The White Board



Printer
Computer w/ Internet Access
Assessment
 Discussions (4)
0-40
 Practice (3)
0-75
 Quizzes (4)
0-40
 Tests (CST)
Test 1 (0-48)
TOTAL
0-203
AP Statistics
Grading Scale
Semester 2 - Unit 1
97 - l00% = A+
93 - 96% = A
90 - 92% = A87 - 89% = B+
83 - 86% = B
80 - 82% = B77 - 79% = C+
73 - 76% = C
70 - 72% = C<69% = INC
Name:
Teacher:
1 of 70
AP Statistics Sem 2 - Unit 1
Materials List
Binomials and Distributions
The White Board
04152013
This Page Left
Intentionally
Blank
2 of 70
AP Statistics
Unit Overview: Binomials and Distributions
Think about It
Page 1 of 4
Unit 1
S2
The following questions will help you study key concepts covered in this unit.
1. You're taking samples of size 10 from a population of 20 marbles and recording the
number of blue marbles in each sample. Why isn't this a binomial situation, and why
would it be almost binomial if you did the same experiment with a population of 200
marbles?
2. Explain the term continuity correction. Why does it work, and how do you use it?
3. Consider a parent population with mean 75 and standard deviation 7. The population
doesn't appear to have extreme skewness or outliers.
A. What are the mean and standard deviation of the distribution of sample means for
n = 40?
B. What's the shape of the distribution? Explain your answer in terms of the central
limit theorem.
C. What proportion of the sample means of size 40 would you expect to be 77 or less?
If you use your calculator, show what you entered.
D. Draw a sketch of the probability you found in part C, and label the horizontal axis.
4. Suppose that after several years of offering AP Statistics, a high school finds that final
exam scores are normally distributed with mean 78 and standard deviation 6.
A. What are the mean, standard deviation, and shape of the distribution of x-bar for
n = 50?
B. What's the probability a sample of scores will have a mean greater than 80?
C. Sketch the distribution curve for part B, showing the area that represents the
probability you found. Be sure to label the horizontal axis.
5. In a survey, 600 mothers and fathers were asked about the importance of sports for
boys and girls. Of the parents interviewed, 70% said the genders are equal and should
have equal opportunities to participate in sports.
A. What are the mean, standard deviation, and shape of the distribution of the sample
ˆ of parents who say the genders are equal and should have equal
proportion p
opportunities? Be sure to justify your answer for the shape of the distribution. Use
n = 600.
B. Using the normal approximation without the continuity correction, sketch the
ˆ . Shade equal areas on both
probability distribution curve for the distribution of p
sides of the mean to show an area that represents a probability of .95, and label the
upper and lower bounds of the shaded area as values of p-hat (not z-scores).
C. In your sketch from part B, the shaded area shows a .95 probability of what
happening? In other words, what does the probability of .95 represent?
______________________________
Copyright © 2011 Apex Learning Inc. (See Terms of Use
3 ofat70www.apexvs.com/TermsOfUse)
AP Statistics
Unit Overview: Binomials and Distributions
Think About It
Page 2 of 4
D. Using the normal approximation, what's the probability that a randomly drawn
sample of 600 parents will have a sample proportion between 67% and 73%?
Draw a sketch of the probability curve, shade the area representing the
probability you're finding, and label the z-scores that represent the upper and
lower bounds of the probability you're finding. Don't use the continuity correction.
E. Now, use the exact binomial calculation to find the probability of getting between,
but not including, 67% and 73% of the respondents in a sample of 600 who say the
genders are equal and should have equal opportunities. To use the exact binomial,
you'll need to convert the proportions to counts by multiplying each proportion by
600.
F. Now try it again, but this time find the probability of getting at least 67% but no
more than 73%. Use the exact binomial calculation.
6. Annie is a basketball player who makes, on average, 65% of her free throws. Assume
each shot is independent and the probability of making any given shot is .65.
A. What's the probability Annie will miss three straight free throws before she makes
one? (If you use a calculator to get your answer, write your answer in standard
notation and show what you do on the calculator as well.)
B. During a season, Annie takes 140 free throws. What's the exact binomial probability
she'll make at least 100 out of 140 of these throws?
C. For part B, are the conditions that permit you to use a normal approximation to the
binomial satisfied? Explain.
D. Redo part B using a normal approximation, without continuity correction, to the
binomial. Draw a sketch of the situation. Then draw the distribution, shade the
area representing the probability you're finding, and label the horizontal
axis approximately.
NOTE: If you use a graphing calculator, show what you do to get the z-scores, and
explain your answer in enough detail to show your instructor you understand what
you're doing.
E. Redo part B using a normal approximation, with continuity correction, to the
binomial. NOTE: If you use a graphing calculator, show what you do to get the
z-scores on your calculator, and explain your answer in enough detail to show your
instructor you understand what you're doing.
______________________________
Copyright © 2011 Apex Learning Inc. (See Terms of Use
4 ofat70www.apexvs.com/TermsOfUse)
AP Statistics
Unit Overview: Binomials and Distributions
Think About It
Page 3 of 4
Discussion
7. Have you ever been stuck in Jail when playing Monopoly? To get out, you need to roll doubles in
three tries or fewer or you have to pay. On average, how many times would people have to roll
before getting doubles, and is that number larger than three? This is an average waiting-time
question, where you're interested in the number of tries you need on average to get the outcome
you want.
You'll learn more about situations like this and the probability distributions they produce in the
next Tutorial. Meanwhile, this Discussion will help you think about waiting-time situations.
Respond to any one (or more) of the following, or respond to another student's posting. As you
explain your reasoning, be sure to use what you know about the laws of probability.
1. If you buy a very large bag of candies colored brown, yellow, green, blue, orange, and red,
and you start eating them, how many candies would you expect to pick until you got a blue
one? Explain your reasoning using what you know about probability.
2. What's the probability you'll get out of Jail in Monopoly without having to pay? In other words,
what are the chances you'll roll doubles on two six-sided dice within three tries?
3. You may be familiar with promotional campaigns where a company's products are marked
with a letter, under the bottle cap of a soft drink, for example, and you're supposed to spell
something to win a prize. In these cases, do you think some game pieces are more common
than others? Describe an experiment you could conduct to see if some game pieces are easier
to get than others.
4. Say you're playing a game like the one described in topic 3. A soft drink company has a letter
printed on each bottle cap and the object is to spell the words I bought a lot of bottles to spell
this. You have all the letters you need except p. The company's disclaimer statement says for
each bottle you buy there's a 1/200 chance of getting a p. On average, how many bottles
would you expect most people to buy in order to get this letter?
5. Create your own question about an average = waiting-time situation, or describe a real one
you've seen or participated in. Explain why it's an average = waiting-time situation, and invite
other students to answer it.
6. Have you ever won anything in a game like the ones described in topics 3 and 4? Describe the
game, and calculate the probability of winning.
8.
A sampling distribution is the distribution of a statistic. In other words, if you took the mean of
many samples from a population, the set of means would form the distribution of .
Some of the questions below have important information you'll need in order to understand the
other questions, so before choosing which question to answer please read them all.
Respond to one or more of the following or respond to another student's posting.
1. Suppose you had a population of fish whose lengths were normally distributed with a mean of
50 centimeters and a standard deviation of 5 centimeters. You draw a simple random sample
of size 10, record the length of each fish, and calculate the mean of the sample lengths. What
do you think your sample mean would most likely be? Explain your answer using what you
know about random samples and sample means.
______________________________
Copyright © 2011 Apex Learning Inc. (See Terms of Use
5 ofat70www.apexvs.com/TermsOfUse)
AP Statistics
Unit Overview: Binomials and Distributions
Think About It
Page 4 of 4
2. For the scenario in topic 1, imagine you draw hundreds of samples of size 10 (replacing each
fish after you record its length) and calculate the mean of each sample. If you kept doing this
until you took every possible sample, you'd get a distribution of sample means called a
sampling distribution. What would the shape of the sampling distribution be? Would the
standard deviation of the sampling distribution be larger or smaller than the population
standard deviation? Explain your answer using what you know about random samples and
sample standard deviations.
3. Why might a sampling distribution be helpful if you wanted to estimate the likelihood of
getting a particular sample value, say x-bar = 48 centimeters? What's your guess about the
likelihood of getting a sample mean of 48 centimeters? Explain your answer using what you
know about probability distributions.
4. Building on items 1-3, how might a sampling distribution be helpful if you wanted to estimate
a population mean with a sample mean? Outline a scenario for using a sampling distribution
for inference.
5. Write a paragraph-long story about a sampling distribution; describe the population, the
sampling method, the mean and standard deviation of the population and the sampling
distribution.
6. There's a case where a sampling distribution qualifies as a binomial setting. Describe this case,
using the criteria for a binomial distribution.
______________________________
Copyright © 2011 Apex Learning Inc. (See Terms of Use
6 ofat70www.apexvs.com/TermsOfUse)
1.1
Key Terms
inferential statistics
7 of 70
S2
6WDWLVWLFV6WXG\*XLGH
,QWURGXFWLRQWR,QIHUHQWLDO6WDWLVWLFV
3DJHRI
1.1.1
'LUHFWLRQV
• 8VHWKLVJXLGHWRWDNHQRWHVZKLOH\RXZDWFKWKH7XWRULDO
• .HHS\RXUFRPSOHWHG6WXG\*XLGHLQ\RXUQRWHERRNIRUODWHUVWXG\
.H\7HUPVDQG&RQFHSWV
• QRQH
7XWRULDO6HFWLRQV
3UREDELOLW\DQG&RQILGHQFH,QWHUYDOV
3UREDELOLW\DQG7HVWVRI6LJQLILFDQFH
6WHSVWR,QIHUHQFH
BBBBBBBBBBBBBBBBBBB
‹&RS\ULJKW$3(;2QOLQH/HDUQLQJ,QF$OOULJKWVUHVHUYHG$GYDQFHG3ODFHPHQWDQG$3DUHUHJLVWHUHG
WUDGHPDUNVRIWKH&ROOHJH%RDUG7KLVPDWHULDOLVLQWHQGHGIRUWKHH[FOXVLYHXVHRI$3(;HQUROOHGVWXGHQWV$Q\
XQDXWKRUL]HGFRS\LQJUHXVHRUUHGLVWULEXWLRQLVSURKLELWHG
8 of 70
S2
6WDWLVWLFV6WXG\*XLGH
,QWURGXFWLRQWR,QIHUHQWLDO6WDWLVWLFV
3DJHRI
&RPPRQ$SSOLFDWLRQV
BBBBBBBBBBBBBBBBBBB
‹&RS\ULJKW$3(;2QOLQH/HDUQLQJ,QF$OOULJKWVUHVHUYHG$GYDQFHG3ODFHPHQWDQG$3DUHUHJLVWHUHG
WUDGHPDUNVRIWKH&ROOHJH%RDUG7KLVPDWHULDOLVLQWHQGHGIRUWKHH[FOXVLYHXVHRI$3(;HQUROOHGVWXGHQWV
$Q\XQDXWKRUL]HGFRS\LQJUHXVHRUUHGLVWULEXWLRQLVSURKLELWHG
9 of 70
1.2
Key Terms
almost binomial
binomcdf on a graphing calculator
almost binomial
binomcdf on a graphing calculator
binomial experiment (setting)
binomial probabilities for a range of outcomes
binomial probabilities for an exact number of outcomes
binompdf on a graphing calculator
continuity correction
geometric setting
normal approximation to the binomial
10 of 70
S2
AP Statistics
Study Guide: Binomial Situations (Events)
3DJHRI4
1.2.1
'LUHFWLRQV
• 8VHWKLVJXLGHWRWDNHQRWHVZKLOH\RXZDWFKWKH7XWRULDO
• .HHS\RXUFRPSOHWHG6WXG\*XLGHLQ\RXUQRWHERRNIRUODWHUVWXG\
.H\7HUPVDQG&RQFHSWV
•
•
•
•
DOPRVWELQRPLDO
ELQRPFGI
ELQRPLDOH[SHULPHQW
ELQRPLDOSUREDELOLWLHVIRUDUDQJHRI
RXWFRPHV
•
•
•
•
ELQRPLDOSUREDELOLWLHVIRUDQH[DFW
QXPEHURIRXWFRPHV
ELQRPLDOVHWWLQJ
ELQRPSGI
LQIHUHQWLDOVWDWLVWLFV
7XWRULDO6HFWLRQV
7KH'HILQLWLRQRID%LQRPLDO6HWWLQJRU(YHQW
&DOFXODWLQJ%LQRPLDO3UREDELOLWLHVIRUDQ([DFW1XPEHURI2XWFRPHV
&DOFXODWLQJ%LQRPLDO3UREDELOLWLHVIRUD5DQJHRI2XWFRPHV
______________________________
Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse)
11 of 70
S2
AP Statistics
Study Guide: Binomial Situations (Events)
$OPRVW%LQRPLDO
______________________________
Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse)
12 of 70
Page 2 of 4
AP Statistics
Study Guide: Binomial Situations (Events)
Page 3 of 4
8VLQJWKH7,/TI-84WR&DOFXODWH%LQRPLDO'LVWULEXWLRQ3UREDELOLWLHV
7R&DOFXODWHD%LQRPLDO3UREDELOLW\IRUD6SHFLILF9DOXHELQRPSGI
7KHSGIVWDQGVIRUSUREDELOLW\GHQVLW\IXQFWLRQ8VHELQRPSGIDQGHQWHUWKHQXPEHUV
DQGFRPPDVLQWKHVDPHRUGHUDVWKHSUREDELOLW\QRWDWLRQ)RUH[DPSOHWRFDOFXODWH
%
•
•
•
•
SUHVVQG >',675@
VFUROOGRZQWR>ELQRPSGI@DQGSUHVV(17(5
.H\LQ<RXUKRPHVFUHHQVKRXOGUHDGELQRPSGI
SUHVV(17(5DQG\RXVKRXOGJHW
7R&DOFXODWHD3UREDELOLW\IRUD5DQJHRI9DOXHVELQRPFGI
7KHFGIVWDQGVIRUFXPXODWLYHGHQVLW\IXQFWLRQ(QWHUWKHQXPEHUVDQGFRPPDVLQWKH
VDPHRUGHUDVWKHSUREDELOLW\QRWDWLRQ)RUH[DPSOHWRFDOFXODWH%
•
•
•
SUHVVQG >',675@
VFUROOGRZQWR>ELQRPFGI@DQGSUHVV(17(5
$VLQELQRPSGIHQWHUWKHQXPEHURIWULDOVWKHSUREDELOLW\IRUHDFKWULDODQGWKH[
YDOXH
)RUH[DPSOHIRU%3;”LVELQRPFGI
______________________________
Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse)
13 of 70
AP Statistics
Study Guide: Binomial Situations (Events)
Page 4 of 4
Using the TI-89 to Calculate Binomial Distribution Probabilities
To Calculate a Binomial Probability for a Specific Value: Binomial Pdf f
Pdf stands for “probability density function.” Use binomPdf and enter the numbers and
commas in the same order as they are given in the probability notation. For example, to
calculate B(20,.3,5):
1. From Stats/List Editor, press [F5]:Distr. Scroll down to [B]: Binomial Pdf.... :Press
[ENTER].
