Download notes Ch 16

AP Statistics
Notes Ch 16
Name
Brashear
Date_________________Period______
Random variable—assumes any of several different values as a result of some random event.
Denoted by a capital letter.
Discrete random variable—a random variable that can take one of a finite number of distinct
outcomes
Continuous random variable—a random variable that can take any numeric value within a range of
values. The range may be infinite or bounded at either or both ends.
Probability model—the collection of all the possible values and the probabilities that they occur
Expected value
Formula for expected value:   E (X )   x P (X  x )
(the sum of the product of all of the possible values and their respective probabilities)
Variance of the model
 2  Var (X )    x    P (X  x )
2
Standard deviation of the model
  SD (X )  Var (X )
Notes:
Random variables have a probability model
Actual sample data have a distribution
A probability model’s mean is expected value or  or E(x) but not x .
A model’s variance is Var(x) or 2 but not s2
Expected value is a theoretical amount, not actual data. In fact, it may be impossible to get the
expected value on any one trial.
Brashear AP Statistics Ch 16 notes-teacher (12/2/04)
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THE VARIANCES ADD IDEA
Think of variance as a measure of uncertainty. In any situation, the more variability there is among
the possible outcomes the less certain we can be of the results. Every time we introduce another
random event we add another layer of uncertainty to the final outcome. This is why we add the
variances even if we are interested in the difference of the random variables.
This add the variances is only true if we have independent variables. If one outcome affects what will
happen next, then the uncertainty (read “variance”) is diminished, so we should not add the
variances.
Suppose you mix a quart of lemonade. There should be 32 fluid ounces, but your measuring process
isn’t perfectly accurate so, while 32 is the expected amount, there remains some variability. You pour
a 12-ounce glass. Of course that measurement is not perfect either, so it’s actually somewhere around
12, give or take a little. How much is left in the pitcher? Should be around 20 ounces, of course, but
given the uncertainty about the initial amount and the variability in how far you filled the glass, you
can’t say exactly how much is left. Indeed, you are less sure about the remaining lemonade because
you removed an unknown amount. Subtracting some lemonade has increased the variability in the
amount that might be in the pitcher. Variances add.
Imagine a survey of the class that asked how many hours a student slept between noon yesterday
and noon today. The second question is how many hours the student was awake during that period
of time. We find the mean and standard deviation for each set of answers. Now, what are the mean
and standard deviation of the total number of hours for each student? Since every total must be 24
hours, the mean is 24 and the standard deviation is 0. We don’t add the variances this time because
the variables “hours asleep” and “hours awake” are not independent.
Class examples:

Greedy Pig
Rules: Everyone stands up. I will roll a die. Each student gets the number of points on the die.
If you want to keep your points, sit down. If you want to try to get more points, remain
standing for another roll. Points accumulate with each roll if you are still standing. The game
continues until all students are sitting, BUT if I roll a 5, everyone still standing loses all of their
points, and the game is over.
What strategy did you use?
Did you have a limit for when you would sit down?
Expected value for this game:
1
1
1
1
1
1
P  1   P  2   P  3   P  4   0  P  6 
6
6
6
6
6
6
5P  16
=
, which is greater than their current score if P<16, so in the long run the
6
optimal strategy is to sit down whenever your point total hits 16 (or more)
Brashear AP Statistics Ch 16 notes-teacher (12/2/04)
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
Ace of Hearts
A player pays $5 to play, and draws a card from a deck. If he draws the ace of hearts, he wins
$100. For any other ace he wins $10, and for any other heart he wins $5. For any other card
he loses.
 Wanna play?
 What if the top prize were $200?
 What is the expected value for each of the situations?
E (X )  95  521   5  523   0  5212   ( 5)  36
 $1.35
52 
E (X )  195  521   5  523   0  5212   ( 5)  36
 $0.58
52 
 Does this ever really happen?
No, no one can ever win or lose these amounts on any one play.
 What is the variance of this situation (with the $100 payoff)?
Var(X)=
 95   1.35     5   1.35     0   1.35     5   1.35     $190.44
2
1
52
2
3
52
2
2
12
52
36
52
 What is the standard deviation?
SD (X )  Var (X )  $13.80

Combining independent random variables
Consider this dice game: no points for rolling a 1, 2, or 3; 5 points for a 4 or 5; 50 points for a 6.
 Find the expected value and standard deviation.
 Imagine doubling the points awarded. What are the new mean and standard
deviation?
Brashear AP Statistics Ch 16 notes-teacher (12/2/04)
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 Now imagine just playing the game twice. What are the mean and the standard
deviation of your total points?
 Suppose you and a friend both play the dice game. What are the mean and the
standard deviation of the difference of your scores?
***Adding or subtracting a constant from the values of a random variable shifts the
mean but doesn’t change the variance or standard deviation.
***In general, multiplying each value of a random variable by a constant multiplies the
mean by that constant and the variance by the square of the constant.
***The mean (expected value) of the sum of two random variables is the sum of the
means (expected values).
***The mean (expected value) of the difference of two random variables is the
difference of the means (expected values).
***The variance of the sum (or difference) of two independent random variables is the
sum (or difference) of their variances.
Brashear AP Statistics Ch 16 notes-teacher (12/2/04)
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
Suppose a used car dealer runs autos through a two-stage process to get them ready to sell.
The mechanical checkup costs $50 per hour and takes an average of 90 minutes with a
standard deviation of 15 minutes. The appearance prep (wash, polish, etc) costs $6 per hour
and takes an average of 60 minutes with a standard deviation of 6 minutes.
 What are the mean and standard deviation of the total time spent preparing a car?
Assuming these events are independent:
E(M+A)=E(M)+E(A)=90+60=150 minutes
SD(M+A)= Var (M ) Var ( A )  152  6 2  16.2 min
 What are the mean and the standard deviation of the total expense to prepare a car?
(Note: working in hours now)
E(50M+6A)=50E(M)+6E(A)=50(1.5)+6(1)=$81
SD(50M+6A)= 502Var (M )  6 2Var ( A )  502 (0.25)2  6 2 (0.1)2  $12.51
 What are the mean and standard deviation of the difference in costs for the two
phases of the operation?
(Note: new variance is same as variance for the sum)
E(50M-6A)=50E(M)-6E(A)=50(1.5)-6(1)=$69
SD(50M-6A)= 502Var (M )  6 2Var ( A )  502 (0.25)2  6 2 (0.1)2  $12.51
 What is the probability that it will take longer to do the appearance prep than the
mechanical checkup?
Note: we cannot answer this question unless we believe that each phase of the process
can be described by a Normal model.
E(M-A)=E(M)-E(A)=90-60=30 minutes
SD(M-A)= Var (M ) Var ( A )  152  6 2  16.2 minutes
0  30 

P(M-A<0)= P  z 
 P  z  1.85   3%
16.2 

***When two random variables have Normal models, so does their sum or
difference.
Brashear AP Statistics Ch 16 notes-teacher (12/2/04)
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