+
Chapter 8
Binomial and Geometric Distributions
 8.1
Binomial Random Variables

Binomial Distribution
+ Binomial and Geometric Random Variables
Learning Objectives
After this section, you should be able to…

DETERMINE whether the conditions for a binomial setting are met

COMPUTE and INTERPRET probabilities involving binomial random
variables

CALCULATE the mean and standard deviation of a binomial random
variable and INTERPRET these values in context

CALCULATE probabilities involving geometric random variables
Random Variables – ESP
•Chance process: choose a card at random
•Outcome of interest: person identifies card
correctly
•Random Variable X: number of correct
identifications
Binomial and Geometric Random Variables
To test whether someone has ESP, choose one of four cards at
random – a star, wave, cross, or Circle. Ask the person to identify
the card without seeing it. Do this a total of 50 times, and see how
many cards the person identifies correctly.
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 Binomial
Random Variables – Shipping Claims
•Chance process: Randomly select a
shipment and check when it arrived
•Outcome of interest: arrived on time
•Random Variable Y: number of on-time
shipments
These are binomial random variables because we are counting the
number of times the outcome of interest occurs in a fixed number of
repetitions.
Binomial and Geometric Random Variables
A shipping company claims that 90% of its shipments arrive on time.
To test this claim, take a random sample of 100 shipments made by
the company last month and see how many arrived on time.
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 Binomial
Random Variables – Pass the Pigs
•Chance process: roll the pig dice
•Outcome of interest: pig out
•Random Variable T: number of rolls until the
player pigs out
This is a geometric random variable because we are counting the
number of repetitions of the chance process needed for the outcome of
interest to occur.
Binomial and Geometric Random Variables
In the game of Pass the Pigs, a player rolls a pair of pig-shaped dice.
On each roll, the player earns points according to how the pigs land.
If the player gets a “pig out” in which the two pigs land on opposite
sides, she loses all points earned in that round and must pass the
pigs to the next player. A player can choose to stop rolling at any
point during her turn and keep the points that she has earned before
passing the pigs.
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 Geometric
Settings
The number of trials is fixed in advance, and the outcome of one trial has no effect
on the outcome of any other trial.
We are interested in the number of times that a specific event occurs.
Our chances of getting a “success” are the same on each trial.
If so, we have a binomial setting.
• Toss a coin 5 times. Count the number of heads.
• Spin a roulette wheel 8 times. Record how many times the ball lands
in a red slot.
• Take a random sample of 100 babies born in U.S. hospitals today.
Count the number of females.
• Look at a packet of 20 tomato plant seeds. Plant them and determine
how many germinate. (Assume independence.)
Binomial and Geometric Random Variables
Sometimes we are performing repeated trials of the same chance process.
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 Binomial
Settings
Definition:
A binomial setting arises when we perform several independent trials of the
same chance process and record the number of times that a particular
outcome occurs. The four conditions for a binomial setting are
B
• Binary? The possible outcomes of each trial can be classified as
“success” or “failure.”
I
• Independent? Trials must be independent; that is, knowing the result
of one trial must not have any effect on the result of any other trial.
N
• Number? The number of trials n of the chance process must be fixed
in advance.
S
• Success? On each trial, the probability p of success must be the
same.
Binomial and Geometric Random Variables
When the same chance process is repeated several times, we are often interested
in whether a particular outcome does or doesn’t happen on each repetition. In
some cases, the number of repeated trials is fixed in advance and we are
interested in the number of times a particular event (called a “success”) occurs.
If the trials in these cases are independent and each success has an equal
chance of occurring, we have a binomial setting.
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 Binomial
Settings
•Put the names of all of the students in your class in a hat. Mix them
up, and draw four names without looking. Let Y = the number whose
last names have more than six letters.
• Exactly 10% of students in a school are left-handed. Select students
at random from the school, one at a time, until you find one who is lefthanded. Let V = the number of students chosen.
•Exactly 10% of students in a school are left-handed. Select 15
students at random from the school and define W = the number who
are left-handed.
Binomial and Geometric Random Variables
Identify whether the given random variable has a binomial distribution. If so,
determine its BINS.