2. For Num Trials, n, enter 20. Arrow down one cell. For Prob Success, p, enter .3.
Arrow down again. For X Value, enter 5. Press [ENTER] to save your settings, then
[ENTER] again to execute the command. You should get .178863 for your answer.
3. You can also access many of the statistics functions by pressing [CATALOG][F3]. This
will produce a list of common functions available on the calculator.
To Calculate a Binomial Probability for a Range of Values: Binomial Cdf
Cdf stands for “cumulative density function.” This function allows you to calculate the
probability of a range of possible outcomes from a binomial setting. You will need to identify
the lower and upper bounds for that range. To calculate the probability of obtaining five or
fewer successes out of 20 trials, with a probability of success of 0.3:
1. From Stats/List Editor, press [F5]:Distr. Scroll down to [C]: Binomial Cdf.... :Press
[ENTER].
2. For Num Trials, n, enter 20. Arrow down one cell. For Prob Success, p, enter .3.
Arrow down again. For the Lower Bound, enter 0. Arrow down. For the Upper Bound,
enter 5. Press [ENTER] to save your settings, then [ENTER] again to execute the
command.
3. You should get .416371 for your answer.
______________________________
Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse)
14 of 70
Page 1 of 3
1.2.3
Why Study Binomial Distributions?
Remember that, although statistical inference takes many forms, it often follows these
phases:
1. Identify what you want to study.
2. Ensure that your study and your sample are valid and useful for drawing conclusions
about what you're studying. This includes understanding how the data were collected
(so you know their limitations).
3. Recognize the type of probability distribution that models your situation. The
probability distribution tells you how likely or unlikely your findings are. You can also
use it to find the range of values your population parameter is likely to have.
4. If you're comparing at least two measures or studying a relationship, conduct
significance tests to compare your results against what you'd expect from chance
alone. If the study's results are highly unlikely, your results are statistically
significant.
This Activity focuses on phase 3, the probability distribution, with particular emphasis on
the binomial distribution. You'll need to know about a few different types of probability
distributions to do inferential statistics.
Review
Distribution is one of the most important concepts in statistics. Every variable has a
distribution, whether the variable is categorical or numerical, continuous or discrete. A
variable's distribution consists of the values it takes and how often it takes each of these
values. Distributions are often shown in graphs so you can see the relative frequency
with which values occur.
A binomial distribution (also referred to sometimes as the binomial setting) is a discrete
probability distribution of the counts of successes and failures, such as the number of
times you get heads when you flip a coin 100 times. Most textbooks list the following
four characteristics for the binomial distribution (or setting):
In the Binomial Setting,
1. There's a set n of identical trials.
2. The outcome of each trial is either success or failure.
3. The probability of success of each trial is p. (The probability of failure of each trial is
q = 1 - p.)
4. Each trial is independent.
15 of 70
S2
Page 2 of 3
Some textbooks list the following as a fifth characteristic, and others use it as an
introduction to the previous four:
5. In the binomial setting, we're interested in x, the number of successes observed
during the n trials, for x = 0, 1, 2, 3 , … , n.
The following are examples of binomial settings:
• What's the probability of getting exactly three heads on five flips of a coin?
• A coin is flipped 100 times. What's the probability of getting 45 heads?
• If the probability of having Rh-negative blood is .4 in a population of 500, what's the
probability 210 have type Rh-negative blood?
These are all probability settings in which:
1. We know the number of repetitions.
2. The outcome of each trial is either success or failure.
3. We know the probability of success or failure of any trial.
4. The probability doesn't change from trial to trial (the trials are independent.)
Other Definitions to Remember
Binomial Event: a set of trials within a binomial setting
Simple Event or Outcome: a single trial
B(n, p): the binomial probability distribution with parameters n, the number of
observations, and p, the probability of success on any one observation. If X is a binomial
 n
random variable with B(n, p), B(n, p, x) = P(X = x) =   (p)x(1 – p)n–x.
x
On your calculator, binompdf(n,p,X) calculates the probability of exactly x successes in n
trials with a probability of p.
With a range of outcomes for X, where X takes on the number of successes 0, 1, 2, 3, …,
n, use binomcdf(n,p,X) on your calculator, which calculates the probability of up to and
including x successes in n trials with a probability of p for the binomial probability
distribution.
Remember, the binomial distribution is discrete: Each single value has a probability. This
means that, unlike a continuous distribution such as the normal distribution, there's a
difference between P(x < 4) and P(x ≤ 4). So for P(x < 4), you add
P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3). But for P(x ≤ 4) you include the 4, so you
add P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4).
Be careful about your endpoints when using binomcdf. With binomcdf on a calculator,
you get the lower-tail probability that includes the upper bound. So, if you enter
binomcdf(20,.3,5), you're finding P(x ≤ 5) for B(20, .3). If you need a range with a
specific lower and upper bound, remember to subtract like this:
(probability below upper bound) – (probability below lower bound).
16 of 70
Page 3 of 3
Be careful about the range you calculate. It may help to draw the range on a number
line. For example, if you want to calculate P(4 < x < 8) for B(10, .3), you want the
range represented by the underlined numbers:
0 1 2 3 4 5 6 7 8 9 10.
So you'd calculate:
binomcdf(10,.3,7) – binomcdf(10,.3,4). Note that this will give you the range from 5 to
7.
If you wanted to calculate P(4 ≤ x < 8), you include the 4 in your range:
0 1 2 3 4 5 6 7 8 9 10.
So you'd calculate:
binomcdf(10,.3,7) – binomcdf(10,.3,3). Note that now you're subtracting the 3 (since
it's included in binomcdf(10,.3,3)) but not the 4.
If you want P(4 ≤ x ≤ 8), you'd include the 8:
0 1 2 3 4 5 6 7 8 9 10.
You'd calculate:
binomcdf(10,.3,8) – binomcdf(10,.3,3).
There are cases where the criteria of the binomial setting might not be satisfied, but
which would still be considered binomial. An event is almost binomial if the population
size of an experiment is at least 20 times larger than the sample size. (NOTE: 20 is a
rule of thumb. Some references use 10 instead of 20.)
17 of 70
Page 1 of 3
1.2.3
Why Study Binomial Distributions?
Remember that, although statistical inference takes many forms, it often follows these
phases:
1. Identify what you want to study.
2. Ensure that your study and your sample are valid and useful for drawing conclusions
about what you're studying. This includes understanding how the data were collected
(so you know their limitations).
3. Recognize the type of probability distribution that models your situation. The
probability distribution tells you how likely or unlikely your findings are. You can also
use it to find the range of values your population parameter is likely to have.
4. If you're comparing at least two measures or studying a relationship, conduct
significance tests to compare your results against what you'd expect from chance
alone. If the study's results are highly unlikely, your results are statistically
significant.
This Activity focuses on phase 3, the probability distribution, with particular emphasis on
the binomial distribution. You'll need to know about a few different types of probability
distributions to do inferential statistics.
Review
Distribution is one of the most important concepts in statistics. Every variable has a
distribution, whether the variable is categorical or numerical, continuous or discrete. A
variable's distribution consists of the values it takes and how often it takes each of these
values. Distributions are often shown in graphs so you can see the relative frequency
with which values occur.
A binomial distribution (also referred to sometimes as the binomial setting) is a discrete
probability distribution of the counts of successes and failures, such as the number of
times you get heads when you flip a coin 100 times. Most textbooks list the following
four characteristics for the binomial distribution (or setting):
In the Binomial Setting,
1. There's a set n of identical trials.
2. The outcome of each trial is either success or failure.
3. The probability of success of each trial is p. (The probability of failure of each trial is
q = 1 - p.)
4. Each trial is independent.
Some textbooks list the following as a fifth characteristic, and others use it as an
introduction to the previous four:
5. In the binomial setting, we're interested in x, the number of successes observed
during the n trials, for x = 0, 1, 2, 3 , … , n.
18 of 70
S1
Page 2 of 3
The following are examples of binomial settings:
• What's the probability of getting exactly three heads on five flips of a coin?
• A coin is flipped 100 times. What's the probability of getting 45 heads?
• If the probability of having Rh-negative blood is .4 in a population of 500, what's the
probability 210 have type Rh-negative blood?
These are all probability settings in which:
1. We know the number of repetitions.
2. The outcome of each trial is either success or failure.
3. We know the probability of success or failure of any trial.
4. The probability doesn't change from trial to trial (the trials are independent.)
Other Definitions to Remember
Binomial Event: a set of trials within a binomial setting
Simple Event or Outcome: a single trial
B(n, p): the binomial probability distribution with parameters n, the number of
observations, and p, the probability of success on any one observation. If X is a binomial
 n
random variable with B(n, p), B(n, p, x) = P(X = x) =  (p)x(1 – p)n–x.
x
On your calculator, binompdf(n,p,X) calculates the probability of exactly x successes in n
trials with a probability of p.
With a range of outcomes for X, use Binomial Cdf on your calculator, which allows you to
select lower and upper values for X, and finds the probability of obtaining a count of successes
that falls within those values.
Remember, the binomial distribution is discrete: Each single value has a probability. This
means that, unlike a continuous distribution such as the normal distribution, there's a
difference between P(x < 4) and P(x ≤ 4). So for P(x < 4), you add
P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3). But for P(x ≤ 4) you include the 4, so you
add P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4).
Be careful about your endpoints when using Binomial Cdf. If you are finding P(X<5), then your
lower value will be 0 and your upper value will be 5. However, if you are finding P(X<5), then
your values will be 0 and 4.
If you need a range with a specific lower and upper value, then be sure to determine whether or
not to include those specific values. Generally, words such as ‘at least’, ‘at most’, ‘no greater/
more/fewer/less than’, ‘including’, and ‘inclusive’ indicate the inclusion of an upper or lower value,
while words such as ‘less than’, ‘more than’, ‘between’, ‘over/under’, and ‘between’ indicate the
omission of that value and the need for the next higher or lower value to be used.
19 of 70
Page 3 of 3
Be careful about the range you calculate. It may help to draw the range on a number
line. For example, if you want to calculate P(4 < x < 8) for B(10, .3), you want the
range represented by the underlined numbers:
0 1 2 3 4 5 6 7 8 9 10.
So you'd calculate:
binomcdf(10,.3,7) – binomcdf(10,.3,4). Note that this will give you the range from 5 to
7.
If you wanted to calculate P(4 ≤ x < 8), you include the 4 in your range:
0 1 2 3 4 5 6 7 8 9 10.
So you'd calculate:
binomcdf(10,.3,7) – binomcdf(10,.3,3). Note that now you're subtracting the 3 (since
it's included in binomcdf(10,.3,3)) but not the 4.
If you want P(4 ≤ x ≤ 8), you'd include the 8:
0 1 2 3 4 5 6 7 8 9 10.
You'd calculate:
binomcdf(10,.3,8) – binomcdf(10,.3,3).
There are cases where the criteria of the binomial setting might not be satisfied, but
which would still be considered binomial. An event is almost binomial if the population
size of an experiment is at least 20 times larger than the sample size. (NOTE: 20 is a
rule of thumb. Some references use 10 instead of 20.)
20 of 70
'LUHFWLRQV
1.2.4
• 8VHWKLVJXLGHWRWDNHQRWHVZKLOH\RXZDWFKWKHtutorial.
• .HHS\RXUFRPSOHWHGstudy guide in your notebook for later study.
