Download Probability Sampling

Probability & Sampling
The Practice of Statistics 4e
Mostly Chpts 5 – 7
Lew Davidson (Dr.D.)
Mallard Creek High School
[email protected]
704-786-0470
Probability & Sampling
The Practice of Statistics 4e
Mostly Chpts 5 – 7
•
•
•
•
Classical Probability
(Ch 5)
Random Variables
(Ch 6)
Sampling Theory
(Ch 7)
Sampling Distributions
(Chpts. 8-12)
(Basis for all of Inference)
Section 5.1 - 5.3
Classical Probability
Learning Objectives
 Probabilities from Two-Way Tables, Venn and Tree diagrams
Probability Rules – Summary - 1
1. Random Process:
P(event) = long term frequency of occurrence
2.
number of weighted outcomes corresponding to event A
P( A) 
total number of weighted outcomes in sample space
3.
A probability model: lists the possible outcomes in the
sample space S and the probability for each (Discrete
Variables).
4.
For any event A:
0 ≤ P(A) ≤ 1  P(sample space) = P(S) = 1
5.
P(complement) = P(AC) = 1 – P(A)
6.
A two-way table or a Venn diagram can be used to
display the sample space for a chance process
Probability Rules – Continued - 2
7. Intersection (A ∩ B) of A & B = all outcomes in both A & B
8. Union (A ∪ B) = all outcomes in event A, event B, or both.
9. P(A or B) = P(A U B) = P(A) + P(B) – P(A ∩ B)
10. Events A and B are mutually exclusive (disjoint) if they
have no outcomes in common. If so:
P(A ∩ B) = 0 & P(A or B) = P(A) + P(B)
11. Two-way tables, Tree & Venn diagrams can be used to
display the sample space for a chance process.
Probability Rules – Continued – 3
12. Intersection (A ∩ B) of A & B = all outcomes in both A & B
13. Union (A ∪ B) = all outcomes in event A, event B, or both.
14. P(A or B) = P(A U B) = P(A) + P(B) – P(A ∩ B)
15. Events A and B are mutually exclusive (disjoint) if they
have no outcomes in common. If so:
P(A ∩ B) = 0
&
P(A or B) = P(A) + P(B)
16. If both A & B are independent:
P(A ∩ B) = P(A|B)P(B) = P(A) • P(B)
17. Check:
a.
b.
c.
d.
Are disjoints events independent?
Yes
No
Not necessarily
Not enough information to say
• Example: Grade Distributions (Two Way Table)
E: the grade comes from an Eng. & Phy Sci course
L: the grade is lower than a B.
Total
6300
1600
2100
Total 3392 2952
Find P(L)
P(L) = 3656 / 10000 = 0.3656
Find P(E | L)
P(E | L) = 800 / 3656 = 0.2188
Find P(L | E)
P(L| E) = 800 / 1600 = 0.5000
3656 10000
• Example: What percent of all adult
Internet users visit video-sharing sites?
P(video yes ∩ 18 to 29) = 0.27 • 0.7 = 0.1890
P(video yes ∩ 30 to 49) = 0.45 • 0.51 = 0.2295
P(video yes ∩ 50 +) = 0.28 • 0.26 = 0.0728
P(video yes) = 0.1890 + 0.2295 + 0.0728 = 0.4913
P(A ∩ B) = P(A and B) = P(A intersect B)
General Multiplication Rule
The probability that events A and B both occur can be
found using the general multiplication rule
P(A ∩ B) = P(A) • P(B | A)
where P(B | A) is the conditional probability that event B
occurs given that event A has already occurred.
• Example: Teens with Online Profiles
93% of teenagers use the Internet,
55% of online teens have posted a profile on a social-networking site.
a) Draw a T diagram for this situation
b) What percent of teens are online and have posted a profile?
P(online)  0.93
P(profile | online)  0.55
51.15% of teensare online and have
posted a profile.
P(online and have profile)  P(online) P(profile | online)
 (0.93)(0.55)
 0.5115


General Multiplication Rule
P(AP(A
∩∩
B) B) =P(A)
P(A) • P(B
P(B || A)
A)
Conditional Probability Formula
To find the conditional probability P(B | A), use the formula
=
What is the probability that a randomly
selected resident who reads USA Today
also reads the New York Times?
P(A  B)
P(B | A) 
P(A)