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 Binomial
Settings
•Genetics says that children receive genes from each of their parents
independently. Each child of a particular pair of parents has probability
0.25 of having type O blood. Suppose these parents have 5 children.
Let X = the number of children with type O blood.
• Shuffle a deck of cards. Turn over the first 10 cards, one at a time.
Let Y = the number of aces observed.
•Shuffle a deck of cards. Turn over the top card. Put the card back in
the deck, and shuffle again. Repeat this process until you get an ace.
Let W = the number of cards required.
Binomial and Geometric Random Variables
Identify whether the given random variable has a binomial distribution. If so,
determine its BINS.
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 Binomial
Settings
•Roll a fair die 10 times and let X = the number of sixes.
• Shoot a basketball 20 times from various distances on the court. Let
Y = number of shots made.
Observe the next 100 cars that go by and let C = color.
Binomial and Geometric Random Variables
Identify whether the given random variable has a binomial distribution. If so,
determine its BINS.
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 Binomial
Settings
•Shuffle a deck of cards. Turn over the top card. Put the card back in
the deck and shuffle again. Repeat this process 10 times. Let X = the
number of hearts you observe.
Choose 3 students from the class. Let Y = the number who are over 6
feet tall.
Flip a coin. If it’s heads, roll a 6-sided die. If it’s tails, roll an 8-sided
die. Repeat this process 5 times. Let W = the number of 5s rolled.
Binomial and Geometric Random Variables
Identify whether the given random variable has a binomial distribution. If so,
determine its BINS.
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 Binomial
Random Variable
The number of heads in n tosses is a binomial random variable X.
The probability distribution of X is called a binomial distribution.
Definition:
The count X of successes in a binomial setting is a binomial random
variable. The probability distribution of X is a binomial distribution with
parameters n and p, where n is the number of trials of the chance process
and p is the probability of a success on any one trial. The possible values of
X are the whole numbers from 0 to n.
Note: When checking the Binomial condition, be sure to check the
BINS and make sure you’re being asked to count the number of
successes in a certain number of trials!
Binomial and Geometric Random Variables
Consider tossing a coin n times. Each toss gives either heads or tails.
Knowing the outcome of one toss does not change the probability of
an outcome on any other toss. If we define heads as a success, then
p is the probability of a head and is 0.5 on any toss.
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 Binomial
Probabilities
having type O blood. Genetics says that children receive genes from
each of their parents independently. If these parents have 5 children,
the count X of children with type O blood is a binomial random
variable with n = 5 trials and probability p = 0.25 of a success on
each trial. In this setting, a child with type O blood is a “success” (S)
and a child with another blood type is a “failure” (F).
What’s P(X = 0)?
P(FFFFF) = (0.75)(0.75)(0.75)(0.75)(0.75) = (0.75)5 = 0.2373
Therefore, P(X = 0) = (0.75)5 = 0.2373
Binomial and Geometric Random Variables
In a binomial setting, we can define a random variable (say, X) as the
number of successes in n independent trials. We are interested in
finding the probability distribution of X.
Example
Each child of a particular pair of parents has probability 0.25 of
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 Binomial
Probabilities
having type O blood. Genetics says that children receive genes from
each of their parents independently. If these parents have 5 children,
the count X of children with type O blood is a binomial random
variable with n = 5 trials and probability p = 0.25 of a success on
each trial. In this setting, a child with type O blood is a “success” (S)
and a child with another blood type is a “failure” (F).
What’s P(X = 1)?
P(SFFFF) = (0.25)(0.75)(0.75)(0.75)(0.75) = (0.25) (0.75)4 = 0.0791
However, there are a number of different arrangements, each of which
have a probability of 0.0791, in which 1 out of the 5 children have type O
blood:
SFFFF
FSFFF
FFSFF
FFFSF
FFFFS
Verify that in each arrangement, P(X = 1) = (0.25)(0.75)4 = 0.0791
Therefore, P(X = 1) = 5(0.25) (0.75)4 = 0.39551
Binomial and Geometric Random Variables
In a binomial setting, we can define a random variable (say, X) as the
number of successes in n independent trials. We are interested in
finding the probability distribution of X.