.H\7HUPVDQG&RQFHSWV
• FRQWLQXLW\FRUUHFWLRQ
• QRUPDODSSUR[LPDWLRQWRWKHELQRPLDO
7XWRULDO6HFWLRQV
7KH5HODWLRQVKLS%HWZHHQWKH%LQRPLDO'LVWULEXWLRQDQGWKH1RUPDO
'LVWULEXWLRQ
8VLQJWKH1RUPDO$SSUR[LPDWLRQWR)LQG3UREDELOLWLHV
,QFUHDVH$FFXUDF\E\8VLQJWKH&RQWLQXLW\&RUUHFWLRQ
______________________________
Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse)
21 of 70
S2
Using the TI-83 or TI-84 to Calculate Binomial Distribution Probabilities
7R&DOFXODWHD%LQRPLDO3UREDELOLW\IRUD6SHFLILF9DOXHELQRPSGI
7KHSGIVWDQGVIRUSUREDELOLW\GHQVLW\IXQFWLRQ8VHELQRPSGIDQGHQWHUWKH
QXPEHUVDQGFRPPDVLQWKHVDPHRUGHUDVWKHSUREDELOLW\QRWDWLRQ)RUH[DPSOHWR
FDOFXODWH%
•
•
•
•
SUHVVQG >',675@
VFUROOGRZQWR>ELQRPSGI@DQGSUHVV(17(5
.H\LQ<RXUKRPHVFUHHQVKRXOGUHDGELQRPSGI
SUHVV(17(5DQG\RXVKRXOGJHW
7R&DOFXODWHD3UREDELOLW\IRUD5DQJHRI9DOXHVELQRPFGI
7KHFGIVWDQGVIRUFXPXODWLYHGHQVLW\IXQFWLRQ(QWHUWKHQXPEHUVDQGFRPPDV
LQWKHVDPHRUGHUDVWKHSUREDELOLW\QRWDWLRQ)RUH[DPSOHWRFDOFXODWH%
•
•
•
SUHVVQG >',675@
VFUROOGRZQWR>ELQRPFGI@DQGSUHVV(17(5
$VLQELQRPSGIHQWHUWKHQXPEHURIWULDOVWKHSUREDELOLW\IRUHDFKWULDODQGWKH
[YDOXH
)RUH[DPSOHIRU%3;”LVELQRPFGI
______________________________
Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse)
22 of 70
Using the TI-83 or TI-84 to Calculate Normal Distribution Probabilities
)LQGLQJ$1RUPDO&XUYH3UREDELOLW\ZLWKQRUPDOFGI
3UHVVQG >',675@7KLVVHOHFWVQRUPDOFGI
(QWHU>ORZHUERXQGXSSHUERXQGµDQGσ@)RUWKHELUGHJJH[DPSOHWKHORZHU
ERXQGLV]HURWKHXSSHUERXQGLVWKHPHDQLVDQGWKHVWDQGDUGGHYLDWLRQ
LV6RSUHVV
3UHVV(17(5WRVHHWKHDQVZHU
______________________________
Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse)
23 of 70
AP Statistics
Study Guide: The Normal Approximation to the Binomial
1.2.4
Directions


Use this guide to take notes while you watch the tutorial.
Keep your completed study guide in your notebook for later study.
Key Terms and Concepts


Page 1 of 3
continuity correction
normal approximation to the binomial
Tutorial Sections
The Relationship Between the Binomial Distribution and the Normal
Distribution
Using the Normal Approximation to Find Probabilities
Increase Accuracy by Using the Continuity Correction
______________________________
Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse)
24 of 70
S2
AP Sta
atistics
Study Guide: The Normal Approx
ximation
n to the B
Binomial
Page 2 of 3
Calculattor Instruc
ctions for the TI-89
These instructions assume
a
thatt you've read
d and follow
wed the dire
ections in the "Getting
ator manuall. The steps covered in this guide a
are also cov
vered
Started" chapter of your calcula
in the "S
Statistics" ch
hapter of your calculato
or manual.


Text in brack
kets, like thiis [HOME], gives
g
the na
ame of a bu
utton on the calculator.
Two sets of brackets
b
like
e [2nd][VAR
R-LINK] get you to a co
olor-coded ffunction abo
ove
each key.

The diamond
d button
and brack
kets get you
u to anotherr color-coded
d function
sh
hown above
e each key.
The white [A
ALPHA] butto
on followed by another bracket getts you a lettter.
Text in bold, like 3:Regr
ressions->
> is a comm and within a menu on tthe calculattor
sc
creen.


Using th
he TI-89 to
o Calculate
e Binomial Distributio
on Probabillities
To Calcu
ulate a Binomial Prob
bability for
r a Specific
c Value:
mial Pdf to calculate th
Pdf stand
ds for “prob
bability dens
sity function.” Use Binom
he probabilitty of
observing a specific number of successes in an exercisse that invo
olves a fixed
d number of
trials. Fo
or example, to calculate
e Binomial Pdf(20,.3,5),
P
, which is th
he probabilitty of seeing
g
exactly five
f
successes in 20 tria
als when the
e probability
y of successs is 0.3:
1.
2.
3.
4.
5.
6.
7.
Enter List Editor.
D
/ [B]: Binomial
B
Pdf....
Press [F5]: Distr
Enter 20 for Num Trials, n:.
0
for Prob Success, p:.
Enter .3 (or 0.3)
Enter 5 for X Value:
R] to store your
y
settings, then [EN TER] again to execute the command.
Press [ENTER
g .178863.
You should get
ulate a Binomial Prob
bability for
r a Range o
of Values:
To Calcu
Cdf stands for “cumulative dens
sity function
n.” Use Bino
omial CDF to
o calculate tthe probabillity
of observ
ving success
ses that ran
nge from a lower bound
d to an uppe
er bound. . F
For example
e, to
, which is th
calculate
e Binomial Cdf(20,.3,5)
C
he probabilitty of seeing
g between ze
ero and five
e
successe
es in 20 trials when the probability of success is 0.3:
1.
2.
3.
4.
5.
6.
Enter List Editor.
B
Cdff....
Press[F5]: Distr / [C]: Binomial
Enter 20 for Num Trials, n:.
0
for Prob Success, p:.
Enter .3 (or 0.3)
Enter 0 for Lower Value::.
U
Value::.
Enter 5 for Upper
_________
____________
___________
Copyright © 2011 Apex Learning
L
Inc. (See
(
Terms of Use at www.ap
pexvs.com/TerrmsOfUse)
25 of 70
AP Statistics
Study Guide: The Normal Approximation to the Binomial
Page 3 of 3
7. Press [ENTER] to store your settings, then [ENTER] again to execute the command.
8. You should get .416371.
If you want to calculate the probability of seeing between 5 and 10 successes under these
same conditions, simply change your Lower Value: to 5 and your Upper Value: to 10. Your
answer should be .745347
Using the TI-89 to Calculate Normal Distribution Probabilities
Finding a Normal Curve Probability
1. From List Editor, press [F5]: Distr / [4]:Normal Cdf....
2. You will be prompted for a Lower Bound, an Upper Bound, the mean ( μ ) and the
standard deviation ( σ ).
For the bird example, enter a Lower Bound of 0.
Enter an Upper Bound of 11.
Enter a mean of 12.
Enter a standard deviation of 0.8.
Press [ENTER] to save your settings, then press [ENTER] again to execute the
command.
8. Your answer should be .10565.
3.
4.
5.
6.
7.
______________________________
Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse)
26 of 70
6WDWLVWLFV
6WXG\6KHHW
%LQRPLDO3UREOHPV
3DJHRI
1.2.6 S2
7KH1RUPDO$SSUR[LPDWLRQWRWKH%LQRPLDO
,QWKLVDFWLYLW\ZH
OOORRNDWWZRNLQGVRIGLVWULEXWLRQQRUPDODQGELQRPLDO:KHQWKH
QXPEHURIVXFFHVVHVDQGIDLOXUHVIRUDJLYHQVLWXDWLRQDUHHDFKHTXDOWRRUJUHDWHUWKDQ
WKHELQRPLDOLVVDLGWRDSSUR[LPDWHWKHQRUPDOGLVWULEXWLRQ0DQ\WH[WERRNVXVH
UDWKHUWKDQEXWERWKDUHYDOLG,QRWKHUZRUGVWKHELQRPLDODSSUR[LPDWHVWKH
QRUPDOZKHQQS≥DQGQ±S≥1RWHWKDW±SRUWKHSUREDELOLW\RIIDLOXUHLV
RIWHQZULWWHQDVTVRQ±SFDQEHZULWWHQDVQT
,QWKHVHFDVHVWKHELQRPLDOUHVHPEOHVWKHQRUPDOVRFORVHO\WKDW\RXFDQXVH]VFRUHV
DQGDQRUPDOFXUYHWDEOHWRILQGDUHDVXQGHUWKHFXUYH$VPRVWVLWXDWLRQVLQLQIHUHQWLDO
VWDWLVWLFVPHHWWKHUHTXLUHPHQWVIRUWKHELQRPLDODSSUR[LPDWLRQUHVHDUFKHUVUDUHO\QHHG
WRILJXUHELQRPLDOSUREDELOLWLHVH[DFWO\WKH\XVXDOO\XVHWKHQRUPDODSSUR[LPDWLRQ
LQVWHDG
$ELQRPLDOGLVWULEXWLRQWKDWPHHWVWKHUHTXLUHPHQWVIRUWKHQRUPDODSSUR[LPDWLRQZLOO
KDYHDPHDQHTXDOWRQSWKHH[SHFWHGYDOXHDQGDVWDQGDUGGHYLDWLRQHTXDOWR
QS − S WKHVTXDUHURRWRIWKHH[SHFWHGYDOXHWLPHVWKHQXPEHURIIDLOXUHV
7KH&RQWLQXLW\&RUUHFWLRQ
7KHELQRPLDOGLVWULEXWLRQPRGHOVWKHGLVWULEXWLRQVRIGLVFUHWHUDQGRPYDULDEOHV6R
ZKHQ\RXIOLSDFRLQWLPHV\RXFDQKDYHKHDGVRUKHDGVEXWQRWKHDGV
7KHQRUPDOGLVWULEXWLRQRQWKHRWKHUKDQGPRGHOVWKHGLVWULEXWLRQVRIFRQWLQXRXV
UDQGRPYDULDEOHV5HFDOOWKDWDFRQWLQXRXVUDQGRPYDULDEOHFDQWDNHRQDQ\YDOXHZLWKLQ
DUDQJH)RUH[DPSOHWHPSHUDWXUHFDQEHGHJUHHVFHQWLJUDGHGHJUHHV
FHQWLJUDGHGHJUHHVFHQWLJUDGHRUDQ\WKLQJLQEHWZHHQ&RQWLQXRXV
YDULDEOHVKROGPHDVXUHGGDWDDQGGLVFUHWHYDULDEOHVKROGFRXQWHGGDWD
:KHQ\RXXVHWKHQRUPDODSSUR[LPDWLRQWRWKHELQRPLDO\RX
UHWUHDWLQJDGLVFUHWH
GLVWULEXWLRQDVWKRXJKLWZHUHFRQWLQXRXVDQGWKLVLQWURGXFHVHUURU,QRUGHUWRPLQLPL]H
HUURUZKHQXVLQJWKHDSSUR[LPDWLRQWRWKHELQRPLDOXVHWKHFRQWLQXLW\FRUUHFWLRQ7KH
FRQWLQXLW\FRUUHFWLRQDOORZV\RXWRWUHDWQXPEHUVDVWKRXJKWKH\RFFXSLHGDUDQJH
DERYHDQGEHORZWKHGLVFUHWHYDOXHLQFOXVLYH
+HUH
VWKHPRVWLPSRUWDQWWKLQJWRUHPHPEHUZKHQ\RX
UHXVLQJWKHFRQWLQXLW\
FRUUHFWLRQ:KHQ\RX
UHILQGLQJWKHSUREDELOLW\IRUDUDQJHRIYDOXHVLQDELQRPLDO
GLVWULEXWLRQDQG\RX
UHXVLQJWKHQRUPDODSSUR[LPDWLRQVXEWUDFWIURPWKHORZHU
ERXQGDQGDGGWRWKHXSSHUERXQG)RULQVWDQFHIRUWKHFORVHGLQWHUYDO
ZKHUHDQGDUHERWKLQFOXGHG\RXILQGWKHQRUPDOFXUYHSUREDELOLW\IRUWKHLQWHUYDO
,I\RXZDQWWKHSUREDELOLW\IRUWKHELQRPLDOLQWHUYDOZKHUHLV
LQFOXGHGDQGLVQ
W\RX
GILQGWKHSUREDELOLW\IRUWKHLQWHUYDOLQDQRUPDO
GLVWULEXWLRQ,I\RXZDQWWKHSUREDELOLW\IRUWKHLQWHUYDOZKHUHLVQ
WLQFOXGHG
DQGLV\RX
GILQGWKHQRUPDOSUREDELOLW\IRUWKHLQWHUYDO)LQDOO\LI\RX
ZDQWHGWKHSUREDELOLW\IRUZKHUHQHLWKHUQRUDUHLQFOXGHG\RX
GILQGWKH
QRUPDOSUREDELOLW\IRU
BBBBBBBBBBBBBBBBBB
‹&RS\ULJKW$3(;2QOLQH/HDUQLQJ,QF$OOULJKWVUHVHUYHG$GYDQFHG3ODFHPHQWDQG$3DUHUHJLVWHUHG
WUDGHPDUNVRIWKH&ROOHJH%RDUG7KLVPDWHULDOLVLQWHQGHGIRUWKHH[FOXVLYHXVHRI$3(;HQUROOHGVWXGHQWV
$Q\XQDXWKRUL]HGFRS\LQJUHXVHRUUHGLVWULEXWLRQLVSURKLELWHG
27 of 70
1.2.7
28 of 70
S2
29 of 70
30 of 70
31 of 70
32 of 70
33 of 70
34 of 70
1.3
Key Terms
average waiting time
geometcdf on a graphing calculator
geometpdf on a graphing calculator
geometric distribution
35 of 70
S2
AP Statistics
Study: Geometic Probability Distributions
1.3.2
'LUHFWLRQV
