# Chapter 16 Random Variables by niusheng11

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```									  Chapter 16 Random Variables
Random Variable is a variable whose
outcome depends on a random event

Notation: Capital letters such as X, Y, etc.

The values X assumes are denoted by little
x
Examples: Random Variables

• The amount of time you wait at a light

• The lifespan of a light bulb

• The amount of money an insurance
company will pay on a policy
Random Variables

• Like the distribution of any quantitative
variable, the distribution has shape, center

• In fact, we can calculate the mean, or
expected value of a random variable and its
standard deviation.
Chapter 16 Random Variables
• Example: An insurance company offers a “death and
disability” policy that pays \$10000 when you die or
\$5000 if you are permanently disabled, within five
years. It charges its customers \$50 to enroll in this
plan.

• Here: X is a random variable that represents how
much the insurance company pays.
• Note: X = \$10000 or X = \$5000 or X = \$0
Chapter 16 Random Variables
• Problem: Suppose the             Policyholder payout P(X=x)
probability of death rate in 5   outcome
years is 1 out of 1000
people, and that 2 out of
1000 suffer some kind of
Death        10000   0.001
disability. Is the company
making profit selling such
plans?                           Disability   5000    0.002

• The answer is obtained by
forming a probability            Neither      0       0.997
model.
Probability Distribution Table for
Insurance Example
Insurance Payment   X= \$10000   X= \$5000   X = \$0

Probability        0.001      0.002     0.997
Chapter 16 Random Variables
• Suppose the insurance company sells the policy for
\$50 each to 1000 customers. According to the model
we expect 1 to die, 2 to be disabled, and the rest 997
will be ok.
• Hence the company
–   Gets \$50000
–   pays \$10000 to the family of the deceased
–   pays \$ 10000 to the two disabled persons
–   and profits \$30000 (\$30 per person profit)
Chapter 16 Random Variables
Chapter 16 Random Variables
Chapter 16 Random Variables
Chapter 16 Random Variables
• Example: Suppose you have a computer sales
things are fine. If the client gets one refurbished
computer, it will be sent back at your expense of \$100
and you can replace it. If both computers are
refurbished, the client will cancel the order and you
lose \$1000. What is the expected value (mean) and
Chapter 16 Random Variables
• Probability Model:        Outcome      x      P(X=x)
• Assumptions: In your
inventory you found out
that someone has          2 new        0      0.524
stocked 4 refurbished
computers with 11 new
computers. The shipped    One Refurb   100    0.419
computers were
randomly selected from
these group               Two Refurb   1000   0.057
Chapter 16 Random Variables
• Expected value of loss

– E(X) = 0 (0.524)+ 100(0.419)+ 1000(0.057) = E(X)=
\$98.90
– Var(X) = (0-98.90)^2*(0.524)+ (100-
98.90)^2(0.419) + (1000-98.90)^2*(0.057 = \$51,
408.79
– Std Dev = Sqrt(51,408.79) = \$226.735
Chapter 16 Random Variables
• Interpretation:
– In the long run it will cost the company an average
of \$98.90

– The large standard deviation reflects the fact that
there is a pretty large range of possible losses.
Chapter 16 Random Variables
• Some properties:

–   E(X+c) = E(X) +c
–   E(X-c) = E(X) – c
–   E(X+Y)=E(X)+E(Y)
–   Var(X+c) = Var(X)
–   Var(X-c) = Var(X)
–   E(aX) = a E(X)
–   Var(aX) = a^2Var(X)
–   Var(X+Y) = Var(X) + Var(Y), provided X, Y are independent
Chapter 16 Random Variables
• Example: You plan to sell your used Isuzu
Trooper and spend your vacation in Kyrgyzstan
where you plan to buy a Honda motor scooter.
Used Isuzu’s sell for a mean price of \$6940
with a standard deviation of \$250. In
Kyrgyzstan, the scooter sells for a mean price
of 65,000 Kyrgyzstan soms with a standard
deviation of 500 soms. 1 USD = 43 Kyrgyzstan
soms. What is your expected profit, if any?
Chapter 16 Random Variables
• A = sale price of Isuzu (in dollars)
• B = price of scooter (in soms)
• D = profit (in soms) = 43A-B
• Prices are assumed to be independent
• E(D) = E(43A-B) = 43E(A)-E(B) = 43(6940)-(65000) = 233,420
soms (expected profit)
• Var(D)= Var(43A-B)= 43^2var(A)+Var(B) = 115, 812, 500
• Std Dev(D) = sqrt(115, 812, 500) = 10,762 soms.
• In dollars, the expected profit is \$5428 with a stad dev of 250.
Chapter 17 Probability Models
• 1. Bernoulli Trials (properties)
– Two possible outcomes (success and failure)
– The probability of success, denoted by p, is the
same on every trial
– The trials are independent
• Example: Suppose 75% of all drivers wear
their seatbelts. Find the probability that four
drivers might be belted among five cars
waiting traffic light?
Chapter 17 Probability Models
• 2. The Geometric Model: How many trials are
needed to observe the first success?

