Chapter 6: Probability Distributions

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Chapter 6: Probability Distributions

Section 6.1: How Can We Summarize Possible Outcomes and Their

Probabilities?

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Learning Objectives

1. Random variable

2. Probability distributions for discrete random variables

3. Mean of a probability distribution

4. Summarizing the spread of a probability distribution

5. Probability distribution for continuous random variables

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Learning Objective 1:Randomness

The numerical values that a variable assumes are the result of some random phenomenon:Selecting a random sample for a

population

orPerforming a randomized experiment

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Learning Objective 1:Random Variable

A random variable is a numerical measurement of the outcome of a random phenomenon.

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Learning Objective 1:Random Variable

Use letters near the end of the alphabet, such as x, to symbolize Variables A particular value of the random variable

Use a capital letter, such as X, to refer to the random variable itself.

Example: Flip a coin three times X=number of heads in the 3 flips; defines the random

variable x=2; represents a possible value of the random

variable

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Learning Objective 2:Probability Distribution

The probability distribution of a random variable specifies its possible values and their probabilities.

Note: It is the randomness of the variable that allows us to specify probabilities for the outcomes

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Learning Objective 2:Probability Distribution of a Discrete Random Variable

A discrete random variable X has separate values (such as 0,1,2,…) as its possible outcomes

Its probability distribution assigns a probability P(x) to each possible value x: For each x, the probability P(x) falls between 0

and 1 The sum of the probabilities for all the possible x

values equals 1

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Learning Objective 2:Example

What is the estimated probability of at least three home runs?P(3)+P(4)+P(5)=0.13+0.03+0.01=0.17

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Learning Objective 3:The Mean of a Discrete Probability Distribution

The mean of a probability distribution for a discrete random variable is

where the sum is taken over all possible values of x.

)(xpx

The mean of a probability distribution is denoted by the parameter, µ.

The mean is a weighted average; values of x that are more likely receive greater weight P(x)

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Learning Objective 3:Expected Value of X

The mean of a probability distribution of a random variable X is also called the expected value of X.

The expected value reflects not what we’ll observe in a single observation, but rather that we expect for the average in a long run of observations.

It is not unusual for the expected value of a random variable to equal a number that is NOT a possible outcome.

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Find the mean of this probability distribution.

The mean:

= 0(0.23) + 1(0.38) + 2(0.22) + 3(0.13) + 4(0.03) + 5(0.01) = 1.38

)(xpx

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Learning Objective 4:The Standard Deviation of a Probability Distribution

The standard deviation of a probability distribution, denoted by the parameter, σ, measures its spread. Larger values of σ correspond to greater

spread. Roughly, σ describes how far the random

variable falls, on the average, from the mean of its distribution

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Learning Objective 5:Continuous Random Variable

A continuous random variable has an infinite continuum of possible values in an interval.

Examples are: time, age and size measures such as height and weight.

Continuous variables are measured in a discrete manner because of rounding.

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Learning Objective 5:Probability Distribution of a Continuous Random Variable A continuous random variable has possible

values that form an interval. Its probability distribution is specified by a curve. Each interval has probability between 0 and 1. The interval containing all possible values has

probability equal to 1.

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Section 6.2: How Can We Find Probabilities for Bell-Shaped

Distributions?

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Learning Objectives

1. Normal Distribution

2. 68-95-99.7 Rule for normal distributions

3. Z-Scores and the Standard Normal Distribution

4. The Standard Normal Table: Finding Probabilities

5. Using the TI-calculator: find probabilities

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Learning Objectives

6. Using the Standard Normal Table in Reverse7. Using the TI-calculator: find z-scores8. Probabilities for Normally Distributed

Random Variables 9. Percentiles for Normally Distributed Random

Variables10. Using Z-scores to Compare Distributions

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Learning Objective 1:Normal Distribution

The normal distribution is symmetric, bell-shaped and characterized by its mean µ and standard deviation . The normal distribution is the most important

distribution in statistics Many distributions have an approximate normal

distribution Approximates many discrete distributions well

when there are a large number of possible outcomes

Many statistical methods use it even when the data are not bell shaped

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Normal distributions are Bell shaped Symmetric around the mean

The mean () and the standard deviation () completely describe the density curve Increasing/decreasing moves the curve

along the horizontal axis Increasing/decreasing controls the spread of

the curve

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Within what interval do almost all of the men’s heights fall? Women’s height?