• 8VHWKLVJXLGHWRWDNHQRWHVZKLOH\RXZDWFKWKH7XWRULDO
• .HHS\RXUFRPSOHWHG6WXG\*XLGHLQ\RXUQRWHERRNIRUODWHUVWXG\
.H\7HUPVDQG&RQFHSWV
• DYHUDJHZDLWLQJWLPH
• JHRPHWFGI
• JHRPHWSGI
Page 1 of 3
• JHRPHWULFGLVWULEXWLRQ
• JHRPHWULFVHWWLQJ
7XWRULDO6HFWLRQV
,QWURGXFWLRQWR*HRPHWULF3UREDELOLW\'LVWULEXWLRQV
3UREDELOLW\&DOFXODWLRQVLQWKH*HRPHWULF6HWWLQJ
$YHUDJH:DLWLQJ7LPH
______________________________
Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse)
36 of 70
S2
AP Statistics
Study: Geometic Probability Distributions
Page 2 of 3
TI-83 and TI-84: Finding a 6LQJOH Probability Value for a Geometric Distribution
7RGRWKLVXVHJHRPHWSGIZKLFKVWDQGVIRUJHRPHWULFSUREDELOLW\GHQVLW\IXQFWLRQ
•
•
3UHVVQG >',675@$/3+$>'@7KLVZLOOSXWWKHFXUVRURQJHRPHWSGI,WZLOODOVRZRUNWR
VFUROOGRZQWKHOLVWPDQXDOO\SXWWKHFXUVRURQJHRPHWSGIWKHQSUHVV(17(5
(QWHU>SUREDELOLW\RIVXFFHVVRQHDFKWULDOS@>@>QXPEHURIWULDOVQ@WKHQFORVHWKH
SDUHQWKHVLV
)RUH[DPSOHLI\RXZDQWWRILQGWKHSUREDELOLW\RIJHWWLQJVXFFHVVLQIRXUWULDOVLIWKH
SUREDELOLW\RIVXFFHVVRQHDFKWULDOLV
(QWHU ÷ 7KLVWHOOVWKHFDOFXODWRUWKDWLQHDFKWULDOWKHUH
VDSUREDELOLW\RI
VXFFHVVRIDQGWKDWWKHUHDUHWULDOV7KHQSUHVV(17(5DQG\RXVKRXOGJHW
VRPHWKLQJFRQVLVWHQWZLWKQXPEHURIGLJLWVDQGURXQGLQJZLOOYDU\GHSHQGLQJRQ
KRZ\RXUFDOFXODWRULVVHW
TI-83 and TI-84: Finding a 5DQJH of Probability Values for a Geometric Distribution
7RGRWKLVXVHJHRPHWFGIZKLFKVWDQGVIRUJHRPHWULFSUREDELOLW\GHQVLW\IXQFWLRQ,W
ILJXUHVWKHSUREDELOLW\IURPWULDOVXSWRDQGLQFOXGLQJQWULDOV
•
•
3UHVVQG >',675@$/3+$>(@7KLVZLOOVHOHFWJHRPHWFGI,WZLOODOVRZRUNWRVFUROOGRZQ
WKHOLVWPDQXDOO\SXWWKHFXUVRURQJHRPHWFGIWKHQSUHVV(17(5
(QWHU>SUREDELOLW\RIVXFFHVVRQHDFKWULDOS@>@>WKHXSSHUERXQGIRUQXPEHURIWULDOV
Q@WKHQFORVHWKHSDUHQWKHVLV
)RUH[DPSOHLI\RXZDQWWRILQGWKHSUREDELOLW\RIVXFFHVVLQWKUHHRUIHZHUWULDOVLIWKH
SUREDELOLW\RIVXFFHVVRQHDFKWULDOLV
(QWHU ÷ 7KLVWHOOVWKHFDOFXODWRUWKDWLQHDFKWULDOWKHUH
VDSUREDELOLW\RI
DQGWKDWWKHUHDUHWULDOV7KHQSUHVV(17(5DQG\RXVKRXOGJHWQXPEHURIGLJLWV
DQGURXQGLQJZLOOYDU\GHSHQGLQJRQKRZ\RXUFDOFXODWRULVVHW
TI-83 and TI-84: Generating Random Numbers Using randInt
•
•
•
•
3UHVV0$7+
6FUROODFURVVWR35%
6FUROOGRZQWRUDQG,QWDQGSUHVV(17(5
(QWHUWKH>PD[LPXPQXPEHU@DQG>PLPLPXPQXPEHU@DQGFORVHWKHSDUHQWKHVHV
7KHQSUHVV(17(5HDFKWLPH\RXZDQWDUDQGRPQXPEHU
)RUH[DPSOHLI\RXZDQWWRJHQHUDWHUDQGRPQXPEHUVEHWZHHQDQGLQFOXVLYHHQWHU
DQGSUHVV(17(5HDFKWLPH\RXZDQWDQXPEHU
______________________________
Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse)
37 of 70
AP Statistics
Study: Geometric Probability Distributions
Page 3 of 3
TI-89: Finding a Single Probability Value for a Geometric Distribution
To do this, use Geometric Pdf….
1. Enter List Editor.
2. Press [F5]:Distr / [F]: Geometric Pdf….
3. You will be prompted for the probability of success (Prob Success, p:) and the trial
on which you want the first success to occur (X Value:).
4. For example, if you were rolling a die and wanted to roll a 6, P(roll a 6) = 1/6, and
you wanted to find the probability of getting your first 6 on your fourth roll, you
would enter 1/6 for Prob Success, p, and 4 for X Value. Press [ENTER] to save your
settings, then press [ENTER] again to execute the command.
5. Your answer should be .096451.
TI-89: Finding a Range of Probability Values for a Geometric Distribution
To do this, use Geometric Cdf.
1. Enter List Editor.
2. Press [F5]:Distr / [G]: Geometric Cdf….
3. To find the probability of rolling your first 6 within the first four rolls, enter 0 for your
Lower Value and 4 for your Upper Value. This will calculate the probability of rolling
your first 6 on the first, second, third, or fourth roll.
4. Press [ENTER] to save your settings then press [ENTER] again to execute the
command.
5. Your answer should be .517747.
If you want to find the probability of rolling your first 6 between the third and fifth rolls,
enter 4 for your Lower Value and 5 for your Upper Value. Your answer should be .292567.
TI-89: Generating Random Numbers
1. From the home screen, press [2nd][5] and scroll to 7:Probability [ENTER].
2. Select 4:rand( [ENTER].
3. This will return you to the command line, where you will enter the upper value for
your random integer. Close the parentheses and press [ENTER].
4. The screen will display a random number between 1 and the number you
entered. Press [ENTER] again to generate additional random numbers.
______________________________
Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse)
38 of 70
AP Statistics
Study: Geometric Distribution Problems
Page 1 of 2
1.3.4
S2
$V\RXPD\UHFDOOWKHELQRPLDOVHWWLQJKDVIRXUFRQGLWLRQV
7KHWULDOVDUHLQGHSHQGHQW
(DFKWULDOKDVMXVWWZRSRVVLEOHRXWFRPHVVXFFHVVDQGIDLOXUH
7KHSUREDELOLW\RIVXFFHVVLVWKHVDPHIRUHDFKWULDOUHIHUUHGWRDVS
7KHQXPEHURIWULDOVLVIL[HGXVXDOO\UHIHUUHGWRDVQ
,QWKHELQRPLDOVHWWLQJWKHUDQGRPYDULDEOH;LVWKHQXPEHURIVXFFHVVHVWKDWRFFXULQQ
WULDOV
7KHJHRPHWULFVHWWLQJLVVLPLODUWRWKHELQRPLDOVHWWLQJ7KUHHRIWKHFRQGLWLRQVDUHWKH
VDPHWKHWULDOVDUHLQGHSHQGHQWHDFKWULDOKDVMXVWWZRRXWFRPHVDQGWKHSUREDELOLW\
RIVXFFHVVLVWKHVDPHIRUHDFKWULDO,QWKHJHRPHWULFVLWXDWLRQKRZHYHUWKHUDQGRP
YDULDEOH;LVWKHQXPEHURIWULDOVUHTXLUHGWRJHWWKHILUVWVXFFHVV
$UDQGRPYDULDEOH;KDVDJHRPHWULFGLVWULEXWLRQLIHDFKWULDOLVLQGHSHQGHQWDQGKDV
SUREDELOLW\RIVXFFHVVS;LVWKHQXPEHURIWULDOVUHTXLUHGWRJHWWKHILUVWVXFFHVV;FDQ
WDNHDQ\LQWHJHUYDOXHIURPRQHWRLQILQLW\DQGLVDGLVFUHWHUDQGRPYDULDEOH
7KHIRUPXODIRUFDOFXODWLQJDJHRPHWULFSUREDELOLW\IRUDJLYHQQXPEHURIWULDOVLV
3 ; = Q = ( − S)Q − S ZKHUHQLVWKHQXPEHURIWULDOVDQGSLVWKHSUREDELOLW\RI
VXFFHVVIRUHDFKWULDO
127(0DQ\WH[WERRNVZULWHWKHSUREDELOLW\RIIDLOXUHRU±SDVT6R\RXPD\VHH
WKHIRUPXODZULWWHQDV 3 ; = Q = T Q − (S) 7KHSUREDELOLW\\RX
OOJHWDVXFFHVVRQRUEHIRUHWKHQWKWULDOLVVLPSO\WKHVXPRIDOO
SUREDELOLWLHVIURPRQHWRQ)RULQVWDQFH3[≤ 3[ 3[ 1RWLFHWKDWZH
GRQ
WLQFOXGH3[ EHFDXVH\RXFDQQHYHUJHWDVXFFHVVZLWKRXWGRLQJDQ\WULDOV
2QDgraphingFDOFXODWRUWKH*HRPHWULF3UREDELOLW\'HQVLW\)XQFWLRQJHRPHWSGI
FDOFXODWHVWKHSUREDELOLW\RIWKHILUVWVXFFHVVRQWKHQWKWULDO
7KH*HRPHWULF&XPXODWLYH'HQVLW\)XQFWLRQJHRPHWFGIFDOFXODWHVWKHSUREDELOLW\RI
JHWWLQJWKHILUVWVXFFHVVRQRUEHIRUHWKHQWKWULDO,WVXPVWKHSUREDELOLWLHVIRUHDFKRI
WKHHYHQWV3ILUVWVXFFHVVRQILUVWWULDO3ILUVWVXFFHVVRQVHFRQGWULDO3ILUVW
VXFFHVVRQWKLUGWULDO«3ILUVWVXFFHVVRQ[WKWULDO
*HRPHWULFGLVWULEXWLRQVDUHDOZD\VVNHZHGWRWKHULJKWEHFDXVHWKHUH
VDVKDUSOLPLWRQ
RQHHQG\RXFDQ
WKDYHIHZHUWKDQQ DQGDORQJWDLORIGLPLQLVKLQJSUREDELOLWLHVRQ
WKHRWKHUHQG
______________________________
Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse)
39 of 70
AP Statistics
Study: Geometric Distribution Problems
Page 2 of 2
7KHH[SHFWHGYDOXHRUDYHUDJHYDOXHLQDJHRPHWULFGLVWULEXWLRQLVFDOOHGWKHDYHUDJH
ZDLWLQJWLPH,W
VDOZD\VS6RLIWKHSUREDELOLW\RIJHWWLQJDIRXUZKHQUROOLQJDVL[
VLGHGGLHLVWKHDYHUDJHZDLWLQJWLPHZRXOGEHWULDOV,WWHOOV\RXWKDWRQDYHUDJH
\RX
OOKDYHWRUROODGLHVL[WLPHVEHIRUHJHWWLQJ\RXUILUVWIRXU7KLVPD\VHHPFRQIXVLQJ
ZKHQ\RXFRQVLGHUWKDWLQDJHRPHWULFSUREDELOLW\GLVWULEXWLRQWKHKLJKHVWIUHTXHQF\LV
DWQ n=1
2
3
4
5
6
7
8
9
10
%XWUHFDOOWKDWDQH[SHFWHGYDOXHWDNHVLQWRDFFRXQWDOOSRVVLEOHRXWFRPHV7KHPRVW
FRPPRQRXWFRPHPD\EHQ EXWWKHUHDUHSOHQW\RIRWKHUWLPHVZKHUHWKHILUVW
VXFFHVVFRPHVDWQ RUPRUH7KHVHKLJKHUYDOXHVSXOOWKHDYHUDJHXSIURPRQHVR
WKDWRQDYHUDJH\RXFDQH[SHFWWRUROOWKHGLHVL[WLPHVEHIRUHJHWWLQJ\RXUILUVW
VXFFHVV
______________________________
Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse)
40 of 70
1.4
Key Terms
central limit theorem
(distribution of)
mean of the sampling distribution of a sample mean
mean of the sampling distribution of a sample proportion
sample proportions
sampling distribution
sampling distribution of a sample mean
standard deviation of the sampling distribution of a sample mean
standard deviation of the sampling distribution of a sample proportion
41 of 70
S2
6WDWLVWLFV6WXG\*XLGH
6DPSOLQJ'LVWULEXWLRQVDQGWKH&HQWUDO/LPLW7KHRUHP
'LUHFWLRQV
3DJHRI
1.4.2
• 8VHWKLVJXLGHWRWDNHQRWHVZKLOH\RXZDWFKWKH7XWRULDO
• .HHS\RXUFRPSOHWHG6WXG\*XLGHLQ\RXUQRWHERRNIRUODWHUVWXG\
.H\7HUPVDQG&RQFHSWV
•
•
•
•
&HQWUDO/LPLW7KHRUHP
PHDQRIWKHVDPSOLQJGLVWULEXWLRQRID
VDPSOHPHDQ
VDPSOLQJGLVWULEXWLRQ
•
VDPSOLQJGLVWULEXWLRQRIDVDPSOH
PHDQ
VWDQGDUGGHYLDWLRQRIWKHVDPSOLQJ
GLVWULEXWLRQRIDVDPSOHPHDQ
7XWRULDO6HFWLRQ
7KH0HDQLQJRID6DPSOLQJ'LVWULEXWLRQ
7KH6DPSOLQJ'LVWULEXWLRQRID6DPSOH0HDQ
7KH&HQWUDO/LPLW7KHRUHP
BBBBBBBBBBBBBBBBBBB
‹&RS\ULJKW$3(;2QOLQH/HDUQLQJ,QF$OOULJKWVUHVHUYHG$GYDQFHG3ODFHPHQWDQG$3DUHUHJLVWHUHG
WUDGHPDUNVRIWKH&ROOHJH%RDUG7KLVPDWHULDOLVLQWHQGHGIRUWKHH[FOXVLYHXVHRI$3(;HQUROOHGVWXGHQWV$Q\
XQDXWKRUL]HGFRS\LQJUHXVHRUUHGLVWULEXWLRQLVSURKLELWHG
42 of 70
S2
1.4.2 S2
AP Statistics
Page 1 of 4
Study Guide: Sampling Distributions and the Central Limit Theorem
Calculator Instructions for the TI-89
These instructions assume that you've read and followed the directions in the "Getting
Started" chapter of your calculator manual. The steps covered in this guide are also covered
in the "Statistics" chapter of your calculator manual.