Example
Each child of a particular pair of parents has probability 0.25 of
+
 Binomial
Probabilities
+
 Binomial
having type O blood. Genetics says that children receive genes from
each of their parents independently. If these parents have 5 children,
the count X of children with type O blood is a binomial random
variable with n = 5 trials and probability p = 0.25 of a success on
each trial. In this setting, a child with type O blood is a “success” (S)
and a child with another blood type is a “failure” (F).
What’s P(X = 2)?
P(SSFFF) = (0.25)(0.25)(0.75)(0.75)(0.75) = (0.25)2(0.75)3 = 0.02637
However, there are a number of different arrangements in which 2 out of
the 5 children have type O blood:
SSFFF
SFSFF
SFFSF
SFFFS
FSSFF
FSFSF
FSFFS
FFSSF
FFSFS
FFFSS
Verify that in each arrangement, P(X = 2) = (0.25)2(0.75)3 = 0.02637
Therefore, P(X = 2) = 10(0.25)2(0.75)3 = 0.2637
Binomial and Geometric Random Variables
In a binomial setting, we can define a random variable (say, X) as the
number of successes in n independent trials. We are interested in
finding the probability distribution of X.
Example
Each child of a particular pair of parents has probability 0.25 of
Probabilities
doubles means that the outcomes of two dice are the same, such as
1 and 1, or 5 and 5. The probability of rolling doubles when rolling
two dice is 6/36, or 1/6. If X = the number of doubles in 4 rolls of two
dice, then X is binomial with n = 4 and p = 1/6.
What’s P(X = 0)?
P(FFFF) = (5/6)(5/6)(5/6)(5/6) = (5/6)4 = 0.48225
Therefore, P(X = 0) = (5/6)4 = 0.48225
Binomial and Geometric Random Variables
In a binomial setting, we can define a random variable (say, X) as the
number of successes in n independent trials. We are interested in
finding the probability distribution of X.
Example
In many games involving dice, rolling doubles is desirable. Rolling
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 Binomial
Probabilities
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 Binomial
doubles means that the outcomes of two dice are the same, such as
1 and 1, or 5 and 5. The probability of rolling doubles when rolling
two dice is 6/36, or 1/6. If X = the number of doubles in 4 rolls of two
dice, then X is binomial with n = 4 and p = 1/6.
What’s P(X = 1)?
P(FFFF) = (1/6)(5/6)(5/6)(5/6) = (1/6)(5/6)3 = 0.09645
However, there are a number of different arrangements in which 1 out of
the 4 rolls is a double:
SFFF
FSFF
FFSF
FFFS
Verify that in each arrangement, P(X = 1) = (1/6) (5/6)3 = 0.09645
Therefore, P(X = 1) = 4(1/6) (5/6)3 = 0.3858
Binomial and Geometric Random Variables
In a binomial setting, we can define a random variable (say, X) as the
number of successes in n independent trials. We are interested in
finding the probability distribution of X.
Example
In many games involving dice, rolling doubles is desirable. Rolling
Coefficient
We can generalize this for any setting in which we are interested in k
successes in n trials. That is,
P(X = k) = P(exactly k successes in n trials)
= number of arrangements× p k (1- p) n-k
Definition:
The number of ways of arranging k successes among n observations is
given by the binomial coefficient
Some people
æ nö
n!
prefer the notation
=
ç ÷
nCk for the binomial
è k ø k!(n - k)!
coefficient.
for k = 0, 1, 2, …, n where
n! = n(n – 1)(n – 2)•…•(3)(2)(1)
and 0! = 1.
This formula will be on the AP Exam formula sheet! 
Binomial and Geometric Random Variables
Note, in the example about the family and Type O blood, any one
arrangement of 2 S’s and 3 F’s had the same probability. This is
true because no matter what arrangement, we’d multiply together
0.25 twice and 0.75 three times.
