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Stats CH 6 Intro Activity
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1. Purpose – can you tell the difference between bottled water and tap water?
You will drink water from 3 samples. 1 of these is bottled water.
2. You must test them in the following order:
First:
Second:
Third:
3. After testing each, circle the letter that you feel is the bottled water and show the teacher.
Do NOT discuss your results with your classmates.
4. Make a chart of the number of people who stated each sample
Sample
Frequency
A
B
C
Total
Relative
Frequency
The true bottled water sample is…
5. What percent correctly identified the bottled water?
6. Assume that no one can distinguish tap water from bottled water. If this is the case, then what is the
probability that a person would guess correctly?
7. What would the percentage that can distinguish tap water from bottled water have to be in order for you to
be convinced that the students are not guessing?
8. Perform a simulation of the 3 cups example to test how many times the correct response was discovered
among 50 students. The strings are 10 digits long. There are 5 strings. None of the strings include the digit 0.
What digits will you consider correct:
total correct:
percent:
What digits will you consider incorrect:
total incorrect:
percent:
5231676879
2781234823
9165396471
1248164528
8981692471
9. Make a histogram for the number of correct responses in class
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Stats CH 6 Intro Activity
2
1. Purpose – can you tell the difference between bottled water and tap water?
You will drink water from 3 samples. 1 of these is bottled water.
2. You must test them in the following order:
First:
Second:
Third:
3. After testing each, circle the letter that you feel is the bottled water and show the teacher.
Do NOT discuss your results with your classmates.
4. Make a chart of the number of people who stated each sample
Sample
Frequency
A
B
C
Total
Relative
Frequency
The true bottled water sample is…
5. What percent correctly identified the bottled water?
6. Assume that no one can distinguish tap water from bottled water. If this is the case, then what is the
probability that a person would guess correctly?
7. What would the percentage that can distinguish tap water from bottled water have to be in order for you to
be convinced that the students are not guessing?
8. Perform a simulation of the 3 cups example to test how many times the correct response was discovered
among 50 students. The strings are 10 digits long. There are 5 strings. None of the strings include the digit 0.
What digits will you consider correct:
total correct:
percent:
What digits will you consider incorrect:
total incorrect:
percent:
8699344889
3373967234
7414855135
5434738399
7973697488
9. Make a histogram for the number of correct responses in class
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AP STATS
Section 6.1: Discrete and Continuous Random Variables
Random Variable: A variable, usually represented by an X, which has a single numerical value (determined by
chance) for each outcome of an experiment.
Examples:
1) X = The number of seniors who get into college early.
2) X = The number of defective tires on a car
3) X = A random number chosen between 0 and 1
4) X = The lifetime of a light bulb.
a) Discrete Random Variable: Has a finite or countable number of values. (Examples 1 and 2)
b) Continuous Random variable: Has infinitely many values and the values can be associated with a
continuous scale so that there are no gaps or interruptions. (Examples 3 and 4)
Probability Distribution: Gives the probability of each value of a random variable.
Example 1 - Eggs-ample: Suppose the random variable X is the number of broken eggs in a randomly selected
carton of one dozen “store brand” eggs at a certain supermarket. Since the number of broken eggs is a discrete
random variable, the probability distribution is a list or a table of the possible values of X and the corresponding
probabilities
Number of Broken Eggs:
Probability:
0
.65
1
.20
2
.08
3
.04
4
.02
5
.01
Requirements for a probability distribution (legitimate):
1)
! ! =1
says all the probabilities must add to 1
2) 0 ≤ p(X) ≤ 1
for all values of x
Here is a probability Histogram that represents the above probability distribution:
Probability
Probability Histogram
P(X > 3) =
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
P(X ≥ 1) =
P(X < 2) =
0
1
2
3
Number of Broken Eggs
4
5
P(X ≥ 4) =
3
Example 2 - Tossing Coins:
What is the probability distribution of the discrete random variable X that counts the number of heads in three
tosses of a coin?
Number of heads
Probability
0
1
2
3
Let’s create a probability histogram for this distribution.
For a continuous random variable we use density curves, not histograms, to graphically represent the
distribution:
Example 3 - Random Number Example: X is a random number between 0 and 2. The distribution is a
continuous distribution and is represented by the uniform density curve below:
P(X)
0.5
0
2
X
The probability of any event is the area under the density curve
P(1 ≤ x ≤ 2) =
P(0 ≤ x ≤ 0.5) =
P(x ≥ 0.3) =
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If the curve is not uniform, you would need to use geometry or
calculus to find the probability of an event
And don’t forget, the most familiar density curve is the standard
normal curve. The Normal Distribution is a probability
distribution.