• Let p = probability of success, q = 1-p be probability
of failure, X= be the number of trials until the first
success occurs.

• Then P(X=x) q^{x-1}p
• Expected value E(X) = 1/p
• Std deviation of X = Sqrt(q/p^2)
Chapter 17 Probability Models
• Example: People with O-negative blood type are
called universal donors. Only about 6% of people
have O-negative blood. If donors line up at random
for blood drive,
– a) how many do you expect to examine before you find
someone who has O-negative blood?
– b) What’s the probability that the first O-negative donor
found is one of the four people in line?
Chapter 17 Probability Models
• a) E(X) = 1/0.06 = 16.6 On average a universal
donor is found by examining 16.7 people.

• b) P(X <= 4) = P(X=1)+ P(X=2)+ P(X=3)+ P(X=4)
0.2193
Chapter 17 Probability Models
• 3. The Binomial Model: The binomial model is
like a Bernoulli trial with a number of trials
being n, n is a number greater than or equal
to 2.

• We look for the number of k successes in n
trial. Here k is less than or equal to n.
• Let X = number of successes in n trials.
Chapter 17 Probability Models
• P(X=x) = C(n,x)*p^x*q^{n-x}
• Here p = probability of success, q = 1-p =
probability of failure.

• E(X) = np

• Std dev of X = Sqrt{npq}
Chapter 17 Probability Models
• Example: Suppose 20 people come to the
blood drive. What is the probability that there
are 2 or 3 universal donors?
Solution:
• P(X=2) = C(20,2)(.06)^{2}(.94)^{18}= 0.2246
• P(X=3) = C(20,3)(.06)^{3}(.94)^{17}= 0.0860
Chapter 17 Probability Models
• To compute the C(n,x) number from the TI-83
do the following

•   2. Press MATH
•   3. Select Prob
•   4. Press nCr
•   5. Type your second number
Examples
shots during the season. Assuming the shots
are independent, find the probability that in
tonight’s game he

a) Misses for the first time on his fifth attempt
b) Makes his first basket on his fourth shot
Examples
• a) success probability = p = 0.8, failure
probability = p = 0.2

• b) Answer: 0.2^3*0.8 = 0.0064
Examples
2. A certain tennis player makes a successful first
serve 70% of the time. Assume that each
serve is independent of the others. If she
serves 6 times, what’s the probability she
gets
a) all 6 serves in
b) exactly 4 serves in?
c) at least four serves in?
d) no more than 4 serves in?
Examples
2 binomial model: p = 0.7, q = .3, n = 6

a) C(6,6)*.7^6*.3^0 = 0.118
b) C(6,4)*.7^4*.3^2 = 15*.7^4*.3^2 = .324
c) C(6,4)*.7^4*.3^2 = 0.324
C(6,5)*.7^5*.3^1 = 0.303
C(6,6)*.7^6*.3^0 = 0.118

d) 1 – (.303+.118) = .575
Probability Models
•   Discrete Models
– Bernoulli
– Geometric
– Binomial

•   Continuous
– Normal Model
– Poisson Model
Why Continuous Models?
• Example: About 6% of people have O-negative
blood. Suppose the Red Cross anticipates the
need for at least 1850 units of O-negative
blood for this year. It is estimated that 32000
donors will give blood this year. How great is
the risk that the Red Cross will fall short of
meeting its need?
Why Continuous Models?
• C(32000,1850)*.06^(1850)*.94^(30150)??

• But the problem asks for at least 1850.

• More numbers need to be computed!!!
• We use the Normal Model for approximation.
CHAPTER 6: STANDARD DEVIATION & THE NORMAL
MODEL
Chapter 6. What is a normal distribution?
• The normal distribution is pattern for the distribution of a set
of data which follows a bell shaped curve.

• This distribution is sometimes called the Gaussian distribution
in honor of Carl Friedrich Gauss, a famous mathematician.

• The bell shaped curve has several properties:

• The curve concentrated in the center and decreases on either
side. This means that the data has less of a tendency to
produce unusually extreme values, compared to some other
distributions.

• The bell shaped curve is symmetric. This tells you that he
probability of deviations from the mean are comparable in
either direction.                                                  33
CHAPTER 6: STANDARD DEVIATION & THE
NORMAL MODEL
• When you want to describe probability for a
continuous variable, you do so by describing a
certain area.