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Learning Objective 2:68-95-99.7 Rule for Any Normal Curve

68% of the observations fall within one standard deviation of the mean

95% of the observations fall within two standard deviations of the mean

99.7% of the observations fall within three standard deviations of the mean

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Learning Objective 2: Example : 68-95-99.7% Rule

Heights of adult women can be approximated by a normal distribution

= 65 inches; =3.5 inches 68-95-99.7 Rule for women’s heights

68% are between 61.5 and 68.5 inches

[ µ = 65 3.5 ]

95% are between 58 and 72 inches

[ µ 2 = 65 2(3.5) = 65 7 ]

99.7% are between 54.5 and 75.5 inches

[ µ 3 = 65 3(3.5) = 65 10.5 ]

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Learning Objective 2:Example : 68-95-99.7% Rule

What proportion of women are less than 69 inches tall?

?

65 68.5 (height values)

+1

? = 84%

68% (by 68-95-99.7 Rule)16%

-1

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Learning Objective 3:Z-Scores and the Standard Normal Distribution

The z-score for a value x of a random variable is the number of standard deviations that x falls from the mean

A negative (positive) z-score indicates that the value is below (above) the mean

z-scores can be used to calculate the probabilities of a normal random variable using the normal tables in the back of the book

zx

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Learning Objective 3:Z-Scores and the Standard Normal Distribution

A standard normal distribution has mean µ=0 and standard deviation σ=1

When a random variable has a normal distribution and its values are converted to z-scores by subtracting the mean and dividing by the standard deviation, the z-scores have the standard normal distribution.

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Learning Objective 4:Table A: Standard Normal Probabilities

Table A enables us to find normal probabilities It tabulates the normal cumulative probabilities

falling below the point +zTo use the table:

Find the corresponding z-score Look up the closest standardized score (z) in

the table. First column gives z to the first decimal place First row gives the second decimal place of z

The corresponding probability found in the body of the table gives the probability of falling below the z-score

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Learning Objective 4:Example: Using Table A

Find the probability that a normal random variable takes a value less than 1.43 standard deviations above µ; P(z<1.43)=.9236

TI Calculator = Normcdf(-1e99,1.43,0,1)= .9236TI Calculator = Normcdf(-1e99,1.43,0,1)= .9236

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Learning Objective 4:Example: Using Table A

Find the probability that a normal random variable takes a value greater than 1.43 standard deviations above µ: P(z>1.43)=1-.9236=.0764

TI Calculator = Normcdf(1.43,1e99,0,1)= 0.0764TI Calculator = Normcdf(1.43,1e99,0,1)= 0.0764

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Learning Objective 4:Example:

Find the probability that a normal random variable assumes a value within 1.43 standard deviations of µ Probability below 1.43σ = .9236 Probability below -1.43σ = .0764 (1-.9236) P(-1.43<z<1.43) =.9236-.0764=.8472

TI Calculator = Normcdf(-1.43,1.43,0,1)= .8472TI Calculator = Normcdf(-1.43,1.43,0,1)= .8472

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Learning Objective 5:Using the TI Calculator

To calculate the cumulative probability 2nd DISTR; 2:normalcdf(lower bound, upper

bound,mean,sd) Use –1E99 for negative infinity and 1E99 for

positive infinity

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Learning Objective 5:Find Probabilities Using TI Calculator

Find probability to the left of -1.64 P(z<-1.64)=normcdf(-1e99,-1.64,0,1)=.0505

Find probability to the right of 1.56 P(z>1.56)=normcdf(1.56,1e99,0,1)=.0594

Find probability between -.50 and 2.25 P(-.5<z<2.25)=normcdf(-.5,2.25,0,1)=.6793

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Learning Objective 6:How Can We Find the Value of z for a Certain Cumulative Probability? To solve some of our problems, we will need

to find the value of z that corresponds to a certain normal cumulative probability

To do so, we use Table A in reverse Rather than finding z using the first column

(value of z up to one decimal) and the first row (second decimal of z)

Find the probability in the body of the table The z-score is given by the corresponding values

in the first column and row

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Learning Objective 6:How Can We Find the Value of z for a Certain Cumulative Probability?

Example: Find the value of z for a cumulative probability of 0.025.

Look up the cumulative probability of 0.025 in the body of Table A.

A cumulative probability of 0.025 corresponds to z = -1.96.

Thus, the probability that a normal random variable falls at least 1.96 standard deviations below the mean is 0.025.

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Learning Objective 6:How Can We Find the Value of z for a Certain Cumulative Probability? Example: Find the value of z for a cumulative

probability of 0.975. Look up the cumulative probability of 0.975 in the

body of Table A. A cumulative probability of 0.975 corresponds to z

= 1.96. Thus, the probability that a normal

random variable takes a value no morethan 1.96 standard deviations above the mean is 0.975.

