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Chapter 8
Random
Variables
Copyright ©2004 Brooks/Cole, a division of Thomson Learning, Inc.
8.1 What is a Random Variable?
Random Variable: assigns
g 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 continuous random variable can take any
1.
value in an interval or collection of intervals.
2 A discrete random variable can take one of a
2.
countable list of distinct values.
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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
ill result
l in
i three
h heads?
h d?
• 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
g 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
“ t
“outcome
off interest”
i t
t” occurs in
i three
th
independent
i d
d t tries.
t i
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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.
Discrete random variable: can only result in a
countable set of possibilities – often a finite number
of outcomes, but can be infinite.
Example 8.3 It’s Possible to Toss Forever
Repeatedly
p
y tossingg a fair coin,, and define:
X = number of tosses until the first head occurs
Any number of flips is a possible outcome.
P(X = k) = (1/2)k
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Probability Distribution of a Discrete R.V.
Using the sample space to find probabilities:
Step
p 1: List all simple
p events in sample
p space.
p
Step 2: Find probability for each simple event.
Step 3: List possible values for random variable X
and identify the value for each simple event.
Step 4: Find all simple events for which X = k, for
each
h possible
ibl value
l k.
k
Step 5: P(X = k) is the sum of the probabilities for
all simple events for which X = k.
k
Probability distribution function (pdf) X is a table or
rule
l that
th t assigns
i
probabilities
b biliti to
t possible
ibl values
l
off X.
X
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Example 8.4 How Many Girls are Likely?
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
g possible
p
arrangements
g
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.
births
For each simple event, the value of X is the number
of G’s listed.
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Example 8.4 How Many Girls? (cont)
Value of X for each simple event:
Probability
y distribution function for Number of Girls X:
Graph of the pdf of X:
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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 pprobabilityy of anyy specific
p
outcome
for a discrete random variable must be
between 0 and 1.
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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
p
( ≤ k)) for anyy real number k.
Cumulative probability = probability that X is less
than or equal to a particular value.
Example 8.4 Cumulative Distribution Function
for the Number of Girls (cont)
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Finding Probabilities for Complex Events
Example 8.4 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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8.3 Expectations for
Random Variables
The expected value of a random variable is the
mean value of the variable X in the sample space,
or population, of possible outcomes.
If X is a random variable with possible values x1, x2, x3, . . . ,
occurring
i with
ith probabilities
b biliti p1, p2, p3, . . . ,
then the expected value of X is calculated as
μ = E ( X ) = ∑ xi pi
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Example 8.6 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
prize. It costs $1.00
$1 00 for each play.
play
How much would you win/lose per ticket over long run?
=> Lose an average of 35 cents per play.
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Standard Deviation for a
Discrete Random Variable
The standard deviation of a random variable is
essentially the average distance the random
variable falls from its mean 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 8.7 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 8.7 Stability or Excitement (cont)
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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8.3 Binomial Random Variables
Class of discrete random variables =
Binomial -- results from a binomial experiment
experiment.
Conditions for a binomial experiment:
p
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.
3.
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.
experiment
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Finding Binomial Probabilities
n!
n−k
k
P( X = k ) =
p (1 − p ) for k = 0, 1, 2, …, n
k!(n − k )!
Example 8.9 Probability of Two Wins in Three Plays
p = pprobabilityy win = 0.2; plays
p y of game
g
are independent.
p
X = number of wins in three plays.
What is P(X = 2)?
3!
3− 2
P( X = 2) =
.2 2 (1 − .2 )
2!(3 − 2 )!
= 3(.2) 2 (.8)1 = 0.096
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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,
p
Mean
μ = E ( X ) = np
Standard deviation σ = np (1 − p )
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Example 8.12 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
i
l a binomial
bi
i l random
d
variable.
i bl
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
ld differ
diff from
f
sample
l to
t sample
l is
i about
b t 5.
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
d density
d i curve over the
h
interval between the values a and b.
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Example 8.13 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
This is an example of a
Uniform
U
o
random
do vvariable
be
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Example 8.13 Waiting for Bus (cont)
What is the probability the waiting time X was in
the interval from 5 to 7 minutes?
Probability = area under curve between 5 and 7
)( g ) = (2)(.1)
( )( ) = .2
= ((base)(height)
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8.6 Normal Random Variables
If a population
l ti off measurements
t ffollows
ll
a normall
curve, and if X is the measurement for a randomly
selected individual from that population,
population then
X is said to be a normal random variable
X 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 8.14 College Women’s Heights
Data suggest the distribution of heights of college
women modeled 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
andd bbelow
l the
th mean.