The Expected Value of a random variable represents the average value of the outcomes. The expected value is
the mean of a probability distribution. It is found by finding the value of:
Expected Value:
E = (! • ! ! ) = !
Mean: µ =
∑ x ⋅ p(x)
Example 4 - Looking back at the Probability distribution from example 1, we can calculate, on average, how
many broken eggs there will be in a carton:
Number of Broken Eggs:
Probability:
0
.65
1
.20
2
.08
3
.04
4
.02
5
.01
This means that “on average” we can expect that _________ eggs will be broken in a randomly selected carton
of eggs.
Example 5 - Getting the Flu: The probability that 0, 1, 2, 3, or 4 people will seek treatment for the flu during
any given hour at an emergency room is show in the probability distribution below.
x
P(x)
0
.12
1
.25
2
?
3
.24
4
.06
a) What is the probability that 2 people will seek treatment for the flu during any given hour at an emergency
room?
b.) What is the probability that at least 1 person will be treated for the flu in the next hour?
c.) What is the probability that 3 or more people will be treated for the flu in the next hour?
d.) What is the average number of people that an emergency room can expect to treat for the flu during any
given hour?
HW Section 6-1 A: 1,5,7,9,13
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More Examples of Expected Value and Games:
•
Expected value can also be used to determine whether or not a particular game is “fair” and therefore worth
playing.
Example 6 - Patrick offers Christine to play a dice game whereby he will pay her $6 if she rolls a six, but
Christine would have to pay Patrick $1 every time she rolls a number other than 6. Should Christine agree to
play this game over a long period of time?
To find the expected value of a game multiply each payoff by its probability and then add.
Based on the above definition, the expected value for Christine is:
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This means that for every roll of the die, Christine will earn, on average, $ , which also means that Patrick will
6
1
lose, on average $ for every roll of the die.
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If Patrick and Christine were to play this game 100 times, how much money (on average) can Christine expect
to make? How much can Patrick expect to lose?
Fair Game: We say that a game is “fair” whenever its expected value is equal to zero. For example, you win a
dollar every time you toss a coin and get heads, and lose a dollar every time you get tails.
Example 7 - Zoot Suit Example: Would you be willing to play this game? You pay me $25 to play. I take a
standard deck of cards. I Shuffle the cards well. You pick 1 card at random. The suit of that card becomes
your winning suit. You do not put the card back. You pick 2 more cards at random. If those 2 cards are both
the winning suit, I will pay you $500. If they are not, I pay you nothing.
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Variance, and standard deviation of a probability distribution
Variance
2)
Standard Deviation
∑ µ
σ
µ
∑
σ =
∑[x ⋅ p(x)] − µ
1) σ 2 =
3)
(the square root of the variance)
2
=
( x − ) 2 ⋅ p ( x)
[ x 2 ⋅ p( x)] −
2
2
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Note: The mean and standard deviation formulas are in the stat formula packet that you get on the AP
Example 8 - Find the mean, variance, and standard deviation of the following probability distribution:
x
0
1
2
3
4
P(x)
.3
.1
.2
.1
.3
Calculator Method:
HW Section 6-1 B: 14,18,19,23,25
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Section 6.2: Transforming and Combining Random Variables
Review of Chapter 2: Operations on statistics:
Addition of constant a
Mean
Increase by a
Subtraction of
constant a
Decrease by a
Multiplication of
constant a
Multiply by a
Division of constant a
Median
Increase by a
Decrease by a
Multiply by a
Divides by a
Standard Deviation
No change
No change
Multiply by |a|
Divides by |a|
Quartiles
Increase by a
Decrease by a
Multiply by a
Divides by a
IQR
No change
No change
Multiply by |a|
Divides by |a|
Min and Max
Increase by a
Decrease by a
Multiply by a
Divides by a
Range
No change
No change
Multiply by |a|
Divides by |a|
Percentiles
Increase by a
Decrease by a
Multiply by a
Divides by a
Divides by a
Rules for Random Variables:
1. If the value “b” is multiplied by each value of a random variable:
Measures of Center (mean, median, quartiles, percentiles) multiply by b
Measures of Spread (range, IQR, standard deviation) multiply by |b|
Shape is NOT changed
2. If the value “b” is divided into each value of a random variable:
Measures of Center (mean, median, quartiles, percentiles) divide by b
Measures of Spread (range, IQR, standard deviation) divide by |b|
Shape is NOT changed
3. If the value “b” is added to each value of a random variable:
Adds b to measures of Center (mean, median, quartiles, percentiles)
Measures of Spread (range, IQR, standard deviation) are NOT changed
Shape is NOT changed
4. If the value “b” is subtracted from each value of a random variable:
Subtracts b from measures of Center (mean, median, quartiles, percentiles)
Measures of Spread (range, IQR, standard deviation) are NOT changed
Shape is NOT changed
5. If Y = a + bX is a linear transformation of the random variable X, then
The probability distribution of Y is the same as X
The mean for Y = a + b•(mean of X)
!! = ! + ! • !!