Text in brackets, like this [HOME], gives the name of a button on the calculator.
Two sets of brackets like [2nd][VAR-LINK] get you to a color-coded function above
each key.

The diamond button
and brackets get you to another color-coded function shown
above each key.
The white [ALPHA] button followed by another bracket gets you a letter.
Text in bold, like 3:Regressions-> is a command within a menu on the calculator
screen.


Section 1: Find the mean of a sampling distribution.
Step 1. Clear a list, generate a sample, and store it in list1.
1. Enter List Editor.
2. Clear List 1.
3. Press[2nd][5] / 3:List / 1:seq( [ENTER] [2nd][5] 7:Probability 5:randNorm(20,3),
x,1,10) [ENTER] [ENTER].
4. This generates 10 random numbers from a normal distribution with mean 20 and
standard deviation 3. These numbers are automatically stored in list1.
Step 2. Find the mean of the sample and store the mean value into a variable.
1. Press [HOME][CATALOG] [ALPHA] M, scroll to mean( , and press [ENTER]. Type
the name list1 followed by [)]. [ENTER]. You will see the mean of list1 on the
screen.
2. If the command mean(list1) is not highlighted, press[ENTER] to highlight it.
Press your right arrow, and your cursor will be placed to the right of your
command.
3. Press [STO->]. This places an arrow after your formula, where you can enter a
name for the value of the mean of list1. Type a name there (do not use the
calculator domain variables like s, y, z, etc., but enter something like a by
pressing [ALPHA] a [ENTER]). The mean of list1 is now stored in the variable a.
Step 3. Generate nine more random samples. Find and store their means in
additional variables.
1. Repeat Steps 1 and 2 to generate additional lists of random values, calculate the
new mean of each list, and store each mean in a new variable. You may continue
to use list1 to store the list of random values, but be sure to store each new
mean before creating the next list of values.
2. For each random sample, store the mean in a new variable. Using the letters a,
b, c, d, e, . . . should help you remember where you put them and make it easy
to retrieve them.
__________________________
Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse)
TI-89 screens are used with the permission of the publisher. Copyright © 2011, Texas Instruments, Incorporated.
43 of 70
AP Statistics
Page 2 of 4
Study Guide: Sampling Distributions and the Central Limit Theorem
3. Continue creating random samples and storing their means in variables until you
have 10 random sample means. The table below shows each random sample and
the variable letter in which its mean could be stored.
Random Sample
Variable
1
a
2
b
3
c
4
d
5
e
6
f
7
g
8
h
9
i
10
j
Step 4: Find the mean of the sampling distribution.
Now that you’ve stored the means of your 10 samples, you need to find the
mean of the sampling distribution.
1. Enter List Editor.
2. Clear list2.
On line 1 of list2 (list2[1]), enter [ALPHA] a to put the first mean (stored in
variable a) in this cell. Then do the same to put the rest of the letters that have
the stored means (b, c, d, etc.) on each line of list2.
3. Press [HOME] and enter mean(list2). Press [ENTER].
4. This will be the mean of your sampling distribution.
Section 2: For a population with N(50,5), what proportion of the sample means of
size 36 would be expect to be less than 51?
Step 1: Solve for P( x < 51) for N(50, (5/√36)).
1. From List Editor, press [F5]:Distr / 4:Normal Cdf….
2. Enter a Lower Value of  , an Upper Value of 51, a mean of 50, and a standard
deviation of 5/√(36).
3. Press [ENTER] to save your settings, then [ENTER] again to execute the
command.
4. You will see the answer to your question, .88493.
Step 2: Graph P( x < 51) for N(50, .83).
1. Press
[F1] and uncheck any plots or functions you have active.
__________________________
Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse)
TI-89 screens are used with the permission of the publisher. Copyright © 2011, Texas Instruments, Incorporated.
44 of 70
AP Statistics
Page 3 of 4
Study Guide: Sampling Distributions and the Central Limit Theorem
2. Enter List Editor.
3. Press [F5]:Distr / 1:Shade  1: Shade Normal… [ENTER]. Enter a Lower Value of
 .
4. Enter an Upper Value of 51.
5. Enter a mean of 50 and a standard deviation of 5/6.
6. Make sure the Auto-scale setting is on YES.
7. Press [ENTER] to save your settings, then [ENTER] again to execute.
8. You should see the graph shown below.
__________________________
Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse)
TI-89 screens are used with the permission of the publisher. Copyright © 2011, Texas Instruments, Incorporated.
45 of 70
AP Statistics
Page 4 of 4
Study Guide: Sampling Distributions and the Central Limit Theorem
Directions


Use this guide to take notes while you watch the tutorial.
Keep your completed study guide in your notebook for later study.
Key Terms and Concepts



sampling distribution of a sample mean
• sampling distribution
mean of the sampling distribution of a sample mean
• Central Limit Theorem
standard deviation of the sampling distribution of a sample mean
Tutorial Sections
The Meaning of a Sampling Distribution
The Sampling Distribution of a Sample Mean
The Central Limit Theorem
__________________________
Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse)
TI-89 screens are used with the permission of the publisher. Copyright © 2011, Texas Instruments, Incorporated.
46 of 70
6WDWLVWLFV,QGHSHQGHQW6WXG\
6WXG\6KHHW
6DPSOLQJ'LVWULEXWLRQV
3DJHRI
1.4.4
7KH0HDQLQJRID6DPSOLQJ'LVWULEXWLRQ
,PDJLQHWDNLQJKXQGUHGVRIVDPSOHVRIVL]HQIURPDSRSXODWLRQ,I\RXWDNHWKHPHDQRI
HDFKVDPSOH\RX
OOKDYHDGLVWULEXWLRQRIVDPSOHPHDQV,I\RXFRXOGVRPHKRZWDNH
HYHU\SRVVLEOHVDPSOHIURP\RXUSRSXODWLRQ\RX
GJHWDGLVWULEXWLRQRIHYHU\SRVVLEOH
VDPSOHPHDQ7KLVLVFDOOHGWKHVDPSOLQJGLVWULEXWLRQIRUWKHVWDWLVWLF
,I\RXURULJLQDOSRSXODWLRQKDVPHDQµDQGVWDQGDUGGHYLDWLRQσWKHVDPSOLQJ
GLVWULEXWLRQRIPHDQVDOVRFDOOHGWKHGLVWULEXWLRQRI [ ZLOOKDYHPHDQ µ [ PXVXE[
EDUDQG σ [ VLJPDVXE[EDUZKLFKDUHFDOFXODWHGDV
µ[
µ
σ[ =
σ
Q
1RWHWKDWWKHIRUPXODIRUWKHVWDQGDUGGHYLDWLRQDVVXPHV\RXNQRZσWKHSRSXODWLRQ
VWDQGDUGGHYLDWLRQ,QUHDOOLIHWKLVLVXQUHDOLVWLFDQGDV\RXFRQWLQXHWRVWXG\VWDWLVWLFV
\RX
OOOHDUQZKDWWRGRZKHQ\RXGRQ
WNQRZσ
$OVRQRWHWKDWLQDVDPSOLQJGLVWULEXWLRQDOOVDPSOHVDUHWKHVDPHVL]HWKHVDPSOLQJ
GLVWULEXWLRQRI[EDULVWKHGLVWULEXWLRQRIDOO[EDUVZLWKWKHVDPHVL]HGQ
7KH6KDSHRIWKH6DPSOLQJ'LVWULEXWLRQ
7KHVKDSHRIWKHVDPSOLQJGLVWULEXWLRQGHSHQGVERWKRQWKHVDPSOHVL]HDQGRQWKH
VKDSHRIWKHSDUHQWSRSXODWLRQ,IWKHRULJLQDOSRSXODWLRQLVQRUPDOWKHVDPSOLQJ
GLVWULEXWLRQZLOODOVREHQRUPDO,IWKHRULJLQDOSRSXODWLRQLVQRQQRUPDOWKHQWKHVKDSH
RIWKHVDPSOLQJGLVWULEXWLRQGHSHQGVRQWKHVDPSOHVL]H,QWKHVHFDVHVWKHVDPSOLQJ
GLVWULEXWLRQZLOOKDYHDVLPLODUVKDSHWRWKHSDUHQWSRSXODWLRQIRUVPDOOQDQGEHFRPH
DSSUR[LPDWHO\QRUPDOIRUODUJHQ,QPRVWFDVHVODUJHLVGHILQHGDVQ!2XWOLHUVRU
H[WUHPHVNHZQHVVZLOOUHTXLUHODUJHUYDOXHVRIQ
7KH&HQWUDO/LPLW7KHRUHP
7KHEHKDYLRURIVDPSOLQJGLVWULEXWLRQVLVVXPPDUL]HGE\WKH&HQWUDO/LPLW7KHRUHP
ZKLFKVWDWHV
µ [ µUHJDUGOHVVRIWKHVDPSOHVL]HRUWKHVKDSHRIWKHRULJLQDOGLVWULEXWLRQ
σ [ =
σ
Q
UHJDUGOHVVRIWKHVDPSOHVL]HQRUWKHVKDSHRIWKHRULJLQDOGLVWULEXWLRQ
,IWKHRULJLQDOGLVWULEXWLRQLVQRUPDOWKHVKDSHRIWKHVDPSOLQJGLVWULEXWLRQRIWKH
VDPSOHPHDQZLOOEHQRUPDOIRUDQ\Q
47 of 70
S2
6WDWLVWLFV,QGHSHQGHQW6WXG\
6WXG\6KHHW
6DPSOLQJ'LVWULEXWLRQV
3DJHRI
)RUVPDOOVDPSOHVL]HVLIWKHRULJLQDOGLVWULEXWLRQLVQRQQRUPDOWKHVKDSHRIWKH
VDPSOLQJGLVWULEXWLRQRIWKHVDPSOHPHDQZLOOEHVLPLODUWRWKHVKDSHRIWKHRULJLQDO
SRSXODWLRQ7KDWLVLIWKHRULJLQDOSRSXODWLRQZDVVNHZHGWRWKHULJKWWKHVDPSOLQJ
GLVWULEXWLRQRIWKHVDPSOHPHDQIRUVPDOOQZLOODOVREHVNHZHGWRWKHULJKWDOWKRXJK
OHVVVRWKDQWKHRULJLQDOSRSXODWLRQ
,IWKHVDPSOHVL]HLVODUJHWKHVDPSOLQJGLVWULEXWLRQRIWKHVDPSOHPHDQZLOOEH
DSSUR[LPDWHO\QRUPDOPDQ\WH[WVGHILQHODUJHDVQ!DOWKRXJKWKHDFWXDOYDOXH
ZLOOGHSHQGRQWKHVKDSHRIWKHRULJLQDOSRSXODWLRQ
6DPSOH3UREDELOLWLHV
6LQFHDVDPSOLQJGLVWULEXWLRQIRUODUJHQLVQRUPDO\RXFDQXVHQRUPDOSUREDELOLWLHVWR
ILQGWKHSUREDELOLWLHVIRUVDPSOHYDOXHV)RUH[DPSOH
$GLVWULEXWLRQRI [ KDVPHDQDQGVWDQGDUGGHYLDWLRQ:KDW
VWKHSUREDELOLW\
RIGUDZLQJDVDPSOHZLWKDPHDQRIOHVVWKDQ"
7RDQVZHUWKHTXHVWLRQILQGWKH]VFRUHIRUDVDPSOHPHDQRIDQGILQGWKH
SUREDELOLW\IRUYDOXHVEHORZWKDW]VFRUH
]
− ±
3]± 48 of 70
6WDWLVWLFV,QGHSHQGHQW6WXG\
6WXG\6KHHW
6DPSOLQJ'LVWULEXWLRQV
3DJHRI
6WXG\4XHVWLRQV
,Q\RXURZQZRUGVGHILQHWKHWHUPVDPSOLQJGLVWULEXWLRQ
,Q\RXURZQZRUGVGHVFULEHWKHILYHPDLQSRLQWVRIWKH&HQWUDO/LPLW
7KHRUHPDVGHILQHGDERYH
-HIILVDQHJJSODQWLQVSHFWRUIRU&HQWUDO0LFKLJDQ&RXQW\%DVHGRQKLV\HDUV
H[SHULHQFHRQWKHMRE-HIIDVVXPHVWKHPHDQZHLJKWRIWKHHJJSODQWSRSXODWLRQLV
JUDPVDQGWKHVWDQGDUGGHYLDWLRQLVJUDPV
$ $VVXPLQJWKDW-HII
VDVVXPSWLRQLVFRUUHFWLW
VDGHEDWDEOHDVVXPSWLRQEXW
JRZLWKLWDQ\ZD\IRUQRZZKDWDUHWKHPHDQDQGWKHVWDQGDUGGHYLDWLRQ
RIWKHVDPSOLQJGLVWULEXWLRQIRUVDPSOHVRIVL]HQ "
% $VVXPLQJ-HII
VDVVXPSWLRQLVFRUUHFWZKDWDUHWKHPHDQDQGWKHVWDQGDUG
GHYLDWLRQRIWKHVDPSOLQJGLVWULEXWLRQIRUVDPSOHVRIVL]HQ "
:HGUDZVDPSOHVRIVL]HIURPDGLVWULEXWLRQWKDW
VVNHZHGVWURQJO\WRWKHULJKW
7KHPHDQRIWKHSRSXODWLRQGLVWULEXWLRQLVDQGWKHVWDQGDUGGHYLDWLRQLV
$ :KDWDUHWKHPHDQDQGWKHVWDQGDUGGHYLDWLRQRIWKHVDPSOLQJGLVWULEXWLRQ"
% :KDW
VWKHVKDSHRIWKHVDPSOLQJGLVWULEXWLRQ"
:HGUDZVDPSOHVRIVL]HIURPDQRUPDOGLVWULEXWLRQ7KHPHDQRIWKHGLVWULEXWLRQLV
DQGWKHVWDQGDUGGHYLDWLRQLV
$ :KDWDUHWKHPHDQDQGVWDQGDUGGHYLDWLRQRIWKHVDPSOLQJGLVWULEXWLRQ"
% :KDW
VWKHVKDSHRIWKHVDPSOLQJGLVWULEXWLRQ"
& )RUWKLVSRSXODWLRQZKDW
VWKHSUREDELOLW\RIJHWWLQJDVDPSOHRIVL]HZLWK
DPHDQOHVVWKDQ"
:HGUDZVDPSOHVRIVL]HIURPDGLVWULEXWLRQWKDW
VELPRGDO7KHPHDQRIWKH
