties of Probability, P( X = xi )

1)(0 (1) ixXP

1)( (2)1

n

iixXP

Example

Find the probability distribution obtained by flipping an unbiased coin three times and counting the number of times heads comes up.

Binomial Experiment

A binomial experiment is one that:

1) Has a fixed number of trials (n)

2) These trials are independent

3) Each trial must have all outcomes classified into two categories (Success or Failure)

4) The probability of success remains constant for all trials.

Notation:

• S = success and P(S) = p

• F = Failure and P(F) = q = 1- p

• n = fixed number of trials

• x = specific number of successes in n trials

• P(x) = the probability of getting exactly x successes among n trials

Example

Shaquille Rashaun O'Neal (Shaq) is a basketball player who takes a lot of free throws. The probability of Shaq making a free throw is 0.60 on each throw.

With 3 free throws what is the probability that he makes 2 shots?

Notation:

• S = success and P(S) = .6

• F = Failure and P(F) = .4

• n = 3

• x = 2

• P(2) = the probability of getting exactly 2 successes (successful free throws) among n=3 trials

Factorials

0! = 1

1! = 1

2! = 2 * 1

3! = 3 * 2 * 1

4! = 4* 3 * 2 * 1

n! = n*(n-1)!

Factorials

0! = 1

1! = 1

2! = 2 * 1=2

3! = 3 * 2 * 1=6

4! = 4* 3 * 2 * 1=24

n! = n*(n-1)!

Binomial Probability Distribution

In a binomial experiment, with constant probability p of success at each trial, the probability of x successes in n trials is given by

xnxqpxxn

nsuccessesxP

!)!(

!) (

ExampleShaq is a basketball player who takes a lot of free throws. The probability of Shaq making a free throw is 0.60 on each throw.


Shaq is a basketball player who takes a lot of free throws. The probability of Shaq making a free throw is 0.60 on each throw.


0.432

)4(.)6(.!2)!23(

!3)2( 232

xP

Example

Example

Flipping a biased coin 8 times. The probability of heads on each trial is 0.4. What is the probability of obtaining at least 2 heads.

Example


)8(...)3()2()2( xPxPxPxP

Example


)1(1

)8(...)3()2()2(

xP

xPxPxPxP

Example


)0()1(1

)1(1

)8(...)3()2()2(

xPxP

xP

xPxPxPxP

ExampleFlipping a biased coin 8 times. The probability of heads on each trial is 0.4. What is the probability of obtaining at least 2 heads.

8071 )6(.)4(.!0 !8

!8)6(.)4(.

!1 !7

!81

)0()1(1

)1(1

)8(...)3()2()2(

xPxP

xP

xPxPxPxP


894.)6(.)4(.!0 !8

!8)6(.)4(.

!1 !7

!81

)0()1(1

)1(1

)8(...)3()2()2(

8071

xPxP

xP

xPxPxPxP

How to use the Binomial Tables

• (see page 885)

• First find the appropriate table for the particular value of n

• then find the value of p in the top row

• Find the row corresponding to k and find the intersection with the column corresponding to the value of p

• The value you obtain is the cumulative probability, that is P(x ≤ k)

• N=10, p = 0.7: P(x ≤ 4) = 0.047

• N=10, p = 0.7: P(x = 4) = P(x ≤ 4) - P(x ≤ 3) = 0.047-0.011=0.036

• N=10, p = 0.7: P(x > 4) = 1- P(x ≤ 4)

= 1 - 0.047 = 0.953


)1(1

)8(...)3()2()2(

xP

xPxPxPxP


894.

106.01

)1(1

)8(...)3()2()2(

xP

xPxPxPxP

pq

npqnp

1

Mean and Standard deviation

Problems

Problems 4.52, 4.56, 4.62, 4.64, 4.66, 4.68

Keys to success

Learn the binomial table.

Be able to recognize binomial distributions and when you do apply the appropriate formulas and tables.

31

Homework

• Review Chapter 4.4

• Read Chapter 5.1-5.3

problems problems 4.17, 4.36, 4.40, (try: 4.43). 4. random variables a random variable is a way of...

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