The st. dev. of Y = |b|•(st. dev. of X)
!! = |!| • !!
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Example 1
Pete owns a Jeep Tour Company. There must be at least 2 passengers for the trip and no more than 6. The
discrete random variable X describes the number of passengers with the following distribution
No. of passengers xi 2
Probability pi
0.15
3
0.25
4
0.35
5
0.20
6
0.05
a) Find the expected value (mean) and the standard deviation for X. Describe it in context.
Example 2
Pete charges $150 per passenger. The discrete random variable C is equal the total amount of money that Pete
collects on a randomly selected trip. Therefore if X = 2 passengers, then C = $300.
Then C = 150X
Total Collected ci
Probability pi
300
0.15
450
0.25
0.35
0.20
0.05
b) Complete the values for the total collected in the table above
c) Find the expected value (mean) and the standard deviation for C. Describe it in context.
d) How do the answers to “c” compare to the answer to “a”?
Example 3
It costs Pete $100 to buy permits, gas, and a ferry pass for each half-day trip. The amount of profit V that Pete
makes from the trip is the total amount of money C that he collects from passengers minus $100.
Then V = C – 100
If Pete has only two passengers on the trip (X = 2), then C = 300 and V = 200.
Total Profit vi
Probability pi
200
0.15
350
0.25
0.35
0.20
0.05
e) Complete the values for the total profit in the table above
f) Find the expected value (mean) and the standard deviation for V. Describe it in context.
g) How do the answers to “F” compare to the answer to “a” and “c”?
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Example 4
At a car dealership, let the discrete random variable X = cars sold on a given day with distribution:
Cars sold
Probability
0
.3
1
.4
2
.2
3
.1
1. Find the mean and standard deviation
2. The owner offers the manager a $500 bonus for each car sold. The manager attempts to attract
customers by offering free coffee and donuts at a cost of $75. Define a profit variable and find the
mean and standard deviation for profit.
HW Section 6-2 C: 27-30,37, 39-41,43,45
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Combining Random Variables Activities
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Rules for means & variances for two random variables
Here is an illustration of these important statistical formulas (which are NOT included on your AP Stat Formula
packet)
If X and Y are any two random variables, then
or if T = X + Y
µ x+ y = µ x + µ y
! ! = !! + !!
If X and Y are independent random variables, then σ 2 x + y = σ 2 x + σ 2 y
or if T = X + Y
µ x− y = µ x − µ y
σ 2 x− y = σ 2 x + σ 2 y
!!! = !!! + !!!
Note: These "nice" rules do not hold for standard deviations
Example 5: Let’s say you started your own business on weekends. The business can bring you a profit on
Saturday and Sunday based on various (independent) scenarios. The profits and their respective probabilities
for Saturday and Sunday are listed below:
X= Saturday profit
x
p(x)
$10.00
$50.00
$100.00
SUM
0.1
0.5
0.4
1
Y= Sunday profit
y
$20.00
$60.00
$80.00
SUM
p(y)
0.2
0.3
0.5
1
Based on the tables to the left please calculate:
µx =
µy =
µx + µ y =
µx − µ y =
σ x2 =
σ y2 =
σ x2 + σ y2 =
σx =
σy =
σ x +σ y =
On the AP Formula Page
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Example 6: A company makes packages containing Oreo cookies. The mean of an Oreo cookie is 40 grams
with standard deviation = 2 grams.
The company rejects cookies weighing less than 36 gram. In a batch of 20,000 cookies, how many will be
discarded?
The wrapper for a package containing 5 cookies has a mean of 6 grams with a standard deviation of 1.3 grams.
– What is the mean weight of the package (cookies and wrapper).
– What is the standard deviation of the package?
– Define the random variable A as the average weight of the 5 cookies. Find the mean and standard
deviation of A.
The Law of Large Numbers:
Draw independent observations at random from any population with finite mean µ. Decide how accurately you
would like to estimate µ. As the number of observations drawn increases, the mean ! of the observed values
eventually approaches the mean µ of the population as closely as you specified and then stays that close.
HW Section 6-2 D: 49,51,57-59,63
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Section 6.3: The Binomial Distribution
There are many experiments and situations that result in what are called dichotomous responses – responses for
which there exist two possible choices (True / False, Yes / No, Defective / Non-defective, Male / Female, etc.).
A simple example of such an experiment is that of tossing a coin, where there are only two possibilities, Heads
or Tails. There are many other types of experiments similar to a coin toss where you are observing the “success”
or “failure” of a certain outcome. Such experiments give us a probability distribution called a binomial
distribution.