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 Binomial
Probability
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 Binomial
Binomial Probability
If X has the binomial distribution with n trials and probability p of success on
each trial, the possible values of X are 0, 1, 2, …, n. If k is any one of
these values,
æ nö k
P(X = k) = ç ÷ p (1- p) n-k
è kø
Number of
arrangements
of k successes
Probability of k
successes
Probability of
n-k failures
Binomial and Geometric Random Variables
The binomial coefficient counts the number of different ways in
which k successes can be arranged among n trials. The
binomial probability P(X = k) is this count multiplied by the
probability of any one specific arrangement of the k successes.
Coefficients on the Calculator
è 2ø
TI-83/84:
Type 5, press MATH, arrow over to PRB, choose 3:nCr, and press ENTER
Then type 2, and press ENTER again.
TI-89:
æ 5ö
ç ÷
è 2ø
From the home screen, press 2nd 5 (MATH), choose 7: Probability, 3: nCr
Complete the command nCr(5, 2) and press ENTER
Binomial and Geometric Random Variables
æ 5ö
Calculate a binomial coefficient like ç ÷ as follows:
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 Binomial
Inheriting Blood Type
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 Example:
Each child of a particular pair of parents has probability 0.25 of having blood
type O. Suppose the parents have 5 children
(a) Find the probability that exactly 3 of the children have type O blood.
Let X = the number of children with type O blood. We know X has a binomial
distribution with n = 5 and p = 0.25.
æ5ö
P(X = 3) = ç ÷(0.25) 3 (0.75) 2 = 10(0.25) 3 (0.75) 2 = 0.08789
è 3ø
(b) Should the parents be surprised if more than 3 of their children have
type O blood?
To answer this, we need to find P(X > 3).
P(X > 3) = P(X = 4) + P(X = 5)
æ 5ö
æ5ö
4
1
= ç ÷(0.25) (0.75) + ç ÷(0.25) 5 (0.75) 0
è 4ø
è5ø
= 5(0.25) 4 (0.75)1 + 1(0.25) 5 (0.75) 0
= 0.01465 + 0.00098 = 0.01563
Since there is only a
1.5% chance that more
than 3 children out of 5
would have Type O
blood, the parents
should be surprised!
Rolling Doubles
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 Example:
When rolling two dice, the probability of rolling doubles is 1/6. Suppose that a
game player rolls the dice 4 times, hoping to roll doubles.
(a) Find the probability that the player gets doubles twice in four attempts.
Let X = number of doubles. X has a binomial distribution with n = 4 and p =
1/6.
æ 4öæ 1 ö2æ 5 ö2 æ 1 ö2æ 5 ö2
P(X = 2) = ç ÷ç ÷ ç ÷ = 6ç ÷ ç ÷ = 0.116
è 6ø è 6ø
è2 øè 6 ø è 6 ø
(b) Should the player be surprised if he gets doubles more than twice in 4
attempts?
To answer this, we need to find P(X > 2).
P(X > 2) = P(X = 3) + P(X = 4)
æ 4 öæ 1 ö3 æ 5 ö1 æ 4 öæ 1 ö 4 æ 5 ö 0
= ç ÷ç ÷ ç ÷ + ç ÷ç ÷ ç ÷
è 3øè 6 ø è 6 ø è 4 øè 6 ø è 6 ø
= 4(.00386) +1(0.001)
= 0.016
Since there is only a
1.6% chance that the
player will roll doubles
more than twice in 4
attempts, he should be
surprised.
Probability on the Calculator
Binomcdf(n,p,k) computes P(X < k)
TI-83/84: These commands are found in the distributions menu (2nd/VARS)
TI-89: These commands are found in CATALOG under Flash Apps
For the parents having n = 5 children, each with the probability p = 0.25 of
type O blood:
P(X = 3) = binompdf(5, 0.25, 3)
= 0.08789
To find P(X > 3) = 1 – P(X < 3)
= 1 – binomcdf(5, 0.25, 3)
= 0.01563
Binomial and Geometric Random Variables
Binompdf(n,p,k) computes P(X = k)
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 Binomial
YOUR UNDERSTANDING
+
 CHECK
To introduce his class to binomial distributions, Mr. Leonard gives a 10-item,
multiple choice quiz. The catch is, students must simply guess an answer (A
through E) for each question. Mr. Leonard uses his computer’s random
number generator to produce the answer key, so that each possible answer
has an equal chance to be chosen. Christiana is one of the students in this
class. Let X = the number of Christiana’s correct guesses.