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Learning Objective 7:Using the TI Calculator to Find Z-Scores for a Given Probability 2nd DISTR 3:invNorm; Enter invNorm(percentile,mean,sd)

Percentile is the probability under the curve from negative infinity to the z-score

Enter

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Learning Objective 7:Examples

The probability that a standard normal random variable assumes a value that is ≤ z is 0.975. What is z? Invnorm(.975,0,1)=1.96

The probability that a standard normal random variable assumes a value that is > z is 0.0275. What is z? Invnorm(.975,0,1)=1.96

The probability that a standard normal random variable assumes a value that is ≥ z is 0.881. What is z? Invnorm(1-.881,0,1)=-1.18

The probability that a standard normal random variable assumes a value that is < z is 0.119. What is z? Invnorm(.119,0,1)= -1.18

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Find the z-score z such that the probability within z standard deviations of the mean is 0.50. Invnorm(.75,0,1)= .67 Invnorm(.25,0,1)= -.67

Probability = P(-.67<Z<.67)=.5

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Learning Objective 8:Finding Probabilities for Normally Distributed Random Variables1. State the problem in terms of the observed

random variable X, i.e., P(X<x)2. Standardize X to restate the problem in

terms of a standard normal variable Z

3. Draw a picture to show the desired probability under the standard normal curve

4. Find the area under the standard normal curve using Table A

P(X x) P Z zx

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Learning Objective 8:P(X<x)

Adult systolic blood pressure is normally distributed with µ = 120 and σ = 20. What percentage of adults have systolic blood pressure less than 100?

P(X<100) =

Normcdf(-1E99,100,120,20)=.1587

15.9% of adults have systolic blood pressure less than 100

P Z 100 120

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P(z 1.00) .1587

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Learning Objective 8:P(X>x)

Adult systolic blood pressure is normally distributed with µ = 120 and σ = 20. What percentage of adults have systolic blood pressure greater than 100?

P(X>100) = 1 – P(X<100)

P(X>100)= 1-.1587=.8413 Normcdf(100,1e99,120,20)=.8413 84.1% of adults have systolic blood pressure greater than

100

P Z 100 120

20

P(Z 1.00) .1587

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Learning Objective 8:P(X>x)

Adult systolic blood pressure is normally distributed with µ = 120 and σ = 20. What percentage of adults have systolic blood pressure greater than 133?

P(X>133) = 1 – P(X<133)

P(X>133)= 1-.7422=.2578 Normcdf(133,1E99,120,20)=.2578 25.8% of adults have systolic blood pressure greater than

133

P Z 133 120

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P(Z .65) .7422

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Learning Objective 8: P(a<X<b)

Adult systolic blood pressure is normally distributed with µ = 120 and σ = 20. What percentage of adults have systolic blood pressure between 100 and 133?

P(100<X<133) = P(X<133)-P(X<100)

Normcdf(100,133,120,20)=.5835 58% of adults have systolic blood pressure between 100

and 133

P Z 133 120

20

P Z

100 120 20

P(Z .65) P(Z 1.00) .7422 .1587 .5835

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Learning Objective 9:Find X Value Given Area to Left

Adult systolic blood pressure is normally distributed

with µ = 120 and σ = 20. What is the 1st quartile? P(X<x)=.25, find x:

Look up .25 in the body of Table A to find z= -0.67 Solve equation to find x:

Check: P(X<106.6) P(Z<-0.67)=0.25 TI Calculator = Invnorm(.25,120,20)=106.6

x z 120 ( 0.67)*20 106.6

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Learning Objective 9:Find X Value Given Area to Right

Adult systolic blood pressure is normally distributed with µ = 120 and σ = 20. 10% of adults have systolic blood pressure above what level?

P(X>x)=.10, find x. P(X>x)=1-P(X<x) Look up 1-0.1=0.9 in the body of Table A to find z=1.28 Solve equation to find x:

Check: P(X>145.6) =P(Z>1.28)=0.10 TI Calculator = Invnorm(.9,120,20)=145.6

x z 120 (1.28)*20 145.6

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Learning Objective 10:Using Z-scores to Compare Distributions

Z-scores can be used to compare observations from different normal distributions

Example: You score 650 on the SAT which has =500 and

=100 and 30 on the ACT which has =21.0 and =4.7. On which test did you perform better?

Compare z-scoresSAT: ACT:

Since your z-score is greater for the ACT, you performed better on this exam

z650 500

1001.5

z30 21

4.71.91

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Section 6.3: How Can We Find Probabilities When Each Observation Has

Two Possible Outcomes?

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Learning Objectives

1. The Binomial Distribution2. Conditions for a Binomial Distribution3. Probabilities for a Binomial Distribution4. Factorials5. Examples using Binomial Distribution6. Do the Binomial Conditions Apply?7. Mean and Standard Deviation of the Binomial

Distribution8. Normal Approximation to the Binomial

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Learning Objective 1:The Binomial Distribution

Each observation is binary: it has one of two possible outcomes.