P(A  B)  0.05
P(A)  0.40
0.05
P(B | A) 
 0.125
0.40
 a randomly selected resident who
There is a 12.5% chance that
reads USA Today also reads the New York Times.
Summary
18. If one event has happened, the chance that another event
will happen is a conditional probability.
P(B|A) represents the probability that event B occurs given
that event A has occurred.
19. Events A and B are independent if the chance that event B
occurs is not affected by whether event A occurs. If 2 events
are mutually exclusive (disjoint), they cannot be independent.
20. When chance behavior involves a sequence of outcomes, a
tree diagram can be used to describe the sample space.
21. The general multiplication rule states that the probability of
events A and B occurring together is:
P(A ∩ B)=P(A) • P(B|A)
22. In the special case of independent events:
P(A ∩ B)=P(A) • P(B)
23. The conditional probability formula is :
P(B|A) = P(A ∩ B) / P(A)
Review
• http://tinyurl.com/DrD-Inference
• http://stattrek.com
/ap-statistics/test-preparation.aspx
• http://tinyurl.com/DrD-MC
• Random Variable and Probability Distribution
Definition:
A random variable takes numerical values that describe the outcomes of
some chance process. The probability distribution of a random variable gives
its possible values and their probabilities.
Example:
Consider tossing a fair coin 3 times.
Define X = the number of heads obtained
X = 0: TTT
X = 1: HTT THT TTH
X = 2: HHT HTH THH
X = 3: HHH
Value
Probability
0
1
2
3
1/8
3/8
3/8
1/8
Discrete and Continuous Random Variables
A probability model describes the possible outcomes of a chance process and
the likelihood that those outcomes will occur.
A numerical variable that describes the outcomes of a chance process is called
a random variable. The probability model for a random variable is its
probability distribution
There are two main types of random variables:
discrete and continuous.
If we can list all possible outcomes for a random
variable and assign probabilities to each one, we
have a discrete random variable.
Discrete Random Variables and Their Probability Distributions
A discrete random variable X takes a fixed set of possible values with gaps between. The
probability distribution of a discrete random variable X lists the values xi and their
probabilities pi:
Value:
Probability:
x1
p1
x2
p2
x3
p3
…
…
The probabilities pi must satisfy two requirements:
1. Every probability pi is a number between 0 and 1.
2. The sum of the probabilities is 1.
To find the probability of any event, add the probabilities pi of the particular values xi that
make up the event.
• Discrete random variables usually by counting
• Continuous random variables usually by measuring
A continuous random variable X takes on all
values in an interval of numbers.
The probability distribution of X is a density
curve. (Area= 1, Curve never negative)
Probability of any event is the area under the
density curve corresponding to the interval
infinitely many possible values
Define Y as the height of a randomly chosen young woman. Y is a continuous random
variable whose probability distribution is N(64, 2.7).
What is the probability that a randomly chosen young
woman has height between 68 and 70 inches?
P(68 ≤ Y ≤ 70) = ???
68  64
2.7
 1.48
z
70  64
2.7
 2.22
z
P(1.48 ≤ Z ≤ 2.22) = P(Z ≤ 2.22) – P(Z ≤ 1.48)