Requirements (Use this when asked if a situation is binomial):
1. Binary – there can only be 2 ways the event can turn out
2. Independent – events must be independent for each trial
3. Fixed – there must be a set number of trials
4. Probability of success is constant for all trials and the we are counting successes
Extra Rule – the sample size must be less than 10% of the entire population size.
Example 1: Tom is about to take a five-question true/false quiz for which he is not prepared. He will be
guessing on all five questions.
What is the probability that:
1) He gets all the answers correct?
2) He gets all the answers wrong?
3) He gets exactly three answers correct?
4) He will pass the quiz?
Solution: First we need to know how many total
answer combinations are there. Using the
counting rule:
2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 = 32
Use ‘c’ to denote a correct answer and ‘w’ denote
a wrong answer. All the possible combinations of
correct and incorrect are shown !
The probability distribution for
this situation is shown below.
(X = number of correct responses):
x
0
1
2
3
4
5
p(x)
1/32
5/32
10/32
10/32
5/32
1/32
This can be done using the AP Stat Program.
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The long process used in example 1 can be avoided by using the notion that this is a binomial probability
distribution. There is a formula we can use to find the probability of any value in a binomial distribution:
n!
P( x) =
⋅ p x ⋅ q n− x for x = 0,1, 2, 3, ...., n
(n − x)! x!
n = # of trials
x = # of successes among n trials
p = probability of success in any one trial
q = probability of failure in any one trial (q = 1 – p)
! ! !!!
This formula can be written as ! ! = ! = !
! !
where the parenthesis is called n
!
choose k. This is the combination formula for choosing k items from n total items. In
combinations, order does NOT matter.
Formulas for The binomial distribution:
µ = np
σ 2 = np (1 − p)
σ = np (1 − p)
Example 2: Suppose Charlie manages to manipulate a coin in such a way that it lands on heads with a .7
probability and lands on tails with a .3 probability. Suppose John then flips the coin 20 times, what is the
probability that it will land on heads for exactly 13 of the twenty flips? On average, how many flips will result
in a heads? What is the standard deviation of this binomial distribution?
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Example 3: Suppose that the probability that any random freshman girl will agree to go with Sam to the senior
prom is 0.1. Suppose Sam asks 20 random freshman girls to the prom, what is the probability that exactly 1 will
say “yes”? That at least 1 will say “yes”? On average, how many freshman girls will agree to go with Sam to
the prom? What is the standard deviation of this binomial distribution?
HW Section 6-3
E: 61,65,66,69, 71,73,75,77
F: 79,81,83,85,87,89
The Geometric Distribution
Let’s start with a simulation: Roll a die until the number six appears and keep a record of how many rolls it
took before the six was obtained.
While in a binomial distribution the random variable was the number of successes in a fixed number of trials, in
a geometric distribution the random variable is the number of trials it takes to achieve a success.
Examples:
1) flip a coin until you get heads
2) Roll a die until you get a 6
3) Throw darts at a dartboard until you hit the bull’s-eye
A geometric distribution must have the following properties:
1) Each trial in the experiment must have only two possible outcomes (success or failure)
2) The probability of success, p, doesn’t change from trial to trial
3) The trials in the experiment are independent
4) The variable of interest is the number of trials required to reach the first success.
P(X = n) = (1-p)n-1p
where p = probability of success
The mean, µ, of a geometric distribution (the average number of times we can expect to repeat the trials before
a success occurs) is simply 1/p where p is the probability of success.
µ = 1/p
The variance of a geometric distribution
variance = (1 - p) / p2
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Example 4: We will take our simulation example to analyze various aspects of the geometric distribution.
Lets first find the various probabilities associated with our dice simulation and come up with a probability
distribution:
a. Find the probability that a 6 will come up on the first roll.
b. Find the probability that a 6 will come up on the second roll.
P(won’t come up on the first roll) =
P(will come up on the second roll =
c. Find the probability that a 6 will come up on the third roll.
d. If we proceed in the manner above will come up with the following probability distribution:
X
P(X)
1
2
3
4
5
1/6
5/6•1/6
5/6•5/6•1/6
5/6•5/6•5/6•1/6
5/6•5/6•5/6•5/6 •1/6
n
(5/6)n-1(1/6)
e. How many rolls can we expect, on average, to roll the die before getting the number 6?
f. The probability that it takes more than n trials to see a success is:
g. Find the variance of a geometric distribution
h. Find the probability that it would take more than 5 rolls for us to get the number 6
HW Section 6-3 G: 93,95,97, 99,101-104
Quiz Review: Complete the AP Prep Questions on pages 409-411
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