• A large area implies a large probability and a small
area implies a small probability. Some people don't
like this, because it forces them to remember a bit of
geometry (or in more complex situations, calculus).
But the relationship between probability and area is
also useful, because it provides a visual
interpretation for probability.

• Here's an example of a bell shaped curve. This
represents a normal distribution with a mean of 50
and a standard deviation of 10.                      34
DESCRIBING DISTRIBUTION NUMERICALLY

35
CHAPTER 6: STANDARD DEVIATION & THE
NORMAL MODEL

36
Formula
• Standardizing normal
variables:

Y   
• Formula:

Z       
  
37
68-95-99.7 Rule
• 68% of the observations are within 1 standard
deviation unit

• 95% of the observations are within 2 standard
deviation unit

• 99.7% of the observations are within 3 standard
deviation unit

• http://davidmlane.com/hyperstat/normal_distribut
ion.html
38
Using a normal model to solve
problems
• An example using height data and U.S. Marine
Corps and Army height requirements

• Global Question: Are the height restrictions
set up by the U.S. Army and U.S. Marine Corps
more restrictive for men or women or are they
roughly the same?

39
Data from a National Health Survey
• Heights of adult women are normally
distributed with a
– mean of 63.6 in.
– standard deviation of 2.5 in.
• Heights of adult men are normally distributed
with a
– mean of 69.0 in.
– standard deviation of 2.8 in.

40
Height Restrictions
Men Minimum   Men Maximum   Women     Women Maximum
Minimum

U.S. Army
60 in         80 in         58 in     80 in
U.S. Marine
Corps
64 in         78 in         58 in     73 in

41
Which is more unusual?

• A 58in tall woman?

• A 60in tall man?

42
Which is more unusual?

A. A 58in tall woman?

B. A 64in tall man?

43
Which is more unusual?

A. A 73in tall woman?

B. A 78in tall man?

44
Which is more unusual?

A. An 80 in tall woman?

B. An 80 in tall man?

45
Approximately what percent of U.S. women do
you expect to be under 66.1 in tall?

A.68%
B.95%
C.16%
D.84%
E. I have no idea how to do this

46
Approximately what percent of U.S. women
Do you expect to be over 68.6in tall?

A.5%
B.2.5%
C.1%
D.0.3%
E. I have no idea how to do this

47
Approximately what percent of U.S. women
Do you expect to be between 66.1in and
68.6in tall?

A.5%
B.10%
C.13.5%
D.16%
E. I have no idea how to do this

48
Approximately what percent of U.S. women
Do you expect to be under 5 feet tall (60in)?

A.3%
B.5%
C.7.5%
D.10%
E. I have no idea how to do this

49
Example
Some IQ tests are standardized to a Normal model with a mean of 100 and a
standard deviation of 16.

a) Describe the 68-95-99.7 rule for this problem

b) About what percent of people should have IQ scores above 116?

c) About what percent of people should have IQ scores between 68 and 84?

d) About what percent of people should have IQ scores above 132?

e) About what percent of people should have IQ scores above 120?

f) About what percent of people should have IQ scores below 90?

g) About what percent of people should have IQ scores between 95 and 130?

h) A person is a genius if his/her IQ belong to the top 10% of the all IQ scores.
What minimum IQ score qualifies you to be a genius?
50
Some IQ tests are standardized to a Normal model with a mean of 100 and a standard deviation of 16.

b) 16%

c) 13.5%

d) 2.5%

e) About what percent of people should have IQ scores above 120?
Z = (120 -100)/16 = 1.25
Find P(Z > 1.25) from standard normal chart or your TI calculator.
f) About what percent of people should have IQ scores below 90?
Z = (90 -100)/16 = -0.625
Find P(Z < -0.625) from standard normal chart or your TI calculator.
g) About what percent of people should have IQ scores between 95 and 130?
Z = (95 -100)/16 = -0.3125
Z = (130 -100)/16 = 1.875

Find P(-0.3125< Z < 1.875) = .9699-.3783 = .5916

h) A person is a genius if his/her IQ belong to the top 10% of the all IQ scores. What
minimum IQ score qualifies you to be a genius?

The top 10% corresponds to the 90th percentile. For the standard normal the 90th
percentile is 1.28. Hnece solve 1.28 = (Y-100)/16. The value of Y is 120.48.
51
Example
• In 2006 combined verbal and math SAT scores
followed a normal distribution with mean 1020 and
standard deviation 240.

• Suppose you know that Peter scored in the top 3% of
SAT scores. What was Peter’s approximate SAT score?