= 0.9868 – 0.9306
= 0.0562
There is about a 5.6% chance that a randomly chosen young woman has a height
between 68 and 70 inches.
Probability of an Exact value in a
continuous distribution?
• P ( x = 3) = ?
Section 6.1
Discrete and Continuous Random Variables
Summary
 The mean of a random variable is the long-run average value of the
variable after many repetitions of the chance process. It is also known
as the expected value of the random variable.
 The expected value of a discrete random variable X is
x   x i pi  x1 p1  x 2 p2  x 3 p3  ...
 The variance of a random variable is the average squared deviation of
the values of the variable from their mean. The standard deviation is
the square root of the variance. For a discrete random variable X,
2
 X  (x i X )2 pi  (x1 X )2 p1  (x 2 X )2 p2  (x 3 X )2 p3  ...
Linear Transformations
1.Adding (or subtracting) a constant, a, to each
observation:
• Adds a to measures of center and location.
• Does not change the shape or measures of
spread.
2.Multiplying (or dividing) each observation by a
constant, b:
• Multiplies (divides) measures of center and
location by b.
• Multiplies (divides) measures of spread by
|b|.
• Does not change the shape of the distribution.
• Linear Transformations
Effect on a Random Variable of Adding (or Subtracting) a Constant
Adding the same number a (which could be negative) to each value of a random
variable:
•
Adds a to measures of center and location (mean, median, quartiles,
percentiles).
•
Does not change measures of spread (range, IQR, standard deviation).
•
Does not change the shape of the distribution.
Transforming and Combining Random Variables
How does adding or subtracting a constant
affect a random variable?
• Linear Transformations
Effect on a Random Variable of Multiplying (Dividing) by a Constant
Multiplying (or dividing) each value of a random variable by a
number b:
• Multiplies (divides) measures of center and location
(mean, median, quartiles, percentiles) by b.
• Multiplies (divides) measures of spread (range, IQR,
standard deviation) by |b|.
• Does not change the shape of the distribution.
Note: Multiplying a random variable by a constant b multiplies the variance by b2.
Transforming and Combining Random Variables
How does multiplying or dividing by a
constant affect a random variable?
• Linear Transformations
Effect on a Linear Transformation on the Mean and Standard Deviation
If Y = a + bX is a linear transformation of the random variable X, then
•
The probability distribution of Y has the same shape as the probability
distribution of X.
•
µY = a + bµX.
•
σY = |b|σX (since b could be a negative number).
Transforming and Combining Random Variables
Whether we are dealing with data or
random variables, the effects of a linear
transformation are the same.
• Combining Random Variables
Variance of the Sum of Random Variables
For any two independent random variables X and Y, if T = X + Y, then the variance of
T is
T2  X2  Y2
In general, the variance of the sum of several independent random variables is the
sum of their variances.
Remember that you can add variances only if the two random variables are

independent, and that you can NEVER add standard deviations!
Transforming and Combining Random Variables
As the preceding example illustrates, when
we add two independent random
variables, their variances add. Standard
deviations do not add.
• Combining Random Variables
Mean of the Difference of Random Variables
For any two random variables X and Y, if D = X - Y, then the expected value of D is
E(D) = µD = µX - µY
In general, the mean of the difference of several random variables is the difference
of their means. The order of subtraction is important!
Variance of the Difference of Random Variables
For any two independent random variables X and Y, if D = X - Y, then the variance of
D is
D2  X2  Y2
In general, the variance of the difference of two independent random variables is
the sum of their variances.
Transforming and Combining Random Variables
We can perform a similar investigation to determine what happens when we
define a random variable as the difference of two random variables. In
summary, we find the following:
• Combining Normal Random Variables
Example
Mr. Starnes likes between 8.5 and 9 grams of sugar in his hot tea. Suppose
the amount of sugar in a randomly selected packet follows a Normal distribution with
mean 2.17 g and standard deviation 0.08 g. If Mr. Starnes selects 4 packets at random,
what is the probability his tea will taste right?
Let X = the amount of sugar in a randomly selected packet.
Then, T = X1 + X2 + X3 + X4. We want to find P(8.5 ≤ T ≤ 9).
µT = µX1 + µX2 + µX3 + µX4 = 2.17 + 2.17 + 2.17 +2.17 = 8.68
Transforming and Combining Random Variables
So far, we have concentrated on finding rules for means and variances of
random variables. If a random variable is Normally distributed, we can use
its mean and standard deviation to compute probabilities.
An important fact about Normal random variables is that any sum or
difference of independent Normal random variables is also Normally
distributed.
T2  X2  X2  X2  X2  (0.08)2  (0.08)2  (0.08)2  (0.08)2  0.0256
1
2
T  0.0256  0.16
3
4
Could we have solved this problem
by saying that T = 4X, since there
were 4 packets of sugar?
µT = µX1 + µX2 + µX3 + µX4 =
2.17 + 2.17 + 2.17 +2.17 = 8.68
µT = µX+X+X+X = 4µX = 4 * 2.17 = 8.68
SO, OK for the Expected Values E(X)!
Could we have solved this problem
by consider T = 4X, since there
were 4 packets of sugar?
 T2   X2   X2   X2   X2  (0.08) 2  (0.08) 2  (0.08) 2  (0.08) 2  0.0256
1
2
3
4
 T2  4 X2  4 * (0.08) 2  0.0256
 T   X  X  X  X  0.0256  0.16
but
 4 X  4 *.08  .32
SO :
 X X X X  4X
Summary
Transforming and Combining Random Variables
 Adding a constant a (which could be negative) to a random variable
increases (or decreases) the mean of the random variable by a but does
not affect its standard deviation or the shape of its probability
distribution.
 Multiplying a random variable by a constant b (which could be negative)
multiplies the mean of the random variable by b and the standard
deviation by |b| but does not change the shape of its probability
distribution.
 A linear transformation of a random variable involves adding a constant
a, multiplying by a constant b, or both. If we write the linear
transformation of X in the form Y = a + bX, the following about are true
about Y:
 Shape: same as the probability distribution of X.
 Center: µY = a + bµX
 Spread: σY = |b|σX
Section 6.2
Transforming and Combining Random Variables
Summary
 If X and Y are any two random variables,
X Y  X  Y
 If X and Y are independent random variables

