Transcript Lecture Notes Chapter 8 - Department of Statistics and Probability

Chapter 8
Random Variables
•
Probability Distributions
• Discrete Random Variables
• Continuous Random Variables
• Binomial Probability Model
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Randomness
A random variable is a numerical measurement of the
outcome of a random phenomenon. It is a quantity that
can take on different values.
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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Probability Distribution of a Discrete
Random Variable
Let
X = the discrete random variable.
k = a number the discrete r.v. could assume.
P(X = k) is the probability that X equals k.
Probability distribution function (pdf) of X is a table or
rule that assigns probabilities to possible values of X.
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Example: Number of Courses
• 35% of students taking four courses,
• 45% taking five,
• and remaining 20% are taking six courses.
• Let X = number of courses a randomly selected
student is taking
• The probability distribution function of X can be given
by:
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Example: Number of Courses
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Example: Number of Girls
Family has 3 children. Probability of a girl is ½.
What are the probabilities of having 0, 1, 2, or 3 girls?
Sample Space: For each birth, write either B or G.
There are eight possible arrangements of B and G for
three births. These are the simple events.
Sample Space and Probabilities: The eight simple
events are equally likely.
Random Variable X: number of girls in three births.
For each simple event, the value of X is the number of
G’s listed.
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Example: Number of Girls
Probability Distribution Table
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Example: Number of Girls
Probability Distribution
Function for
Number of Girls X:
Graph of the pdf of X:
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Example: Number of Home Runs in a
Game
What is the estimated probability of at least three home
runs?
Table 8.1 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 8.1 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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Cumulative Distribution Function for a
Discrete Random Variable
Cumulative distribution function (cdf) for a random
variable X is a rule or table that provides the probabilities
P(X ≤ k) for any real number k.
Cumulative probability = probability that X is less than
or equal to a particular value.
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Cumulative Distribution Function for Number of
Girls
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Example: A Mixture of Children
What is the probability that a family with 3 children
will have at least one child of each sex?
If X = Number of Girls then either family has one girl and two
boys (X = 1) or two girls and one boy (X = 2).
P(X = 1 or X = 2) = P(X = 1) + P(X = 2)
= 3/8 + 3/8 = 6/8 = 3/4
Probability distribution for number of girls:
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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 8.1 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 Discrete
Random Variable
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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Standard Deviation of a Discrete Random
Variable
The standard deviation of a random variable is roughly
the average distance the random variable falls from its
mean, or expected value, over the long run.
If X is a random variable with possible values x1, x2, x3, . . .
, occurring with probabilities p1, p2, p3, . . . , and expected
value E(X) = , then
Varianceof X  V  X    2   xi    pi
2
Standard Deviationof X   
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 x   
2
i
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pi
Example: Stability or Excitement?
Two plans for investing $100
– which would you choose?
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Example: Stability or Excitement?
Expected Value For Each Plan:
Plan 1:
E(X ) = $5,000(.001) + $1,000(.005) + $0(.994) =
$10.00
Plan 2:
E(Y ) = $20(.3) + $10(.2) + $4(.5) = $10.00
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Example: Stability or Excitement?
Variability for Each Plan:
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Example: Stability or Excitement?
Plan 1:
V(X ) = $29,900.00 and
Plan 2:
V(X ) = $48.00 and
 = $172.92
 = $6.93
The possible outcomes for Plan 1 are much more variable.
If you wanted to invest cautiously, you would choose
Plan 2, but if you wanted to have the chance to gain a
large amount of money, you would choose Plan 1.
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More Examples From Midterm II Practice
Sheet
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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 8.2 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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Useful Probability Relationships
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Useful Probability Relationships
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Useful Probability Relationships
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Useful Probability Relationships
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Bell-Shaped Distributions
Probabilities for Bell – Shaped
Distributions or Continuous
Random Variables
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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 8.4 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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Empirical Rule or 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 8.5 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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Empirical Rule or 68 – 95 – 99.7% 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 the
back of the book.
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Standard Score (z – score)
The formula for converting any value x to a z-score is
.
Value  Mean
x
z