GLVWULEXWLRQLVDQGWKHVWDQGDUGGHYLDWLRQLV
$ :KDWDUHWKHPHDQDQGWKHVWDQGDUGGHYLDWLRQRIWKHVDPSOLQJGLVWULEXWLRQ"
% :KDW
VWKHVKDSHRIWKHVDPSOLQJGLVWULEXWLRQ"
& )RUWKLVSRSXODWLRQZKDW
VWKHSUREDELOLW\RIJHWWLQJDVDPSOHRIVL]H
ZLWKDPHDQOHVVWKDQ"
49 of 70
6WDWLVWLFV,QGHSHQGHQW6WXG\
6WXG\6KHHW
6DPSOLQJ'LVWULEXWLRQV
3DJHRI
$VWKHVDPSOHVL]HJHWVODUJHUZKDWKDSSHQVWRWKHVWDQGDUGGHYLDWLRQRI
WKHVDPSOLQJGLVWULEXWLRQ":K\"
BBBBBBBBBBBBBBBBBB
‹&RS\ULJKW$3(;2QOLQH/HDUQLQJ,QF$OOULJKWVUHVHUYHG$GYDQFHG3ODFHPHQWDQG$3DUHUHJLVWHUHG
WUDGHPDUNVRIWKH&ROOHJH%RDUG7KLVPDWHULDOLVLQWHQGHGIRUWKHH[FOXVLYHXVHRI$3(;HQUROOHGVWXGHQWV
$Q\XQDXWKRUL]HGFRS\LQJUHXVHRUUHGLVWULEXWLRQLVSURKLELWHG
50 of 70
Page 1 of 3
1.4.5
S2
1. Consider a parent population with mean 75 and a standard deviation 7. The population
doesn't appear to have extreme skewness or outliers.
A. What are the mean and standard deviation of the distribution of sample means
for n = 40? (2 points)
B. What's the shape of the distribution? Explain your answer in terms of the
Central Limit Theorem. (1 point)
C. What proportion of the sample means of size 40 would you expect to be 77 or
less? If you use your calculator, show what you entered. (1 point)
D. Draw a sketch of the probability you found in part C. (1 point)
2. Suppose over several years of offering AP Statistics, a high school finds that final exam
scores are normally distributed with a mean of 78 and a standard deviation of 6.
A. What are the mean, standard deviation, and shape of the distribution of x-bar
for n = 50? (2 points)
B. What's the probability a sample of scores will have a mean greater than 80?
(1 point)
C. Sketch the distribution curve for part B, showing the area that represents the
probability you found. (1 point)
3. Suppose college faculty members with the rank of professor at two-year institutions earn
an average of $52,500 per year with a standard deviation of $4,000. In an attempt to
verify this salary level, a random sample of 60 professors was selected from a personnel
database for all two-year institutions in the United States.
A. What are the mean and standard deviation of the sampling distribution for
n = 60? (2 points)
B. What's the shape of the sampling distribution for n = 60? (1 point)
C. Calculate the probability the sample mean x-bar is greater than $55,000.
(1 point)
D. If you drew a random sample with a mean of $55,000, would you consider this
sample unusual? What conclusions might you draw? (1 point)
51 of 70
Page 2 of 3
4. A manufacturer of paper used for packaging requires a minimum strength of 20 pounds
per square inch. To check the quality of the paper, a random sample of ten pieces of
paper is selected each hour from the previous hour's production and a strength
measurement is recorded for each. The distribution of strengths is known to be normal
and the standard deviation, computed from many samples, is known to equal 2 pounds
per square inch. The mean is known to be 21 pounds per square inch.
A. What are the mean and standard deviation of the sampling distribution for
n = 10? (2 points)
B. What's the shape of the sampling distribution for n = 10? (1 point)
C. Draw a random sample of size 10 from this distribution and compute the mean
of the sample, using the randNorm function on your calculator. On a TI-83/84
store the values in L1: MATH [PRB] randNorm(21,2,10) STO [L1]. On a TI-89
Store the values in list1: seq(randNorm(21,2),x,1,10,1) -> list1.
On the TI-83/84 you can then get the mean of the values by doing 1–Var stats
on the values in L1. Then Press STAT, arrow to calc, select 1-Var stats, Press
ENTER, then 2nd [L1] ENTER.
On the TI-89 you can then get the mean of the values by doing 1-Var Stats on
the values in list1. Enter List Editor, then press F4: 1:1-Var Stats.... Enter list1
as your list and press ENTER twice to do the calculation.
Write down all your sample values. What are the mean and sample standard
deviation of this sample? Compare them with the mean and standard deviation
of the parent population, and explain why they're different or the same.
(2 points)
D. Using the mean and standard deviation of the sampling distribution (from part
A), calculate the probability of getting a sample mean less than what you got.
If you use a calculator, show what you entered. (2 points)
5. Suppose a random sample of n = 25 observations is selected from a normal population,
with mean 106 and standard deviation 12.
A. Find the probability that x-bar exceeds 110. Show the mean and standard
deviation you used. (2 points)
B. Find the probability that the sample mean deviates from the population mean
= 106 by no more than 4. (1 point)
C. Sketch the probability distribution from part B, showing the area that
represents the probability you found. (1 point)
52 of 70
Page 3 of 3
Acknowledgements
Question 3:
This question is based on question 7.24 (c & d) from page 261 of Introduction to Probability and Statistics, Tenth
Edition, by W. Mendenhall, R. Beaver, and B. Beaver. Copyright © 1999 by Brooks Cole, division of Thompson
Learning Incorporated. Further reproduction is prohibited without permission of the publisher.
Question 4:
This question is based on question 7.26 from page 262 of Introduction to Probability and Statistics, Tenth Edition,
by W. Mendenhall, R. Beaver, and B. Beaver. Copyright © 1999 by Brooks Cole, division of Thompson Learning
Incorporated. Further reproduction is prohibited without permission of the publisher.
Question 4:
This question is based on question 7.20 from page 261 of Introduction to Probability and Statistics, Tenth Edition,
by W. Mendenhall, R. Beaver, and B. Beaver. Copyright © 1999 by Brooks Cole, division of Thompson Learning
Incorporated. Further reproduction is prohibited without permission of the publisher.
53 of 70
6WDWLVWLFV6WXG\*XLGH
6DPSOH3URSRUWLRQV
3DJHRI
1.4.6
'LUHFWLRQV
• 8VHWKLVJXLGHWRWDNHQRWHVZKLOH\RXZDWFKWKH7XWRULDO
• .HHS\RXUFRPSOHWHG6WXG\*XLGHLQ\RXUQRWHERRNIRUODWHUVWXG\
.H\7HUPVDQG&RQFHSWV
•
pˆ GLVWULEXWLRQRI
• PHDQRIWKHVDPSOLQJGLVWULEXWLRQRIDVDPSOHSURSRUWLRQ
• VDPSOHSURSRUWLRQV
• VWDQGDUGGHYLDWLRQRIWKHVDPSOLQJGLVWULEXWLRQRIDVDPSOHSURSRUWLRQ
7XWRULDO6HFWLRQV
7KH0HDQLQJRID6DPSOLQJ'LVWULEXWLRQRID6DPSOH3URSRUWLRQ
([HUFLVHV8VLQJWKH6DPSOH3URSRUWLRQ
BBBBBBBBBBBBBBBBBBB
‹&RS\ULJKW$3(;2QOLQH/HDUQLQJ,QF$OOULJKWVUHVHUYHG$GYDQFHG3ODFHPHQWDQG$3DUHUHJLVWHUHG
WUDGHPDUNVRIWKH&ROOHJH%RDUG7KLVPDWHULDOLVLQWHQGHGIRUWKHH[FOXVLYHXVHRI$3(;HQUROOHGVWXGHQWV$Q\
XQDXWKRUL]HGFRS\LQJUHXVHRUUHGLVWULEXWLRQLVSURKLELWHG
54 of 70
S2
Statistics Assignment
Sampling Distribution of p-hat
The Sampling Distribution of p-hat
Page 1 of 5
1.4.8
S2
ˆ is calculated from a count of successes to failures, the distribution of p
ˆ is
Because p
binomial. You may recall that a binomial distribution of counts has a mean np and a
standard deviation
np(1 − p) . You may also recall that to turn a count of successes (x) into
ˆ ), you divide the count x by the sample size n. Similarly, to turn np and
a proportion ( p
ˆ , divide each term by
np(1 − p) into the mean and standard deviation of the proportion p
n:
µ pˆ =
np
= p , σ pˆ =
n
np(1 − p)
n
2
=
p(1 − p)
.
n
p(1 − p)
. Note
n
that the term (1 – p), which is the proportion or probability of failure, is written as q in
pq
many textbooks. So you may see the standard deviation written as
.
n
ˆ is p, and the standard deviation is
So, the mean of the distribution of p
Using the Normal Approximation
When the population is large relative to the sample, and when np ≥ 10 and
n(1 – p) ≥ 10, a binomial distribution can be approximated by a normal distribution. This
ˆ . When conditions are met, it has a distribution
rule applies to the distribution of p
N(p,
p(1 − p)
).
n
NOTE: Some textbooks say you can use the normal approximation if np >5 and nq (or n(1p)) is > 5. These are also valid criteria, but for this Assignment use the criteria np ≥ 10 and
n(1-p) ≥ 10.
The Continuity Correction
ˆ uses a continuous distribution to model a
The normal approximation to the distribution of p
discrete one. Use the continuity correction to offset the error inherent in these situations. To
do this, simply proceed as though the interval you're finding also occupies the space .5
below the lower bound and .5 above the upper bound of the interval.
ˆ , you need to convert your sample proportions to
When using the continuity correction for p
counts, and then use the normal approximation on these counts. For example, let's say the
proportion of people who will vote for candidate A is .7. You draw a sample of 100 people,
and your expected count of successes in a sample (which is the mean of your distribution of
counts) is .7(100) = 70. You want the probability that you'd get a sample with between 66
and 68 successes. To use the continuity correction in this scenario, you'd find the normal
probability for the interval from 65.5 to 68.5.
_____________
© Copyright 2000 Apex Learning Inc. All rights reserved. This material is intended for the exclusive use of
registered users only. No portion of these materials may
be70
reproduced or redistributed in any form without the
55 of
express written permission of Apex Learning Inc.
Statistics Assignment
Sampling Distribution of p-hat
Page 2 of 5
NOTE: As you advance into doing statistical inference, you may not see the continuity
correction very often. In many cases it doesn't make enough of a difference to change your
result.