X2 Y  X2  Y2
 The sum or difference of independent Normal random variables follows
a Normal distribution.

• Binomial Settings
Definition:
A binomial setting arises when we perform several independent trials of the same
chance process and record the number of times that a particular outcome occurs.
The four conditions for a binomial setting are
B
• Binary? The possible outcomes of each trial can be classified as “success” or
“failure.”
I
• Independent? Trials must be independent; that is, knowing the result of one
trial must not have any effect on the result of any other trial.
N
• Number? The number of trials n of the chance process must be fixed in
advance.
S
• Success? On each trial, the probability p of success must be the same.
Binomial and Geometric Random Variables
When the same chance process is repeated several times, we are often interested in
whether a particular outcome does or doesn’t happen on each repetition. In some
cases, the number of repeated trials is fixed in advance and we are interested in the
number of times a particular event (called a “success”) occurs. If the trials in these
cases are independent and each success has an equal chance of occurring, we have a
binomial setting.
• Binomial Random Variable
Definition:
The count X of successes in a binomial setting is a binomial random variable. The
probability distribution of X is a binomial distribution with parameters n and p,
where n is the number of trials of the chance process and p is the probability of a
success on any one trial. The possible values of X are the whole numbers from 0 to
n.
Note: When checking the Binomial condition, be sure to check the BINS and make
sure you’re being asked to count the number of successes in a certain number of
trials!
Binomial and Geometric Random Variables
Consider tossing a coin n times. Each toss gives either heads or tails. Knowing
the outcome of one toss does not change the probability of an outcome
on any other toss. If we define heads as a success, then p is the
probability of a head and is 0.5 on any toss.
The number of heads in n tosses is a binomial random variable X. The
probability distribution of X is called a binomial distribution.
• Binomial Probabilities
Example
Each child of a particular pair of parents has probability 0.25 of having
type O blood. Genetics says that children receive genes from each of their
parents independently. If these parents have 5 children, the count X of
children with type O blood is a binomial random variable with n = 5 trials
and probability p = 0.25 of a success on each trial. In this setting, a child
with type O blood is a “success” (S) and a child with another blood type is
a “failure” (F).
What’s P(X = 2)?
P(SSFFF) = (0.25)(0.25)(0.75)(0.75)(0.75) = (0.25)2(0.75)3 = 0.02637
However, there are a number of different arrangements in which 2 out of the 5
children have type O blood:
SSFFF
SFSFF
SFFSF
SFFFS
FSSFF
FSFSF
FSFFS
FFSSF
FFSFS
FFFSS
Verify that in each arrangement, P(X = 2) = (0.25)2(0.75)3 = 0.02637
Therefore, P(X = 2) = 10(0.25)2(0.75)3 = 0.2637
Binomial and Geometric Random Variables
In a binomial setting, we can define a random variable (say, X) as the number
of successes in n independent trials. We are interested in finding the
probability distribution of X.
• Binomial Coefficient
P(X  k)  P(exactly k successes in n trials)
= number of arrangements  p k (1 p) nk
Definition:
The number
 of ways of arranging k successes among n observations is given by the
binomial coefficient
n 
n!