(a) Show that X is a binomial random variable.
Check the BINS. This is a binomial setting. X is a binomial random
variable with n = 10 and p = 0.20.
(b) Find P(X = 3). Explain what this means.
P(X = 3) = 0.2013. There is a 20% chance that Christiana will answer
exactly 3 questions correctly.
(c) To get a passing score on the quiz, a student must guess correctly at
least 6 times. Would you be surprised if Christiana earned a passing
score? Compute the appropriate probability to support your answer.
P(X > 5) = 0.0064. Since there is only a 0.64% chance that a student
will pass, we would be surprised if Christiana passed.
EXAM ERROR ALERT!
Binomial and Geometric Random Variables
When using the complement rule for discrete probability distributions such
as the binomial distribution, students often have trouble identifying the
correct complementary event. For example, if students are asked to
find P(X ≥ 2), many will calculated 1 – P(X ≤ 2) rather than 1 – P(X ≤ 1).
To avoid this mistake, write out the possible values of X, circle the ones
you want to find the probability of, and cross out the remaining values
that make up the complementary event.
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 AP
EXAM ERROR ALERT!
Binomial and Geometric Random Variables
Don’t use “calculator speak” when showing your work on free-response
questions. Writing binompdf(5, 0.25, 3) = 0.08789 will not earn you full
credit for a binomial probability calculation. At the very least, you must
indicate what each of those calculator inputs represents. For example,
“I used binompdf(5, 0.25, 3) on my calculator with n = 3, p = 0.25, and k
= 3.” Better yet, show the binomial probability formula (it will be on your
formula sheet!) with those numbers plugged in.
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 AP
and Standard Deviation of a Binomial
Distribution
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 Mean
xi
0
1
2
3
4
5
pi
0.2373
0.3955
0.2637
0.0879
0.0147
0.00098
Shape: The probability distribution of X is skewed to
the right. It is more likely to have 0, 1, or 2 children
with type O blood than a larger value.
Center: The median number of children with type O
blood is 1. Based on our formula for the mean:
m X = å x i pi = (0)(0.2373) + 1(0.39551) + ...+ (5)(0.00098)
= 1.25
2
Spread: The variance of X is s X =
å(x - m
i
X
Binomial and Geometric Random Variables
We describe the probability distribution of a binomial random variable just like
any other distribution – by looking at the shape, center, and spread. Consider
the probability distribution of X = number of children with type O blood in a
family with 5 children.
) 2 pi = (0 -1.25) 2 (0.2373) + (1-1.25) 2 (0.3955) + ...+
(5 -1.25) 2 (0.00098) = 0.9375
The standard deviation of X is s X = 0.9375 = 0.968
Notice, the mean µX = 1.25 can be found another way. Since each
child has a 0.25 chance of inheriting type O blood, we’d expect
one-fourth of the 5 children to have this blood type. That is, µX
= 5(0.25) = 1.25. This method can be used to find the mean of
any binomial random variable with parameters n and p.
Mean and Standard Deviation of a Binomial Random Variable
If a count X has the binomial distribution with number of trials n and
probability of success p, the mean and standard deviation of X are
m X = np
s X = np(1- p)
Also included in
the AP Exam
formula sheet.
Note: These formulas work ONLY for binomial distributions.
They can’t be used for other distributions!
Binomial and Geometric Random Variables
and Standard Deviation of a Binomial
Distribution
+
 Mean
Bottled Water versus Tap Water
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 Example:
The 21 teachers at Ms. Raskin’s Nerd Camp did the Bottled Water activity. If we assume
the teachers in this class cannot tell tap water from bottled water, then each has a 1/3
chance of correctly identifying the different type of water by guessing. Let X = the
number of teachers who correctly identify the cup containing the different type of water.
Find the mean and standard deviation of X.