Using Your Calculator for the Normal Approximation to p-hat
Use normalcdf(lowerbound,upperbound,mean,standard deviation).
ˆ > p) = normalcdf(p,E99,µ,σ)
• P( p
ˆ < p) = normalcdf(–E99,p,µ,σ) Note that you can use any large number in place of
• P( p
E99.
• P(p1 < pˆ < p2) = normalcdf(p1,p2,µ,σ)
NOTE: If you don't enter the mean and standard deviation, the calculator assumes a
standard normal distribution with mean 0 and standard distribution 1. In that case, you can
enter z-scores. On your calculator, that'd be normalcdf(lower z,upper z).
_____________
© Copyright 2000 Apex Learning Inc. All rights reserved. This material is intended for the exclusive use of
registered users only. No portion of these materials may
be70
reproduced or redistributed in any form without the
56 of
express written permission of Apex Learning Inc.
Statistics Assignment
Sampling Distribution of p-hat
Page 3 of 5
Assignment Questions
Answer each question completely and send your work back to your instructor. Show all
work. Justify your answers clearly and completely with formulas and numerical solutions.
Showing your work will allow your instructor to give you partial credit if you make a mistake
but show that you still understand the problem.
1. In a survey, 600 mothers and fathers were asked about the importance of sports for
boys and girls. Of the parents interviewed, 70% said the genders are equal and should
have equal opportunities to participate in sports.
A. What are the mean, standard deviation, and shape of the distribution of the
ˆ of parents who say the genders are equal and should have
sample proportion p
equal opportunities? Be sure to justify your answer for the shape of the
distribution. Use n = 600. (1 point)
B. Using the normal approximation without the continuity correction, sketch the
ˆ . Shade equal areas on
probability distribution curve for the distribution of p
both sides of the mean to show an area that represents a probability of .95,
and label the upper and lower bounds of the shaded area as values of p-hat
(not z-scores). Show your calculations for the upper and lower bounds. (2
points)
C. Considering the sketch in part B, the shaded area shows a .95 probability of
what happening? In other words, what does the probability of .95 represent? (2
points)
D. Using the normal approximation, what's the probability a randomly drawn
sample of parents of size 600 will have a sample proportion between 67% and
73%? Draw a sketch of the probability curve, shade the area representing the
probability you're finding, and label the z-scores that represent the upper and
lower bounds of the probability you're finding. Don't use the continuity
correction. (2 points)
E. Now, use the exact binomial calculation to find the probability of getting
between, but not including, 67% and 73% of the respondents in a sample of
600 who say the genders are equal and should have equal opportunities. To use
the exact binomial, you'll need to convert the proportions to counts by
multiplying each proportion by 600. (1 point)
F. Now try it again, but this time find the probability of getting at least 67% but
no more than 73%. Use the exact binomial calculation. (1 point)
_____________
© Copyright 2000 Apex Learning Inc. All rights reserved. This material is intended for the exclusive use of
registered users only. No portion of these materials may
be70
reproduced or redistributed in any form without the
57 of
express written permission of Apex Learning Inc.
Statistics Assignment
Sampling Distribution of p-hat
Page 4 of 5
2. Random samples of size n = 500 were selected from a binomial population with
p = .1.
A. Is it appropriate in this case to use the normal distribution to approximate the
ˆ ? What are the mean, standard deviation, and shape of the
distribution of p
distribution? (1 point)
B. Using the normal approximation without the continuity correction, find the
ˆ < .12. (1 point)
probability that p
ˆ < .12). This time use the continuity
C. Using the normal approximation, find P( p
correction. (3 points)
Hint: To use the continuity correction you'll need to use the binomial distribution of
counts:
• Convert .12 to a count by multiplying it by 500.
•
Use np and
np(1 − p) to find the mean and standard deviation for the normal
•
•
approximation to the binomial for counts.
You're looking for the probability of getting a count below a certain number.
To use the continuity correction, add .5 to your true upper bound. (You don't need to
increase your lower bound because the lower bound is essentially infinity.)
ˆ < .12). Compare your answer
D. Using the exact binomial calculation, find P( p
with the answers you got for parts B and C. Why are they different? (1 point)
3. One of the ways Americans relieve stress is to reward themselves with sweets. Suppose
a study claims 52% of Americans admit to overeating sweets when stressed. Suppose
also that the 52% figure is correct for the population and that random samples of size n
= 100 Americans are selected.
ˆ have an approximately normal distribution? If so,
A. Does the distribution of p
what are its mean and standard deviation? (1 point)
ˆ without the continuity correction, what's
B. Using the normal approximation of p
ˆ greater than .6? (1 point)
the probability of getting a sample (n = 100) with p
C. Using the normal approximation of the binomial distribution with the continuity
ˆ greater
correction, what's the probability of getting a sample (n = 100) with p
than .6? (3 points)
D. Using the exact binomial calculation, what's the probability of getting sample
ˆ greater than .6? (1 point)
(n = 100) with p
_____________
© Copyright 2000 Apex Learning Inc. All rights reserved. This material is intended for the exclusive use of
registered users only. No portion of these materials may
be70
reproduced or redistributed in any form without the
58 of
express written permission of Apex Learning Inc.
Statistics Assignment
Sampling Distribution of p-hat
Page 5 of 5
4. In 1996 there was a battle in the courts and in the marketplace between Intel and
Digital Equipment Corp. about the technology behind Intel's Pentium microprocessing
chip. Digital accused Intel of willful infringement on Digital's patents. Although Digital's
Alpha microprocessor chip was the fastest on the market at the time, its speed fell
victim to Intel's marketing clout. That same year, Intel shipped 76% of the
microprocessor market. Suppose a random sample of n = 1,000 personal computer (PC)
ˆ be the
sales is monitored and the type of microprocessor installed is recorded. Let p
proportion of personal computers in the sample with a Pentium microprocessor.
ˆ ? (1
A. What are the mean, standard deviation, and shape of the distribution of p
point)
B. Using the normal approximation without the continuity correction, what's the
probability you'd draw a random sample of 1,000 PCs with a proportion of
Pentium chips exceeding 80%? (1 point)
C. Looking at the answer you got for part B, if you got a simple random sample
with
ˆ > .8, would you conclude the population proportion may be higher than .76?
p
Why? (2 points)
Acknowledgements
Question 1:
This question is based on question 7.35 from page 267 of Introduction to Probability and Statistics, Tenth Edition,
by W. Mendenhall, R. Beaver, and B. Beaver. Copyright © 1999 by Brooks Cole, division of Thompson Learning
Incorporated. Further reproduction is prohibited without permission of the publisher.
Question 3:
This question is based on question 7.36 from page 267 of Introduction to Probability and Statistics, Tenth Edition,
by W. Mendenhall, R. Beaver, and B. Beaver. Copyright © 1999 by Brooks Cole, division of Thompson Learning
Incorporated. Further reproduction is prohibited without permission of the publisher.
_____________
© Copyright 2000 Apex Learning Inc. All rights reserved. This material is intended for the exclusive use of
registered users only. No portion of these materials may
be70
reproduced or redistributed in any form without the
59 of
express written permission of Apex Learning Inc.
AP Statistics
Review: Binomial Situations and Sampling Distributions
Page 1 of 11
1.5.2
S2
Key Terms and Concepts
Before taking the Quiz, you need to be able to explain the meanings (and recognize symbols
in cases where there is an associated symbol) of each of these terms or concepts. You
should also know when and how to use them in statistics problems.
These terms and concepts are defined in Key Terms.
almost binomial
average waiting time
binomcdf
binomial experiment
binomial probabilities for a range of outcomes
binomial probabilities for an exact number of outcomes
binomial setting
binompdf
Central Limit Theorem
continuity correction
geometcdf
geometpdf
geometric distribution
geometric setting
inferential statistics
mean of the sampling distribution of a sample mean
mean of the sampling distribution of a sample proportion
normal approximation to the binomial
ˆ (distribution of)
p
sample proportions
sampling distribution
sampling distribution of a sample mean
standard deviation of the sampling distribution of a sample mean
standard deviation of the sampling distribution of a sample proportion
_____________
Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse)
60 of 70
AP Statistics
Review: Binomial Situations and Sampling Distributions
Page 2 of 11
Objectives, Example Problems, and Study Tips
Introduction to Inferential Statistics
Objective
Define inferential statistics, and give at least two examples.
Example
1. Define inferential statistics. Give two examples.
Answer
1. Inferential statistics is the process of estimating population parameters from sample
statistics. Examples include predicting population proportions, such as the percentage of
voters who will vote for a particular candidate, and estimating the mean of a population
parameter, such as the mean age of juniors at a high school.
Objective
Give one reason why probability distributions are important in inferential statistics.
Example
1. Give one reason why probability distributions are important in inferential statistics.
Tip
The answer to this question will become much clearer as you learn more about inferential
statistics. For now, it's helpful to have a general idea of why you're learning so much about
normal, binomial, and geometric distributions.
Answer
1. Probability distributions are used to calculate probabilities so that a researcher can
determine the likelihood of an outcome. If a particular outcome is shown to be highly
unlikely, then it's statistically significant and something other than chance is probably
influencing the results.
Objective:
Outline the process that an inferential study often takes.
Example
1. Outline the process that an inferential investigation often takes.
Tip
Don't memorize these steps word for word. They're only general guidelines, and you won't
have to list them in the Unit Quiz. Once you learn how to do inferential statistics in different
settings, you'll find that different cases require different methods.
Answer
1. Inference takes many forms, but it often includes the following phases:
A.
B.
C.
D.
Identify what you want to study.
Ensure that the study and the sample data are valid.
Use a probability distribution to calculate the likelihood of getting the same results.
If you're comparing at least two measures, do significance testing. This involves
writing hypotheses and testing them.
_____________
Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse)
61 of 70
AP Statistics
Review: Binomial Situations and Sampling Distributions
Page 3 of 11
Binomial Distributions
Objective
List the four conditions that define a binomial setting, and use those conditions to determine
whether a situation is binomial.
Examples
1. What are the conditions that define a binomial setting?
2. Suppose we know that 30% of all college students nationwide graduated from private
high schools. A researcher wants to find the number of freshmen at a small college in
Kansas who graduated from private high schools. The college has 300 freshmen. The
researcher randomly selects a student from the freshman class and determines whether
the student graduated from a private high school. The researcher repeats this process
until 50 different students have been selected. Is this a binomial setting?
3. Consider the situation in the previous question. If the 50 students were sampled from
the entire population of 1,250 students at the college, would this be a binomial setting?
Tip
In a sample-drawing situation, each trial may not be strictly independent. That's because
the ratio of the population size to sample size will change each time a sample is drawn.
However, we can still consider the setting binomial in certain situations. For instance, the
setting is said to be binomial when the population is large compared to the sample.
Although we've said in this course that a population 20 times larger than the sample is
sufficient, some textbooks say that 10 times is acceptable.
Answers
1. A setting is binomial if:
A. There are only two possible outcomes for each trial (success or failure).
B. Each trial is independent.
C. The probability of success for each trial is the same.
D. There are a set number of trials.
For example, you roll a die 100 times and count the number of times you get a 6. This is
a success/failure situation, since the die either lands on 6 or it doesn't. Each trial is
independent, and the probability of success on each trial is 1/6. The set number of trials
is 100.
2. This situation isn't binomial, since the probability for each trial isn't independent. Within
the population of 300 freshmen, the proportion of unsampled students from public and
private high schools will change slightly each time a student is drawn. This means that
the probability of success for each trial is different. We'd still consider a setting to be
binomial, even if each trial isn't independent, if the population is very large compared to
the sample size. In this case, the population is only six times as large as the sample
size, so we'd probably not consider this to be a binomial setting.
3. Since the population is more than 20 times as large as the sample, we'd consider this a
binomial setting. In this course we said that the population must be at least 20 times
larger than the sample. Some textbooks say it only needs to be 10 times greater.
_____________
Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse)
62 of 70
AP Statistics
Review: Binomial Situations and Sampling Distributions
Page 4 of 11
Objective
Solve binomial problems involving at least, at most, and between.
Examples
1. You have an octahedral die with faces A, B, C, D, E, F, G, and H. It's a fair diethat is,
each face is equally likely to appear on top. You roll the die 75 times. What's the
probability of getting exactly 9 A's?
2. For the die in question 1, what's the probability of getting at least 7 A's but less than 12
A's?
3. Suppose we know that 30% of all college students graduated from private high schools.
A sample of 50 students is taken from the 1,200 students at a small college, and it's
determined whether each student in the sample went to a private high school. Assuming
that the 30% proportion applies to your population of interest, what's the probability
that at least 18 of the students in the sample went to a private high school.
Tips
• At least 7 includes 7; less than 12 doesn't include 12.
• Remember, the population size should be at least 20 times the sample size to use the
binomial distribution.
• Although you can solve most (but not all) binomial probability problems using your
calculator, you'll have a much better grasp of the concept if you remember the formula:
æn ö
P( X = x) = çç ÷÷ p x (1 − p) n − x
èxø
where
n = number of trials
p = the probability of success for each trial
x = number of successes
Answers
( )( )
9
æ 75 ö
7 66 = .139 . You can also use
1. Solve using n = 75, p = 1/8, x = 9. Thus, çç ÷÷ 1
8
è9 ø 8
binompdf(75,1/8,9).