 
k  k!(n  k)!
for k = 0, 1, 2, …, n where
n! = n(n – 1)(n – 2)•…•(3)(2)(1)

and 0! = 1.
Binomial and Geometric Random Variables
Note, in the previous example, any one arrangement of 2 S’s and 3 F’s had the
same probability. This is true because no matter what arrangement, we’d
multiply together 0.25 twice and 0.75 three times.
We can generalize this for any setting in which we are interested in k
successes in n trials. That is,
•
Binomial Probability
Binomial Probability
If X has the binomial distribution with n trials and probability p of success on each
trial, the possible values of X are 0, 1, 2, …, n. If k is any one of these values,
n k
P(X  k)   p (1 p) nk
k 
Number of
arrangements
of k successes
Probability of k
successes
Probability of nk failures
Binomial and Geometric Random Variables
The binomial coefficient counts the number of different
ways in which k successes can be arranged among n
trials. The binomial probability P(X = k) is this count
multiplied by the probability of any one specific
arrangement of the k successes.
Mean and Standard Deviation of a Binomial Random Variable
If a count X has the binomial distribution with number of trials n and probability of
success p, the mean and standard deviation of X are
 X  np
 X  np(1 p)
• As a rule of thumb, we will use the Normal

approximation
when n is so large that np ≥ 10
and n(1 – p) ≥ 10. That is, the expected number
of successes and failures are both at least 10.
• NORMALICY CONDITION
• Geometric Random Variable
.
The number of trials Y that it takes to get a success in a geometric setting is a
geometric random variable. The probability distribution of Y is a geometric
distribution with parameter p, the probability of a success on any trial. The possible
values of Y are 1, 2, 3, ….
Note: Like binomial random variables, it is important to be able to distinguish
situations in which the geometric distribution does and doesn’t apply!
Binomial and Geometric Random Variables
Definition:
• Geometric Settings
In a binomial setting, the number of trials n is fixed and the binomial random variable X
counts the number of successes. In other situations, the goal is to repeat a chance
behavior until a success occurs. These situations are called geometric settings.
Definition:
A geometric setting arises when we perform independent trials of the same chance
process and record the number of trials until a particular outcome occurs. The four
conditions for a geometric setting are
B
• Binary? The possible outcomes of each trial can be classified as “success” or
“failure.”
I
• Independent? Trials must be independent; that is, knowing the result of one
trial must not have any effect on the result of any other trial.
T
• Trials? The goal is to count the number of trials until the first success occurs.
S
• Success? On each trial, the probability p of success must be the same.
• Example: The Birthday Game
Read the activity on page 398. The random variable of interest in this game is Y = the number of
guesses it takes to correctly identify the birth day of one of your teacher’s friends. What is the
probability the first student guesses correctly? The second? Third? What is the probability the
kth student guesses corrrectly?
Verify that Y is a geometric random variable.
B: Success = correct guess, Failure = incorrect guess
I: The result of one student’s guess has no effect on the result of any other guess.
T: We’re counting the number of guesses up to and including the first correct guess.
S: On each trial, the probability of a correct guess is 1/7.
Calculate P(Y = 1), P(Y = 2), P(Y = 3), and P(Y = k)
P(Y 1) 1/7
P(Y  2)  (6/7)(1/7)  0.1224
P(Y  3)  (6/7)(6/7)(1/7)  0.1050
Notice the pattern?
TREE !!!!
Geometric Probability
If Y has the geometric distribution with probability p of
success on each trial, the possible values of Y are 1, 2,
3, … . If k is any one of these values,
P(Y  k)  (1 p)k1 p
• Mean of a Geometric Distribution
The table below shows part of the probability distribution of Y. We can’t show the entire
distribution because the number of trials it takes to get the first success could be an
incredibly large number.
1
2
3
4
5
6
pi
0.143
0.122
0.105
0.090
0.077
0.066
…
Shape: The heavily right-skewed shape is characteristic of
any geometric distribution. That’s because the most likely
value is 1.
Center: The mean of Y is µY = 7. We’d expect it to take 7
guesses to get our first success.
Spread: The standard deviation of Y is σY = 6.48. If the class played the Birth Day game many
times, the number of homework problems the students receive would differ from 7 by an
average of 6.48.
Mean (Expected Value) of Geometric Random Variable
If Y is a geometric random variable with probability p of success on each trial,
then its mean (expected value) is E(Y) = µY = 1/p.
Binomial and Geometric Random Variables
yi
Binomial and Geometric Random Variables
Summary
 A binomial setting consists of n independent trials of the same chance
process, each resulting in a success or a failure, with probability of
success p on each trial. The count X of successes is a binomial random
variable. Its probability distribution is a binomial distribution. BINS
 The binomial coefficient counts the number of ways k successes can be
arranged among n trials. Think n choose k
 If X has the binomial distribution with parameters n and p, the possible
values of X are the whole numbers 0, 1, 2, . . . , n. The binomial
probability of observing k successes in n trials is (USE CALC)
n k
P( X  k )    p (1  p)n k  binomialpdf (n, p, k )
k 
Section 6.3
Binomial and Geometric Random Variables
Summary
In this section, we learned that…
 The mean and standard deviation of a binomial random variable X are
 X  np
 X  np(1 p)
 The Normal approximation to the binomial distribution says that if X is a
count having the binomial distribution with parameters n and p, then

