Download Chapter 8 - Dana E. Madison, Ph.D.

Chapter 8
Random
Variables
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8.1 What is a Random Variable?
Random Variable: assigns a number to
each outcome of a random circumstance, or,
equivalently, to each unit in a population.
Two different broad classes of random variables:
1. A discrete random variable can take one of a
countable list of distinct values.
2. A continuous random variable can take any
value in an interval or collection of intervals.
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Example 8.1 Random Variables at an
Outdoor Graduation or Wedding
Random factors that will determine how
enjoyable the event is:
Temperature: continuous random variable
Number of airplanes that fly overhead:
discrete random variable
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Example 8.2 Probability an Event Occurs
Three Times in Three Tries
• What is the probability that three tosses of a fair coin
will result in three heads?
• Assuming boys and girls are equally likely, what is the
probability that 3 births will result in 3 girls?
• Assuming probability is 1/2 that a randomly selected
individual will be taller than median height of a
population, what is the probability that 3 randomly
selected individuals will all be taller than the median?
Answer to all three questions = 1/8.
Discrete Random Variable X = number of times the
“outcome of interest” occurs in three independent tries.
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8.2 Discrete Random Variables
X the 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) X is a table or
rule that assigns probabilities to possible values of X.
Example 8.5 Number of Courses
35% of students taking four courses, 45% taking five,
and remaining 20% are taking six courses.
X = number of courses a randomly selected student is taking
The probability distribution
function of X can be given by:
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Conditions for Probabilities
for Discrete Random Variables
Condition 1
The sum of the probabilities over all
possible values of a discrete random
variable must equal 1.
Condition 2
The probability of any specific outcome
for a discrete random variable must be
between 0 and 1.
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Example 8.6 PDF for 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 8.6 & 8.7
Number of Girls
Value of X for each simple event:
Probability Distribution
Function for
Number of Girls X:
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Graph of the pdf of X:
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Cumulative Distribution Function
of 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.
Example 8.8 Cumulative Distribution Function
for the Number of Girls
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Finding Probabilities for Complex Events
Example 8.9 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
pdf for Number of Girls X:
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Example 8.12 California Decco Lottery
Player chooses one card from each of four suits. Winning
card drawn from each suit. If one or more matches the
winning cards  prize. It costs $1.00 for each play.
How much would you win/lose per ticket over long run?
Lose an average of 35 cents per play. This is called
the expected value of playing the lottery.

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Standard Deviation for 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) = µ, then
Variance of X = V ( X ) = σ 2 = ∑ ( xi − µ ) pi
2
Standard Deviation of X = σ =
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∑ (x − µ )
2
i
pi
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Example 8.13 Stability or Excitement
Two plans for investing $100
– which would you choose?
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 8.13 Stability or Excitement
Variability for each plan:
Plan 1: V(X ) = $29,900.00 and
σ = $172.92
Plan 2: V(X ) = $48.00
σ = $6.93
and
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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8.4 Binomial Random Variables
Class of discrete random variables =
Binomial -- results from a binomial experiment.
Conditions for a binomial experiment:
1. There are n “trials” where n is determined in advance
and is not a random value.
2. Two possible outcomes on each trial, called “success”
and “failure” and denoted S and F.
3. Outcomes are independent from one trial to the next.
4. Probability of a “success”, denoted by p, remains same
from one trial to the next. Probability of “failure” is 1 – p.
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Examples of Binomial Random Variables
A binomial random variable is defined as X=number
of successes in the n trials of a binomial experiment.
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n
Ck - The number of ways
of choosing k items from a group of n.
543 60
C
=
=
=
10
5 3 321
6
876543 56
28
C
=
=
=
8 6 654321
2
87 56
28
C
=
=
=
8 2 21
2
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Finding Binomial Probabilities
P ( X= k=
)
Ck p (1 − p )
k
n
n−k
for k = 0, 1, 2, …, n
Example 8.15 Probability of Two Wins in Three Plays
p = probability win = 0.2; plays of game are independent.
X = number of wins in three plays.
What is P(X = 2)?
32
3− 2
2
P ( X= 2=
(.2) (1 − .2 )
)
21
2
(.8)1 0.096
= 3(.2)
=
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Expected Value and Standard Deviation
for a Binomial Random Variable
For a binomial random variable X based
on n trials and success probability p,
Mean
µ = E ( X ) = np
Standard deviation σ = np(1 − p )
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Example 8.18 Extraterrestrial Life?
50% of large population would say “yes” if asked,
“Do you believe there is extraterrestrial life?”
Sample of n = 100 is taken.
X = number in the sample who say “yes” is
approximately a binomial random variable.