Since X is a binomial random variable with parameters n = 21 and p = 1/3, we can
use the formulas for the mean and standard deviation of a binomial random
variable.
m X = np
= 21(1/3) = 7
We’d expect about one-third of his
21 students, about 7, to guess
correctly.
s X = np(1- p)
= 21(1/3)(2 /3) = 2.16
If the activity were repeated many
times with groups of 21 students
who were just guessing, the
number of correct identifications
would differ from 7 by an average of
2.16.
Diet Soda vs. Regular Soda
+
 Example:
The makers of a diet soda claim that its taste is indistinguishable from the full-calorie
version of the same soda. To investigate, Sarah prepared small samples of each type
of soda in identical cups. Then she had volunteers taste each soda in a random order
and try to identify which was the diet soda and which was the regular. Overall, 23 of the
30 subjects made the correct identification.
Find the mean and standard deviation of X.
Since X is a binomial random variable with parameters n = 30 and p = 0.5, we can
use the formulas for the mean and standard deviation of a binomial random
variable.
m X = np
= 30(0.5) = 15
We’d expect about 15 people to
guess correctly.
P(Xthe
Of
≥ 23)
30 =
volunteers,
1 – P(X ≤ 22
23 made
correct
=
1 – binomcdf(30,
identifications.
0.5, 22)
Does this
give
=
1 –convincing
0.9974
evidence that the
volunteers
=
0.0026 There
canistaste
a verythe
small
difference
chance that
between
there
woulddiet
be 23
and
or more
regular
correct
soda?
guesses.
sX = np(1 - p)
= 30(0.5)(1 - 0.5) = 2.74
If the activity were repeated many
times with groups of 30students
who were just guessing, the
number of correct identifications
would differ from 15 by an average
of 2.74.
Diet Soda vs. Regular Soda
Binomial and Geometric Random Variables
Here is the probability distribution for the number for correct guesses in Ms. Raskin’s
summer class. Where have you previously seen this graph?
+
 Example:
YOUR UNDERSTANDING
+
 CHECK
To introduce his class to binomial distributions, Mr. Leonard gives a 10-item,
multiple choice quiz. The catch is, students must simply guess an answer (A
through E) for each question. Mr. Leonard uses his computer’s random
number generator to produce the answer key, so that each possible answer
has an equal chance to be chosen.
(a) Find μx. Interpret this value in context.
μx = 2. We would expect the average student to get 2 answers correct.
(b) Find σx. Interpret this value in context.
σx = 1.265. We would expect individual students’ scores to vary from a
mean of 2 by an average of 1.265 correct answers.
(c) What’s the probability that the number of Christiana’s correct guesses
is more than 2 standard deviations above the mean? Show your method.
P(X > 4.53) = 1 – P(X ≤ 4) = 0.0328
Distributions in Statistical Sampling
Suppose 10% of CDs have defective copy-protection schemes that can harm
computers. A music distributor inspects an SRS of 10 CDs from a shipment of
10,000. Let X = number of defective CDs. What is P(X = 0)? Note, this is not
quite a binomial setting. Why?
The actual probability is
9000 8999 8998
8991
×
×
× ...×
= 0.3485
10000 9999 9998
9991
æ10ö
P(X = 0) = ç ÷(0.10) 0 (0.90)10 = 0.3487
è0ø
P(no defectives) =
Using the binomial distribution,
In practice, the binomial distribution gives a good approximation as long as we don’t
sample more than 10% of the population.
Sampling Without Replacement Condition
When taking an SRS of size n from a population of size N, we can use a
binomial distribution to model the count of successes in the sample as
long as
1
n£
10
N
Binomial and Geometric Random Variables
The binomial distributions are important in statistics when we want to
make inferences about the proportion p of successes in a population.
+
 Binomial
+
2. Normal Approximation
* Check SHOW PROBABILITY and set
values from 3-5
* Change values for N and notice the
difference in actual versus approximated
probability
* Change values for p and notice the
difference in actual versus approximated
probability
When is this approximation accurate
enough to be useful???