when n is large, X is approximately Normally distributed. We will use this
approximation when np ≥ 10 and n(1 - p) ≥ 10.
Section 6.3
Binomial and Geometric Random Variables
Summary
In this section, we learned that…
 A geometric setting consists of repeated trials of the same chance process
in which each trial results in a success or a failure; trials are independent;
each trial has the same probability p of success; and the goal is to count the
number of trials until the first success occurs. If Y = the number of trials
required to obtain the first success, then Y is a geometric random variable.
Its probability distribution is called a geometric distribution.
 If Y has the geometric distribution with probability of success p, the possible
values of Y are the positive integers 1, 2, 3, . . . . The geometric probability
that Y takes any value is USE Calc
P(Y  k)  (1 p)k1 p
 The mean (expected value) of a geometric random variable Y is 1/p.
SAMPLING DISTRIBUTIONS
Population
Sample
Collect data from
a representative
Sample...
Make an
Inference about
the Population.
SAMPLING DISTRIBUTIONS
Different random samples yield different
statistics. Thus a statistic is a random
variable
The sampling distribution shows all possible
statistic values
sampling variability: the value of a statistic
varies in repeated random sampling.
To make sense of sampling variability, we
ask, “What would happen if we took
many samples?”
Sample
Population
Sample
Sample
Sample
Sample
Sample
Sample
Sample
?
Population Distributions vs. Sampling Distributions
There are actually three distinct distributions involved when
we sample repeatedly and measure a variable of interest.
1) The population distribution gives the values of the variable
for all the individuals in the population.
2) The distribution of sample data shows the values of the
variable for all the individuals in the sample.
3) The sampling distribution shows the statistic values from
all the possible samples of the same size from the
population.

• The Sampling Distribution of
pˆ
We can summarize the facts about the sampling distribution
of pˆ as follows:
 of a Sample Proportion
Sampling Distribution
The mean of the sampling distribution ofpˆ is  pˆ  p
The standard deviation of the sampling distribution ofpˆ is
p(1 p)
 pˆ 
n
as long as the10% condition is satisfied: n  (1/10)N.
As n increases, the sampling distribution becomes approximately Normal. Before you perform
Normal calculations, check that the Normal condition is satisfied: np ≥ 10 and n(1 – p) ≥
10.
Sample Proportions
Choose an SRS of size n from a population of size N with proportion p
of successes. Let pˆ be the sample proportion of successes. Then
:
• The Sampling Distribution of
pˆ
In Chapter 6, we learned that the mean and standard deviation of a binomial
random variable X are

X  np
X  np(1  p)

1
 pˆ  (np)  p
n

pˆ is an unbiased estimator orp
1
 np(1  p)
 pˆ 
np(1  p) 

2
n
n
p(1  p)
n
As sample size increases, the spread decreases.
Sample Proportions
Since pˆ  X /n  (1/n)  X, we are just multiplying the random variableX
by a constant (1/n) to get the random variable pˆ . Therefore,
• Using the Normal Approximation for
pˆ
Inference about a population proportionp is based on the sampling distribution
of pˆ . When the sample size is large enough fornp and n(1 p) to both be at
least 10 (the Normal condition),the sampling distribution ofpˆ is
approximately Normal.