Mean
µ = E ( X ) = 100(.5) = 50
Standard deviation σ = 100(.5)(.5) = 5
In repeated samples of n=100, on average 50 people
would say “yes”. The amount by which that number
would differ from sample to sample is about 5.
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8.5 Continuous
Random Variables
Continuous random variable: the outcome can
be any value in an interval or collection of intervals.
Probability density function for a continuous random
variable X is a curve such that the area under the curve over
an interval equals the probability that X is in that interval.
P(a ≤ X ≤ b) = area under density curve over the
interval between the values a and b.
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Example 8.19 Time Spent Waiting for Bus
Bus arrives at stop every 10 minutes. Person arrives at
stop at a random time, how long will s/he have to wait?
X = waiting time until next bus arrives.
X is a continuous random variable over 0 to 10 minutes.
Note: Height is 0.10
so total area under the
curve is (0.10)(10) = 1
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Example 8.20 Probability about Wait Time
What is the probability the waiting time X
is in the interval from 5 to 7 minutes?
Probability = area under curve between 5 and 7
= (base)(height) = (2)(.1) = .2
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8.6 Normal Random Variables
Most commonly encountered type of continuous
random variable is the normal random variable,
which has a specific form of a bell-shaped
probability density curve called a normal curve.
A normal random variable is also said to have a
normal distribution
Any normal random variable can be completely
characterized by its mean, µ, and standard deviation, σ.
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Example 8.21 College Women’s Heights
Data suggest the distribution of heights of college
women described well by a normal curve with mean
µ = 65 inches and standard deviation σ = 2.7 inches.
Note: Tick marks given
at the mean and at 1, 2, 3
standard deviations above
and below the mean.
Empirical Rule are exact
characteristics of a normal
curve model.
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Useful Probability Relationships
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Useful Probability Relationships
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Example 8.22 Prob for Math SAT Scores
Math SAT scores have a normal distribution with mean
µ = 515 and standard deviation σ = 100.
Q: Probability a score is
less than or equal to 600?
A: (using Minitab) .8023.
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Example 8.22 Prob for Math SAT Scores
Math SAT scores have a normal distribution with mean
µ = 515 and standard deviation σ = 100.
Q: Probability a score is
greater than 600?
A: By complement,
1 – .8023 = .1977
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Example 8.22 Prob for Math SAT Scores
Math SAT scores have a normal distribution with mean
µ = 515 and standard deviation σ = 100.
Q: Probability a score is
between 515 and 600?
A: (since 50% of scores are
larger than the mean of 515),
.8023 – .5000 = .3023
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Example 8.22 Prob for Math SAT Scores
Math SAT scores have a normal distribution with mean
µ = 515 and standard deviation σ = 100.
Q: Probability a score is
more than 85 points away
from the mean in either
direction?
A: (since 600 is 85 points
above the mean)
.1977 + .1977 = .3954.
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Using a Table to Find Probabilities
Table A.1 = Standard Normal (z) Probabilities
• Body of table gives cumulative probabilities P(Z ≤ z).
• First column gives digit before the decimal place, the
first decimal place for z.
• Second decimal place of z is in column heading.
P(Z ≤ 1.82) = .9656
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More Finding Probabilities for z-scores
Table A.1 = Standard Normal (z) Probabilities
P(Z ≤ -2.5) =.0048 (see in portion above)
Verify on your own:
P(Z ≤ 1.31) = .9049
P(Z ≤ -2.00) = .0228
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Calculating a Standard Score
The formula for converting any value x to
a z-score is
Value − Mean
x−µ
z=
=
Standard deviation
σ
A z-score measures the number of standard
deviations that a value falls from the mean.
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Example 8.23 z-Score for Height
For a population of college women, the z-score
corresponding to a height of 62 inches is
62 − 65
Value − Mean
= −1.11
=
z=
2.7
Standard deviation
This z-score tells us that 62 inches is 1.11 standard
deviations below the mean height for this population.
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Use z-scores to Solve General Problems
Example 8.24 Probability of Smaller Height
What is the probability that a randomly selected
college woman is 62 inches or shorter?
62 − 65
Z≤
= −1.11
2.7
P( X ≤ 62 ) = P(Z ≤ −1.11) = .1335
About 13% of college women
are 62 inches or shorter.
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Finding Percentiles
If 25th percentile of pulse rates is
64 bpm, then 25% of pulse rates
are below 64 and 75% are above 64.
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Example 8.26 75th Percentile of
Systolic Blood Pressure
Blood Pressures are normal with mean 120 and
standard deviation 10. What is the 75th percentile?
Step 1: Find closest z* with area of 0.7500 in body
of Table A.1.
z* = 0.67
Step 2: Compute x = z*σ + µ
x = (0.67)(10) + 120 = 126.7 or about 127.
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