Examples: Accept, or decline an offer from a bank for a credit

card. Have, or do not have, health insurance. Vote yes or no on a referendum.

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Learning Objective 2:Conditions for the Binomial Distribution

Each of n trials has two possible outcomes: “success” or “failure”.

Each trial has the same probability of success, denoted by p.

The n trials are independent.

The binomial random variable X is the number of successes in the n trials.

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Learning Objective 3:Probabilities for a Binomial Distribution

Denote the probability of success on a trial by p.

For n independent trials, the probability of x successes equals:

P(x) n!

x!(n - x)!px (1 p)n x, x 0,1,2,...,n

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Learning Objective 4:Factorials

Rules for factorials: n!=n*(n-1)*(n-2)…2*1 1!=1 0!=1

For example, 4!=4*3*2*1=24

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Learning Objective 5:Example: Finding Binomial Probabilities

John Doe claims to possess ESP. An experiment is conducted:

A person in one room picks one of the integers 1, 2, 3, 4, 5 at random.

In another room, John Doe identifies the number he believes was picked.

Three trials are performed for the experiment. Doe got the correct answer twice.

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Learning Objective 5:Example 1

If John Doe does not actually have ESP and is actually guessing the number, what is the probability that he’d make a correct guess on two of the three trials?

The three ways John Doe could make two correct guesses in three trials are: SSF, SFS, and FSS.

Each of these has probability: (0.2)2(0.8)=0.032.

The total probability of two correct guesses is 3(0.032)=0.096.

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Learning Objective 5:Example 1

The probability of exactly 2 correct guesses is the binomial probability with n = 3 trials, x = 2 correct guesses and p = 0.2 probability of a correct guess.

P(2) 3!

2!1!(0.2)2(0.8)1 3(0.04)(0.8) 0.096

2nd Vars0:binampdf(n,p,x)Binampdf(3,.2,2)=0.096

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Learning Objective 5:Binomial Example 2

1000 employees, 50% Female None of the 10 employees chosen for management

training were female.

The probability that no females are chosen is:

Binompdf(10,.5,0)=9.765625E-4 It is very unlikely (one chance in a thousand) that none of the

10 selected for management training would be female if the employees were chosen randomly

P(0) 10!

0!10!(0.50)0(0.50)10 0.001

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Learning Objective 6:Do the Binomial Conditions Apply?

Before using the binomial distribution, check that its three conditions apply:

Binary data (success or failure). The same probability of success for each

trial (denoted by p). Independent trials.

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Learning Objective 6:Do the Binomial Conditions Apply to Example 2?

The data are binary (male, female). If employees are selected randomly, the

probability of selecting a female on a given trial is 0.50.

With random sampling of 10 employees from a large population, outcomes for one trial does not depend on the outcome of another trial

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Learning Objective 7:Binomial Mean and Standard Deviation

The binomial probability distribution for n trials with probability p of success on each trial has mean µ and standard deviation σ given by:

p)-np(1 np,

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Learning Objective 7: Example: Racial Profiling?

Data: 262 police car stops in Philadelphia in 1997. 207 of the drivers stopped were African-American. In 1997, Philadelphia’s population was 42.2%

African-American.

Does the number of African-Americans stopped suggest possible bias, being higher than we would expect (other things being equal, such as the rate of violating traffic laws)?

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Learning Objective 7:Example: Racial Profiling?

Assume: 262 car stops represent n = 262 trials. Successive police car stops are

independent. P(driver is African-American) is p = 0.422.

Calculate the mean and standard deviation of this binomial distribution:

8(0.578)262(0.422)

111 262(0.422)

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Learning Objective 7: Example: Racial Profiling?

Recall: Empirical Rule When a distribution is bell-shaped, close to

100% of the observations fall within 3 standard deviations of the mean.

1353(8)111 3

87 3(8) - 111 3 -u

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Learning Objective 7:Example: Racial Profiling?

If there is no racial profiling, we would not be surprised if between about 87 and 135 of the 262 drivers stopped were African-American.

The actual number stopped (207) is well above these values.

The number of African-Americans stopped is too high, even taking into account random variation.

Limitation of the analysis:Different people do different amounts of driving, so we don’t really know that 42.2% of the potential stops were African-American.

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Learning Objective 8:Approximating the Binomial Distribution with the Normal Distribution

The binomial distribution can be well approximated by the normal distribution when the expected number of successes, np, and the expected number of failures, n(1-p) are both at least 15.

Chapter 6: Probability Distributions

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