STATE: We want to find the probability that the sample proportion falls between 0.33 and
0.37 (within 2 percentage points, or 0.02, of 0.35).
PLAN: We have an SRS of size n = 1500 drawn from a population in which the proportion p
= 0.35 attend college within 50 miles of home.
 pˆ  0.35
 pˆ 
(0.35)(0.65)
 0.0123
1500
Sample Proportions
A polling organization asks an SRS of 1500 first-year college students how far away their home
is. Suppose that 35% of all first-year students actually attend college within 50 miles of home.
What is the probability that the random sample of 1500 students will give a result within 2
percentage points of this true value?
DO: Since np = 1500(0.35) = 525 and n(1 – p) = 1500(0.65)=975 are
both greater than 10, we’ll standardize and then use Table A to find

the desired probability.
 0.35
0.37  0.35
0.33
z
 1.63
 1.63
0.123
0.123
P(0.33  pˆ  0.37)  P(1.63  Z 1.63)  0.9484  0.0516  0.8968
z
CONCLUDE: About 90% of all SRSs of size 1500 will give a result within 2
 the population.

percentage
points of the truth about

x
When we choose many SRSs from a population, the sampling distribution of the
sample mean is centered at the population mean µ and is less spread out
than the population distribution. Here are the facts.

Mean and Standard Deviation of the Sampling Distribution of Sample Means
Suppose that x is the mean of an SRSof size n drawn from a large population
with mean  and standard deviation  . Then :
The mean of the sampling distribution ofx is x  
The standard deviation of the sampling distribution of x is
x 

n
as long as the 10% condition is satisfied: n ≤ (1/10)N.
Sample Means

• The Sampling Distribution of
Note : These facts about the mean and standard deviation ofx are true
no matter what shape the population distributionhas.
• Sampling from a Normal Population
We have described the mean and standard deviation of the sampling
distribution of the sample meanx but not its shape. That's because the
shape of the distribution of x depends on the shape of the population
distribution.
Sampling Distribution of a Sample Mean from a Normal Population
Sample Means
In one important case,there is a simple relationship between the two
distributions. If the population distribution is Normal,
then so is the
sampling distribution ofx. This is true no matter what the sample size is.
Suppose that a population is Normally distributed with mean and standard deviation
 . Then the sampling distribution ofx has the Normal distribution with mean and
standard deviation  / n, provided that the 10% condition is met.
Example: Young Women’s Heights
Sample Means
The height of young women follows a Normal
distribution with mean µ = 64.5 inches and
standard deviation σ = 2.5 inches.
Find the probability that a randomly selected
Let Xyoung
= the heightwoman
of a randomly is
selected
youngthan
woman. X66.5
is N(64.5,inches.
2.5)
taller
66.5  64.5
P(X  66.5)  P(Z  0.80) 1 0.7881 0.2119
 0.80
2.5
The probability of choosing a young woman at random whose height exceeds 66.5 inches is about
0.21.

z
 Find the probability that the mean height of an SRS of 10 young women exceeds 66.5
inches.

For an SRS of 10 young women, the sampling Since the population distribution is Normal, the
distribution of their sample mean height will sampling distribution will follow an N(64.5, 0.79)
distribution.
have a mean and standard deviation
P(x  66.5)  P(Z  2.53)
66.5  64.5
z
 2.53
 1 0.9943  0.0057

2.5
0.79
x    64.5 x 

 0.79
n
10
It is very unlikely (less than a 1% chance) that we
would choose an SRS of 10 young women whose

average height exceeds
66.5 inches.

• The Central Limit Theorem
Draw an SRS of size n from any population with mean  and finite
standard deviation  . The central limit theorem (CLT) says that when n
is large, the sampling distribution of the sample meanx is approximately
Normal.
Note: How large a sample size n is needed for the sampling distribution to be close to
Normal depends on the shape of the population distribution. More observations are
required if the population distribution is far from Normal.
• The Central Limit Theorem
Consider the strange population distribution from the Rice University sampling
distribution applet.
Describe the shape of the sampling
distributions as n increases. What do
you notice?
Sample Means
Normal Condition for Sample Means
If the population distribution is Normal,then so is the
sampling distribution ofx. This is true no matter what
the sample size n is.
If the population distribution is not Normal,the central
limit theorem tells us that the sampling distribution
of x will be approximately Normal in most cases if
n  30.

SAMPLING DISTRIBUTIONS
Different random samples yield different
statistics. Thus a statistic is a random
variable
The sampling distribution shows all possible
statistic values
Review
• http://tinyurl.com/DrD-Inference
• http://stattrek.com
/ap-statistics/test-preparation.aspx
• http://tinyurl.com/DrD-MC