Mathshepherd.com
1. Normal Approximation of Binomial Distribution
Approximation for Binomial Distributions
Normal Approximation for Binomial Distributions
Suppose that X has the binomial distribution with n trials and success
probability p. When n is large, the distribution of X is approximately
Normal with mean and standard deviation
mX = np
s X = np(1- p)
As a rule of thumb, we will use the Normal approximation when n is so
large that np ≥ 10 and n(1 – p) ≥ 10. That is, the expected number of
successes and failures are both at least 10.
Binomial and Geometric Random Variables
As n gets larger, something interesting happens to the shape of a
binomial distribution. The figures below show histograms of
binomial distributions for different values of n and p. What do
you notice as n gets larger?
+
 Normal
Attitudes Toward Shopping
+
 Example:
Sample surveys show that fewer people enjoy shopping than in the past. A survey asked a nationwide
random sample of 2500 adults if they agreed or disagreed that “I like buying new clothes, but
shopping is often frustrating and time-consuming.” Suppose that exactly 60% of all adult US
residents would say “Agree” if asked the same question. Let X = the number in the sample who
agree. Estimate the probability that 1520 or more of the sample agree.
1) Verify that X is approximately a binomial random variable.
B: Success = agree, Failure = don’t agree
I: Because the population of U.S. adults is greater than 25,000, it is reasonable to assume the
sampling without replacement condition is met.
N: n = 2500 trials of the chance process
S: The probability of selecting an adult who agrees is p = 0.60
2) Check the conditions for using a Normal approximation.
Since np = 2500(0.60) = 1500 and n(1 – p) = 2500(0.40) = 1000 are both at least 10, we may use
the Normal approximation.
3) Calculate P(X ≥ 1520) using a Normal approximation.
m = np = 2500(0.60) = 1500
s = np(1 - p) = 2500(0.60)(0.40) = 24.49
z=
1520 -1500
= 0.82
24.49
P(X ³1520) = P(Z ³ 0.82) =1- 0.7939 = 0.2061
Teens and debit cards
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 Example:
In a survey of 506 teenagers aged 14 to 18, subjects were asked if they had a debit card. Suppose that
exactly 10% of teens aged 14 to 18 have debit cards. Let X = the number of teens in a random
sample of size 506 who have a debit card.
1) Verify that the distribution of X is approximately binomial.
B: Yes
I: No, since we are sampling without replacement. However, since the sample size (n = 506) is
much less than 10% of the population, the responses will be very close to independent.
N: n = 506
S: The probability of selecting a teen with a debit card is p = 0.60
2) Check the conditions for using a Normal approximation.
Since np = 506(0.10) = 50.6 and n(1 – p) = 506(0.90) = 455.5 are both at least 10, we may use
the Normal approximation.
3) Calculate P(X ≤ 40) using a Normal approximation.
m = np = 506(0.10) = 50.6
s = np(1 - p) = 506(0.1)(0.9) = 6.75
P(X ≤ 40) = normalcdf(-9999, 40, 50.6, 6.75) = 0.058
Dead Batteries
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 Example:
Almost everyone has a drawer that holds miscellaneous batteries of all sizes. Suppose
that your drawer contains 8 AAA batteries but only 6 of them are good. You need to
choose 4 for your graphing calculator. If you randomly select 4 batteries, what is the
probability that all 4 of the batteries you choose will work? The actual probability is
0.2143, not 0.3164.
P(X = 4) = æç 4 ö÷ (0.75)4(0.25)4 = 0.3164.
è 4ø
Why isn’t the answer 0.3164?
Since we are sampling without replacement, the selections of batteries aren’t
independent. We can ignore this problem if the sample we are selecting is less than
10% of the population. However, in this case, we are sampling 50% of the population,
so it is not reasonable to ignore the lack of independence and use the binomial
distribution. This explains why the binomial probability is so different from the actual
probability.
President Bush’s morning security briefing is wrapping up.
Defense Secretary Donald Rumsfeld is concluding his part and
says, "Finally, three Brazilian soldiers were killed yesterday
near Baghdad."

"OH MY GOD!" shrieks Bush, and he buries his head in his
hands for a seemingly interminable 30 seconds.

Stunned at the unexpected display of emotion, the President's
staff sits speechless, not sure how to react. Finally, Bush looks
up and asks Rumsfeld, "How many is a brazillion?”
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