Empirical
p
Rule values
are exact characteristics
of a normal curve model
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Standard Scores
The formula for converting any value x to
a z-score is
Value − Mean
x−μ
z=
=
σ
Standard deviation
A z-score measures the number of standard
deviations that a value falls from the mean.
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Example 8.14
Height (cont)
For a population of college women, the z-score
corresponding to a height of 62 inches is
Value − Mean
62 − 65
z=
=
= −1.11
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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Finding Probabilities for z-scores
Table A.1 = Standard Normal (z) Probabilities
• Body of table contains P(Z ≤ z*).
z*)
• Left-most column of table shows algebraic sign,
digit before the decimal place,
place the first decimal
place for z*.
• Second
S
d ddecimal
i l place
l
off z** is
i in
i column
l
heading.
h di
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More Finding Probabilities for z-scores
Table A.1 = Standard Normal (z) Probabilities
P(Z
( ≤ -3.00)) =.0013 ((see in pportion above))
P(Z ≤ −2.59) = .0048
P(Z ≤ 1.31)
1 31) = .9049
9049
P(Z ≤ 2.00) = .9772
P(Z ≤ -4.75)
4 75) = .000001
000001 (from in the extreme)
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Example 8.15
Probability Z > 1.31
P(Z > 1.31) = 1 – P(Z ≤ 1.31)
= 1 – .9049 = .0951
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Example 8.16 Probability Z is
b
between
–2.59
2 59 andd 1.31
1 31
P(-2.59 ≤ Z ≤ 1.31)
= P(Z ≤ 1.31) – P(Z ≤ -2.59)
= .9049
9049 – .0048
0048 = .9001
9001
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Use z-scores to Solve General Problems
Example 8.14 Height (cont)
What is the probability that a randomly selected
college woman is 62 inches or shorter?
62 − 65 ⎞
⎛
P ( X ≤ 62 ) = P⎜ Z ≤
⎟
2 .7 ⎠
⎝
= P (Z ≤ −1.11) = .1335
About 13% of college women
are 62 inches or shorter.
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Use z-scores to Solve General Problems
Example 8.14 Height (cont)
What proportion of college woman are taller than 68 inches?
68 − 65 ⎞
⎛
P ( X > 68) = P⎜ Z >
⎟ = P (Z > 1.11) = 1 − P (Z ≤ 1.11)
2 .7 ⎠
⎝
= 1 − .8665 = .1335
About 13% of
college women
are taller than
68 inches.
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Finding Percentiles
If 25th percentile of pulse rates is
p , then 25% of pulse
p
rates
64 bpm,
are below 64 and 75% are above 64.
The percentile is 64 bpm, and the
percentile ranking is 25%.
Step 1: Find z-score
z score that has specified cumulative
probability. Using Table A.1, find percentile rank
in body of table and read off the z-score.
z-score
Step 2: Calculate the value of variable that has the
z score found in step 1: x = zz*σ + μ.
z-score
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Example 8.17 75th Percentile of
S
Systolic
li Blood
Bl d Pressure
P
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
p 2: Calculate x = z*σ + μ
x = (0.67)(10) + 120 = 126.7 or about 127.
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8.7 Approximating Binomial
Distribution Probabilities
If X is
i a binomial
bi
i l random
d
variable
i bl based
b d on n trials
ti l
with success probability p, and n is large, then the
random variable X is also approximately normal,
with mean and standard deviation given as:
Mean
μ = E ( X ) = np
Standard deviation σ = np (1 − p )
Conditions: Both np and n(1 – p) are at least 10.
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Example 8.18
Number of H in 30 Flips
X = number of heads in n = 30 flips of fair coin
Binomial distribution with n = 30 and p = .5.
Distribution is bell-shaped
p
and can be approximated
by a normal curve.
Mean
μ = E ( X ) = 30(.5) = 15
Standard deviation σ = 30(.
( 5)(.5) = 2.74
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Example 8.19
Political Woes
Poll: n = 500 adults; 240 supported position
pp pposition,, what is the
If 50% of all adults support
probability that a random sample of 500 would find
240 or fewer holding this position? P(X ≤ 240)
X is approximately normal with
Mean
μ = E ( X ) = 500(.5) = 250
Standard deviation σ = 100(.5)(.5) = 11.2
240 − 250 ⎞
⎛
P ( X ≤ 240 ) ≈ P⎜ Z ≤
⎟ = P (Z ≤ −.89 ) = .1867
11.2 ⎠
⎝
Not unlikelyy to see 48% or less,, even if 50% in ppopulation
p
favor.
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