52
Some TI-83/84 Commands

•       Press STAT, Enter (for EDIT)
–     If there are old data under L1:

–     Press the up arrow, then CLEAR, ENTER

•       Enter data values in L1 one at a time, pressing
ENTER after each
–     If you make an error, use the up or down arrows to highlight the
error, then enter the correct value. Use the arrows to get to the
bottom of the list for the next value, if necessary.

–     Be sure to press ENTER after the last data value.                   53
Some TI-83/84 Commands
To Find One Variable Statistics (mean, median, standard
deviation, etc)

• Press STAT, Right Arrow (for CALC), ENTER

• Press ENTER (for 1-Var Stats)

• Press ENTER again

– The Standard Deviation is labeled Sx

54
Using the TI-83 to Find a Normal Percentage
Always draw a
• The TI-83 provides a function named normalcdf
picture!
– Press 2nd, DISTR (found above VARS)
– Scroll to normalcdf ( and press ENTER, or press 2.
• If z has a standard normal distribution:
– Percent(a < z < b) = normalcdf ( a , b )                           ?
– Example: to find P( -1.2 < z < .8 ),                    -1.2           .8
press 2nd, DISTR, 2, then -1.2 , .8 )
– Note that the comma between -1.2 and .8 must be entered
?
• To find Percent( z < a ), enter normalcdf ( -5 , a )
– Example: normalcdf( -5 , 1.96 ) gives .9750                                 1.96

?
• To find Percent( z > a ), enter normalcdf ( a , 5 )
– Example: normalcdf( -1.645 , 5 ) gives .9500         -1.645
55
Using the TI-83/84 for Normal Percentages Without
Computing z-Scores
We can let the TI find its own z-scores:
– Find Percent(90 < x < 105) if x follows the normal model with mean 100 and
standard deviation 15:
• Percent(90 < x < 105) = normalcdf( 90 , 105 , 100 , 15)
= .378

Notice that this is a time-saver for this type of problem, but that you may still need
to be able to compute z-scores for other types of problems!

x1            x2

56
Suppose We’re Given a normal
Percentage and Need A z-score?
• IQ scores are distributed normally with a
mean of 100 and a standard deviation of 15.
What score do you need to capture the
bottom 2%?
– That is, we must find a so that Percent(x < a) = 2% when x
has a normal distribution with a mean of 100 and a
standard deviation of 15.
– With the TI 83/84:
a = invNorm( .02, 100 , 15) = 69.2

x

57
Why Continuous Models?
• Example: About 6% of people have O-negative
blood. Suppose the Red Cross anticipates the
need for at least 1850 units of O-negative
blood for this year. It is estimated that 32000
donors will give blood this year. How great is
the risk that the Red Cross will fall short of
meeting its need?
Why Continuous Models?
• The Normal Model comes to the rescue!
• Idea: Approximate the binomial model with a
normal model!

• The Binomial Model has
– mean = np = 1920
– std dev. = sqrt(np(1-p))= 42.48
• Let X = # of O-blood donors
Why Continuous Models?
• The problem now is to find

P(X < 1850) = P (Z < (1850-1920)/42.28) ~

P(Z < 1.65) ~ 0.05
• Hence there is a 5% chance that the Red Cross
will not meet its goal.
Why Continuous Models?
• Question: Can we always approximate the
binomial model by a normal model?

Why Continuous Models?
• Example: Suppose there is a 20% chance that cereal
boxes contain pictures of Tiger woods. You buy 5
boxes. Let X = # of Tiger Woods Pictures you get.
What is the distribution of X?

X           0     1     2     3     4     5

P(X=x)        .33   .41   .20   .05   .01   .03
Why Continuous Models?
• Simulation at

– http://www.stat.wvu.edu/SRS/Modules/NormalApprox/no
rmalapprox.html

• A Normal Model is a close approximation of binomial
for a large number of trials
– “Large” is explained by the
• Success/failure Condition: A Binomial Model is approximately
normal if we expect at least 10 successes and 10 failures,
i. e. np is at least 10 and nq = n(1-p) is at least 10
Why Continuous Models?
• Example: A communication monitoring
company reports that 91% of e-mail messages
are spam. Recently you installed a spam filter.
You observe that over the past week it okayed
only 151 of 1422 e-mails you received,
classifying the rest as junk. What is the
probability that no more than 151 of 1422 e-
mails is a real message?
Why Continuous Models?
Solution:

Let X = # of real messages
p = .09, q = 1-p = .91, n = 1422, x = 151
np = 127.98, sqrt(npq) = 10.79

P(X < 151) = P( Z < (151-127.98)/10.79) =
= P( Z < 2.13) = .98

There is a 98% chance that no more than 151 messages
among the 1422 receives are real messages. Filter is
working properly

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