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Chapter 6
Probability Distributions
Section 6.1
Summarizing Possible Outcomes and Their
Probabilities
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Randomness
A random variable is a numerical measurement of the
outcome of a random phenomenon.
Often, the randomness results from
 selecting a random sample for a population
or
 performing a randomized experiment
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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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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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Probability Distribution of a Discrete
Random Variable
A discrete random variable X takes a set of 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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Example: Number of Home Runs in a
Game
What is the estimated probability of at least three home
runs?
Table 6.1 Probability Distribution of Number of Home Runs in a Game for
San Francisco Giants
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Example: Number of Home Runs in a
Game
Table 6.1 Probability Distribution of Number of Home Runs in a Game for
San Francisco Giants
The probability of at least three home runs in a game is
P(3)+P(4)+P(5 or more)= 0.0556 + 0.0185 + 0 = 0.0741
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Mean of a Discrete Probability Distribution
The mean of a probability distribution for a discrete
random variable is:
   x  p(x)
where the sum is taken over all possible values of x.
The mean of a probability distribution is denoted by the
parameter,  .
The mean is a weighted average; values of x that are
more likely receive greater weight P(x).
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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 what 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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Example: Number of Home Runs in a
Game
Find the mean of this probability distribution.
Table 6.1 Probability Distribution of Number of Home Runs in a Game for
San Francisco Giants
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Example: Number of Home Runs in a
Game
The mean:
   x  p(x)
= 0(0.3889) + 1(0.3148) + 2(0.2222) + 3(0.0556) + 4(0.0185)
= 0 * P(0) + 1 * P(1) + 2 * P(2) + 3 * P(3) + 4 * P(4)
=1
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The Standard Deviation of a Probability
Distribution
The standard deviation of a probability distribution,
denoted by the parameter,  , measures variability from
the mean.
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
Larger values of  correspond to greater spread.

Roughly,  describes how far the random variable
falls, on the average, from the mean of its
distribution.
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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 usually measured in a
discrete manner because of rounding.
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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.
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
Each interval has probability between 0 and 1.

The interval containing all possible values has
probability equal to 1.
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Probability Distribution of a Continuous
Random Variable
Smooth curve approximation
Figure 6.2 Probability Distribution of Commuting Time. The area under the curve for
values higher than 45 is 0.15. Question: Identify the area under the curve represented by
the probability that commuting time is less than 15 minutes, which equals 0.29.
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Chapter 6
Probability Distributions
Section 6.2
Probabilities for Bell-Shaped Distributions
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Normal Distribution
The normal distribution is symmetric, bell-shaped and
characterized by its mean  and standard deviation  .
 The normal distribution is the most important
distribution in statistics.
 Many distributions have an approximately normal
distribution.
 The normal distribution also can approximate
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 Distribution
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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Normal Distribution
Within what interval do almost all of the men’s heights
fall? Women’s height?
Figure 6.4 Normal Distributions for Women’s Height and Men’s Height. For each different
combination of  and  values, there is a normal distribution with mean  and standard
deviation  . Question: Given that  = 70 and  = 4, within what interval do almost all of the
men’s heights fall?
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Normal Distribution: 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.
Figure 6.5 The Normal Distribution. The probability equals approximately 0.68 within
1 standard deviation of the mean, approximately 0.95 within 2 standard deviations,
and approximately 0.997 within 3 standard deviations. Question: How do these
probabilities relate to the empirical rule?
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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
 [     65  3.5]
95% are between 58 and 72 inches
 [   2  65  2(3.5)  65  7]
99.7% are between 54.5 and 75.5 inches
 [   3  65  3(3.5)  65  10.5]
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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.
z
x 

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 Table A
in the back of the book.
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Z-Scores and the Standard Normal
Distribution
A standard normal distribution has mean
 0
standard deviation   1 .
and
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
follow the standard normal distribution.
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Table A: Standard Normal Probabilities
Table A enables us to find normal probabilities.
 It tabulates the normal cumulative probabilities
falling below the point   z .
To 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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Example: Using Table A
Find the probability that a normal random variable takes
a value less than 1.43 standard deviations above ;
P( z  1.43)  0.9236
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Example: Using Table A
Figure 6.7 The Normal Cumulative Probability, Less than z Standard Deviations
above the Mean. Table A lists a cumulative probability of 0.9236 for z  1.43, so
0.9236 is the probability less than 1.43 standard deviations above the mean of any
normal distribution (that is, below   1.43 ). The complement probability of 0.0764 is
the probability above   1.43 in the right tail.
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Example: Using Table A
Find the probability that a normal random variable
assumes a value within 1.43 standard deviations of  .
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
Probability below 1.43  0.9236