Standard deviation

A z-score measures the number of standard
deviations that a value falls from the mean
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Shifting and Rescaling Data
Shifting data:
 Adding (or subtracting) a constant to every data
value adds (or subtracts) the same constant to
measures of position.
 Adding (or subtracting) a constant to each value will
increase (or decrease) measures of position:
center, percentiles, max or min by the same
constant.
 Its shape and spread - range, IQR, standard
deviation - remain unchanged.
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Shifting and Rescaling Data
The following histograms show a shift from men’s
actual weights to kilograms above recommended
weight:
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Shifting and Rescaling Data
Rescaling data:
 When we multiply (or divide) all the data values
by any constant, all measures of position (such
as the mean, median, and percentiles) and
measures of spread (such as the range, the IQR,
and the standard deviation) are multiplied (or
divided) by that same constant.
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Shifting and Rescaling Data
The men’s weight data set measured weights in
kilograms. If we want to think about these weights in
pounds, we would rescale the data:
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Examples: Midterm Exam II Practice Sheet
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Z-Scores and the Standard Normal
Distribution
A standard normal distribution has mean   0 and
standard deviation   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 follow the standard normal
distribution.
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Table Z: Standard Normal Probabilities
Table Z 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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Finding Probabilities Using The Standard
Normal Table (Table Z)
The figure shows us how to find the area to the left when
we have a z-score of 1.80:
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Example: Using Table Z
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 Z
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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Example: Using Table Z
Figure 8.7 The Normal Cumulative Probability, Less than z Standard Deviations
above the Mean. Table Z 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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From Percentiles to z - Scores
Sometimes we start with areas and need to find the
corresponding z-score or even the original data value.
Example: What z-score represents the first quartile in
a Normal model?
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From Percentiles to z - Scores
Look in Table Z for an area of 0.2500.
The exact area is not there, but 0.2514 is pretty close.
This figure is associated with z = -0.67, so the first
quartile is 0.67 standard deviations below the mean.
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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 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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Are You Normal? Normal Probability Plots
When you actually have your own data, you must
check to see whether a Normal model is reasonable.
Looking at a histogram of the data is a good way to
check that the underlying distribution is roughly
unimodal and symmetric.
A more specialized graphical display that can help you
decide whether a Normal model is appropriate is the
Normal probability plot.
If the distribution of the data is roughly Normal, the
Normal probability plot approximates a diagonal
straight line. Deviations from a straight line indicate
that the distribution is not Normal.
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Are You Normal? Normal Probability Plots
Nearly Normal data have a histogram and a Normal
probability plot that look somewhat like this example:
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Are You Normal? Normal Probability Plots
A skewed distribution might have a histogram and
Normal probability plot like below. In such cases it is
unwise to use the Normal Model.
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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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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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Examples: Midterm Exam II Practice Sheet
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Binomial Probability Model
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: Midterm II Exam Practice Sheet
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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.
2
 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!
P(2) 
(0.2)2 (0.8)1  3(0.04)(0.8) 0.096
2!1!
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Warning!!!!!!!!!!!!!!!!!!
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: Midterm Exam II Practice Sheet
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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: Midterm II Exam Practice Sheet
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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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Extra – Credit Work Follows
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Section 8.8 Sums, Differences of Discrete RVs
A linear combination of random variables, X, Y, . . . is a
combination of the form:
L = aX + bY + …
where a, b, etc. are numbers – positive or negative.
Most common:
Sum = X + Y Difference = X – Y
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Section 8.8 Means of Linear Combination
L = aX + bY + …
The mean of L is:
Mean(L) = a Mean(X) + b Mean(Y) + …
Most common:
Mean( X + Y) = Mean(X) + Mean(Y)
Mean(X – Y) = Mean(X) – Mean(Y)
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Section 8.8 Variances of Linear Combination
If X, Y, . . . are independent random variables, then
Variance(L) = a2 Variance(X) + b2 Variance(Y) + …
Most common:
Variance( X + Y) = Variance(X) + Variance(Y)
Variance(X – Y) = Variance(X) + Variance(Y)
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Extra – Credit Exercises
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