2. This is difficult to do using the formula, but easy on a graphing calculator. If B(n, p, x)
represents the probability of getting exactly x successes out of n occurrences of an event
that occurs with probability p, then P(7 ≤ X < 12) = B(75, 1/8, 7) + B(75, 1/8, 8) + . . . +
B(75, 1/8, 11) = binomcdf(75, 1/8, 11) – binomcdf(75, 1/8, 6), = .62
3. It's okay to use a binomial distribution here, since the population is at least 20 times
greater than the sample size. Solve using P(X ≥ 18) = 1 – binomcdf(50,.3,17) = .21.
Objective:
Use the normal approximation in binomial setting problems.
Examples
1. Out of 100 specimens of a certain type of rose bush, 85 will produce flowers annually.
Use the normal approximation to calculate the probability that at least 90% of a plot of
250 plants will flower this year.
_____________
Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse)
63 of 70
AP Statistics
Review: Binomial Situations and Sampling Distributions
Page 5 of 11
2. Can we use a normal approximation to the binomial in the following situation: Statistics
indicate roughly 20% of all adults jog more than twice a week. A sample of 18 adults is
asked if they jog more than twice a week. What's the probability only three of these 18
adults jog more than twice a week?
Tips
• When using the normal approximation, don't forget about the continuity correction.
• There are some rules of thumb for deciding if you can use a normal approximation.
Some textbooks say that the expected numbers of successes and failures should be at
least 5, or np ≥ 5 and n(1 – p) ≥ 5. But the Tutorials in this course use 10 instead of 5,
or np ≥ 10 and n(1 – p) ≥ 10. Either rule will work, though in this course it's preferable
to use 10.
Answers
1. The answer is: µ x = 250(.85) = 212.5, σ x = 250(.85)(.15) = 5.65 . Thus .9(250) = 225,
224.5 − 212.5
= 2.12 .
5.65
So P(z > 2.12) = .017. Note that the exact answer is given by 1 – binomcdf(250, .85,
224) = .013.
so now we find P(X ≥ 225), which is z225 =
2. You can't use a normal approximation here. Some texts will tell you that, to use a
normal approximation, both np and n(1 – p) must be at least 10 (though some books
say 5). Here np = 18(.2) = 3.6 < 10.
Note: Your selections from the Mendenhall textbook may tell you that np and n(1 – p)
must be at least 5, but in this course (and on the AP Exam) it's safer to use the rule that
they should be at least 10.
Geometric Distributions
Objective
List the four conditions that define a geometric setting. Use these conditions to determine
whether a situation is geometric.
Examples
1. Describe a geometric setting and explain how, if at all, it differs from a binomial setting.
2. An event occurs with probability p. On average, how long will we have to wait for the
first success of the event?
3. A sack contains eight red dice and nine green dice. We remove one die at a time (and do
not replace it) until we get a red die. Is this a geometric situation?
Tip
The expected value of X in a geometric setting is the average waiting time and is always
1/p.
Answers
1. A random variable is binomial if the only two possible outcomes are success or failure, if
it occurs a set number of times with each occurrence being independent of the others,
and if each occurrence has the same probability of success. In a geometric setting, there
_____________
Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse)
64 of 70
AP Statistics
Review: Binomial Situations and Sampling Distributions
Page 6 of 11
aren't a set number of trials. Rather, our interest is in the number of trials until the first
success (or in some cases, a certain number of successes) occurs. However, like the
binomial, each occurrence must be independent of the others and occur with the same
probability each time.
2. The answer is: 1/p.
3. No, this isn't geometric. The probability of getting a red die when the first marble is
drawn is 8/17. But the probability of getting a red die on the second draw is either 7/16
or 8/16 depending on the color of the die drawn first.
Objective
Solve problems involving the geometric distribution.
Examples
1. There are 10 different prizes in boxes of Googily-Snaps. Prize #4 is the most valuable to
collectors. What's the probability that you'll get prize #4 without having to buy more
than four boxes of Googily-Snaps? (Assume that all prizes are equally likely in each
box.)
2. Given the situation in question 1, suppose you're unlucky and don't get your prize in one
the first four boxes. On average, how long will you have to wait to get prize #4?
Tips
While you can solve most geometric probability problems using your calculator, you'll have
a much better grasp of the concept if you remember the formula:
P( X = n) = (1 − p )
⋅p
where
n = the number of the trial that yields the first success
p = the probability of success for each trial
n −1
Some geometric probability problems require you to know and understand the formula.
Don't assume you can do it all on your calculator.
Geometric distributions are skewed right. In other words, the probability of getting your first
success decreases with each trial. The expected value for the number of trials is the average
waiting time, or 1/p, where p is the probability of success for each trial.
Answers
1. The answer is: P(X ≤ 4) = G(1) + G(2) + G(3) + G(4) = geometcdf(.1,4) = .3439.
2. This is asking for the average waiting time, which is 1/p. Thus 1/.1 = 10 boxes.
Sampling Distributions: Means and Proportions
Objective
Define sampling distribution and use the definition to answer questions about sampling
distributions.
_____________
Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse)
65 of 70
AP Statistics
Review: Binomial Situations and Sampling Distributions
Page 7 of 11
Examples
1. True or False: A sampling distribution of the mean retirement age in North America can
consist of as few as 100 sample values.
2. Consider drawing samples of size 2 from {A,B,C,D,E} and computing the mean of each
sample. The sampling distribution would consist of how many values?
Tips
If you're not sure what a sampling distribution is, look it up in the Glossary.
Remember, a distribution is composed of all possible values rather than a subset of all
possible values. In most cases, all possible values is a theoretical limit, and so the
distribution is a theoretical construct rather than an actual set of values.
Answers
1. False. A sampling distribution is composed of all possible values of a sample statistic
using samples of the same size. Since the population of retirees in North America is
composed of millions of individuals, 100 sample values is only a simulation of a
distribution, it isn't the actual distribution. While 1,000 values may give a pretty good
idea of what the distribution looks like, it's still just a simulation or an approximation
unless it's composed of all possible values.
2. The sampling distribution would consist of 10 values, since there are 10 ways to select a
æ5 ö
sample of size 2 from a set with 5 elements ( çç ÷÷ = 10 ). Even though there are only 10
è2ø
values in this distribution, it's a sampling distribution, since it's composed of all possible
samples of a given size drawn from the population.
Objective:
Determine the shape, mean, and standard deviation of a sampling distribution.
Examples
1. Samples of size 10 are drawn from a large (N > 10,000), symmetric population with a
mean of 45 and a standard deviation of 9. What are the mean and standard deviation of
the sampling distribution of the mean for samples of size 10, and what's the shape of
the distribution?
2. A population has proportion .35 of some characteristic of interest. What are the mean
ˆ for samples of size 50? What's
and standard deviation of the sampling distribution of p
ˆ
the shape of the sampling distribution of p ?
3. A distribution is strongly skewed to the left. Samples of size n are drawn and the
sampling distribution of the sample mean is constructed. What's the shape of the
sampling distribution of x if n = 10, if n = 25, and if n = 100?
Tips
• The shape of the sampling distribution tends to resemble the shape of the parent
population for small samples.
ˆ , we must have np ≥
• To use the normal approximation to the sampling distribution of p
10 and n(1 – p) ≥ 10.
_____________
Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse)
66 of 70
AP Statistics
Review: Binomial Situations and Sampling Distributions
Page 8 of 11
• Remember that the Central Limit Theorem doesn't apply until the sample size (n) is
large. In many cases, a sample size of 30 is large enough to make the sampling
distribution close to normal.
Answers
1. The answer is: µ x = 45, σ x =
9
= 2.85 . The sample size is small, so the shape will
10
resemble that of the parent population, that is, it will be symmetric about its mean.
(.35)(1 − .35)
= .067 . Samples of size 50 are relatively
50
large, so we'd expect the shape of the sampling distribution to be approximately normal.
(Note that the assumptions we must make to be able to use the normal approximation
ˆ are satisfied in this case.)
to p
2. The answer is: µ pˆ = .35, σ pˆ =
3. For n = 10, we'd expect the sampling distribution to resemble the parent population.
That is, it would still be skewed to the left, but would bunch more about the population
mean than the original population. For n = 25, the distribution would be more normal
than the original, but the extent to which it would resemble the original population
depends on how severe the skewness is. We'd usually expect samples of size 25 to
produce sampling distributions that are approximately normal. If n = 100, we'd certainly
expect an approximately normal sampling distribution regardless of the shape of the
original population.
Objective
Summarize the Central Limit Theorem and use elements of the theorem to answer
questions about sampling distributions.
Examples
1. True or False: The Central Limit Theorem tells us that the sampling distribution of a
sample mean will be approximately normal regardless of the shape of the parent
population.
2. We create a sampling distribution of x by taking samples of size 70 from a population
whose mean is known to be 5 and whose standard deviation is 1. What can we say
about the sampling distribution of x ?
Tips
Remember that the Central Limit Theorem doesn't apply until the sample size (n) is large.
In many cases, a sample size of 30 is large enough to make the sampling distribution close
to normal. For symmetric distributions with no outliers, samples smaller than 30 may be
sufficient.
If you can't easily summarize the Central Limit Theorem in your own words, look it up in the
Glossary and review the Tutorial.
_____________
Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse)
67 of 70
AP Statistics
Review: Binomial Situations and Sampling Distributions
Page 9 of 11
Answers
1. False. The Central Limit Theorem tells us that the shape of the sampling distribution of a
sample mean will be approximately normal for large samples. Just how large is large
enough depends on the shape of the parent population, but samples of size 30 or
greater are often large enough. For symmetric distributions with no outliers, even small
sample sizes will yield approximately normal sampling distributions.
2. Because the sample size is large, the Central Limit Theorem applies, and you can say
that the shape of the sampling distribution of x will be approximately normal even
though you're given no information about the shape of the parent population. You also
1
= .12 . Note: The last two facts would be true
know that µ x = 5 and that σ x =
70
regardless of the shape of the original distribution or the sample size.
Objective
Solve problems involving the sampling distribution of a sample mean or of a sample
proportion.
Examples
1. A popular soda comes in 12-oz cans. However, the actual volume of soda in the can
varies normally with a mean of 11.9 oz and a standard deviation of .3 oz. What's the
probability that the mean amount of soda in a six-pack is less than 12 oz?
2. A popular soda comes in 12-oz cans. However, the actual volume of soda in the can
varies normally with a mean of 11.9 oz and a standard deviation of .3 oz. What's the
probability that the mean amount of soda in a six-pack is between 11.7 oz and 12 oz?
3. The probability of winning at roulette is about .474. Suppose you bet 50 times. What's
your probability of being even or ahead after 50 bets and after 1,000 bets?
Tips
• When you calculate the standard deviation of the sampling distribution, don't forget to
divide by the square root of the sample size.
Answers
1. You need to find the probability of getting a mean less than or equal to 12.
12 − 11.9
= .82 è Area = .793.
P( x ≤ 12) = z12 =
.3
6
_____________
Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse)
68 of 70
AP Statistics
Review: Binomial Situations and Sampling Distributions
Page 10 of 11
2. You need to find the probability of getting a mean between 11.7 and 12, inclusive.
) = .742.
P(11.7 ≤ x ≤ 12) = normalcdf(11.7,12,11.9, .3
6
3. You need to find the probability of getting a proportion greater than or equal to .5, which
ˆ ≥ .5). Because 50(.474) ≥ 10 and 50(1 – .474) ≥ 10, we can use the normal
is P( p
approximation to the sampling distribution of a sample proportion. This would be as
follows:
(.474)(1 − .474)
For n = 50: µ pˆ = .474, σ pˆ =
= .071 , normalcdf(.5,10,.474,.071)
50
= .36. For n = 1000: σ pˆ =
(.474)(1 − .474)
= .016 , normalcdf(.5,10,.474,.016)
1000
= .05.
___________
TI-83 screens are used with the permission of the publisher. Copyright ã 1996, Texas Instruments, Incorporated.
_____________
Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse)
69 of 70
AP Statistics
Review: Binomial Situations and Sampling Distributions
Page 11 of 11
About the Unit Quiz
What to Bring
• Scratch paper
• Calculator
• Approved formula sheet
• Approved tables
You can't have any reference materials other than those specifically mentioned above. You
won't be able to ask for help during the Quiz.
Hints and Tips for the Free-Response Portion
•
Show your work. The test corrector won't assume you used proper set up and
methods if you reach the correct answer. It's up to you to communicate the methods
that you used. Answers alone, without appropriate justification, will receive no credit.
•
Take your time reading the question. Since we want to see how well you can apply
your knowledge to new and somewhat unfamiliar situations, take some time to think
about the question. If you don't understand the question, you're unlikely to find the right
answer. Read the entire question before beginning to answer.
•
Most questions will be given in several parts. The answers from one section will
often be used in subsequent sections. Missing points in an early section does not mean
you'll lose points in subsequent sections. Again, read the entire question to see how the
different sections connect to each other.
•
The calculator. As in the AP Exam, this Quiz will test you on how well you know
statistics, not on how well you can use your calculator. Be sure you understand the
concepts behind the calculator operations. Don't use "calculator-speak" in your answer—
the instructor doesn't want to read a set of steps for your graphing calculator! Use your
calculator for doing the mechanics, but be sure to clearly communicate your process for
solving the problem.
•
Use Units. If units are given in the problem, make sure that you give them in your
answer.
•
Answer the Question. Finally, be very careful to answer the question asked. Before
you move on, read over your answer to make sure you're providing exactly what the
question asks for. Generally, an answer to a question you weren't asked will receive no
credit.
_____________
Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse)
70 of 70