Probability below 1.43  0.0764

P(1.43  z  1.43)  0.9236  0.0764  0.8472
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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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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 zto 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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SUMMARY: Using Z-Scores to Find Normal
Probabilities or Random Variable x Values
 If we’re given a value x and need to find a probability,
convert x to a z-score using z  ( x   ) /  , use a table of
normal probabilities (or software, or a calculator) to get a
cumulative probability and then convert it to the probability
of interest
 If we’re given a probability and need to find the value of
x , convert the probability to the related cumulative
probability, find the z-score using a normal table (or
software, or a calculator), and then evaluate x    z .
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Example: Comparing Test Scores That
Use Different Scales
Z-scores can be used to compare observations from
different normal distributions.
Picture the Scenario:
There are two primary standardized tests used by college
admissions, the SAT and the ACT.
You score 650 on the SAT which has   500 and   100
and 30 on the ACT which has   21.0 and   4.7.
How can we compare these scores to tell which score is
relatively higher?
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Using Z-scores to Compare Distributions

Compare z-scores:
SAT: z 
650  500
1.5
100
30  21
 1.91
ACT: z 

4.7
Since your z-score is greater for the ACT, you

performed
relatively better on this exam.
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Chapter 6
Probability Distributions
Section 6.3
Probabilities When Each Observation Has
Two Possible Outcomes
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The Binomial Distribution: Probabilities
for Counts with Binary Data
We use the binomial distribution when 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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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, so the probability of a failure is denoted by 1  p .

The n trials are independent, That is, the result for one
trial does not depend on the results of other trials.
The binomial random variable X is the number of successes
in the n trials.
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Probabilities for a Binomial Distribution
When the number of trials n is large, it’s tedious to write
out all the possible outcomes in the sample space. There is
a formula you can use to find binomial probabilities for any
n.
Denote the probability of success on a trial by p. For n
independent trials, the probability of x successes equals:
n!
P(x) 
p x (1 p) nx , x  0,1,2,...,n
x!(n - x)!
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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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Example: An ESP Experiment
John Doe claims to possess extrasensory perception
(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.
 John Doe got the correct answer twice.
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Example: An ESP Experiment
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.
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 Each of these has probability: (0.2) (0.8)  0.032 .
 The total probability of two correct guesses is
3(0.032)  0.096.
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Example: An ESP Experiment
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.
3!
2
1
P(2) 
(0.2) (0.8)  3(0.04)(0.8)  0.096
2!1!
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Example: Testing for Gender Bias in
Promotions
A group of female employees has claimed that female
employees are less likely than male employees of similar
qualifications to be promoted.
 Of 1000 employees, 50% are female.
 None of the 10 employees chosen for management
training were female.
 The probability that no females are chosen is:
10!
P(0) 
(0.50)0 (0.50)10  0.001
0!10!
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.
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Example: Testing for Gender Bias in
Promotions
Before using the binomial distribution, check that its three
conditions apply:
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
Binary data (success or failure)

The same probability of success for each trial
(denoted by p)

Independent trials
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Example: Testing for Gender Bias in
Promotions
Do the Binomial Conditions Apply?
 The data are binary (male, female).
 If employees are selected randomly, the probability of
selecting a female on any given trial is 0.50.
 With random sampling of 10 employees from a large
population, the outcome for one trial doesn’t depend on
the outcome of another trial.
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Mean and Standard Deviation of the
Binomial Distribution
Binomial Mean and Standard Deviation:
The binomial probability distribution for n trials with
probability p of success on each trial has mean  and
standard deviation  given by:
  np,  
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np(1-p)
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Example: Checking for Racial Profiling
In 2006, the New York City Police Department (NYPD)
confronted approximately 500,000 pedestrians for
suspected criminal violations.
 88.9% were non-white.
 Meanwhile, according to the 2006 American
Community Survey conducted by the U.S. Census
Bureau, of the more than 8 million individuals living
in New York City, 44.6% were white.
Are the data presented above evidence of racial profiling
in police officers’ decisions to confront particular
individuals?
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Example: Checking for Racial Profiling
Assume:
 500,000 confrontations as n = 500,000 trials
 P(driver is non-white) is p = 0.554
Calculate the mean and standard deviation of this binomial
distribution:
  500,000(0.554)  277,000
  500,000(0.554)(0.446)  351
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Example: Checking for 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:
u - 3  277,000 - 3(351)  275,947
  3  277,000  3(351)  278,053
If no racial profiling is taking place, we would not be surprised if between
about 275,947 and 278,053 of the 500,000 people stopped were nonwhite. However, 88.9% of all stops, or 500,000(0.889) = 444,500
involved non-whites. This suggests that the number of non-whites
stopped is much higher than we would expect if the probability of
confronting a pedestrian were the same for each resident, regardless of
their race.
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