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Chapter 6
Modeling
Random
Events: The
Normal and
Binomial
Models
Copyright © 2017, 2014 Pearson Education, Inc.
Slide 1
Chapter 6 Topics
• Distinguish between discrete and continuous
random variables
• Apply a Normal model to find probabilities
• Apply a binomial model to find probabilities
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Slide 2
Section 6.1
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PROBABILITY DISTRIBUTIONS
• What Is a Probability Distribution
• Distinguish Between Discrete and Continuous
Random Variables
• Using Probability Distributions to Compute
Probabilities
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Slide 3
Probability Distribution
A table or graph that tells us:
1. All the possible outcomes of a random
experiment
2. The probability of each outcome
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Slide 4
Probability Distribution: Table
Experiment: Rolling a die
x
1
2
3
4
5
6
P(X)
1/6
1/6
1/6
1/6
1/6
1/6
What is the probability
of rolling a 5 or a 6?
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Slide 5
Probability Distribution: Graph
Estimate the probability that a randomly selected
reviewer gave this book a three star rating.
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Slide 6
Two Types of Numerical Variables
1. Discrete Numerical values that you can list or
count
Example: Number of siblings, number of
songs on an MP3 player
2. Continuous Cannot be listed or counted
because they occur over a range of values
Example: Height of basketball players,
weight of a sandwich
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Slide 7
Discrete or Continuous
Label each of these variables as discrete or
continuous:
1. Number of cars owned by a household
2. Number of pets owned by a student
3. Time it takes a worker to commute to a job
site
4. Height of a building in downtown San
Francisco
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Slide 8
Discrete or Continuous
Label each of these variables as discrete or
continuous:
1. Number of cars owned by a household - discrete
2. Number of pets owned by a student - discrete
3. Time it takes a worker to commute to a job site continuous
4. Height of a building in downtown San Francisco continuous
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Slide 9
Two Features of a Probability
Distribution
• All P(X) values must be between 0 and 1.
• The sum of all P(X) values must equal 1.
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Slide 10
Creating a Probability Distribution
Suppose you roll a fair 6-sided die. You win $10
if you roll a 5 or a 6. You lose $5 if you roll a 1.
For any other outcome you win or lose nothing.
Create a table showing the probability
distribution for the amount of money you win
playing this game.
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Slide 11
Probability Distribution: Example
There are three possible outcomes: Win $10,
lose $5, or win/lose nothing.
X = 10, -5, or 0
The probability of rolling a 5 or 6 (and winning
$10) is 2/6.
The probability of rolling a 1 (and losing $5) is
1/6.
You win/lose nothing by rolling a 2, 3, or 4
(probability = 3/6).
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Slide 12
Probability Distribution: Example
X
-5
0
10
P(X)
1/6
3/6
2/6
Note: All P(X) values are between 0 and 1, and the
sum of all P(X) values = 1.
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Slide 13
Probability Distributions for
Continuous Random Variables
Because you cannot list all outcomes for a
continuous random variable, we cannot
represent the probability distribution using a
table.
Graphs, called probability density curves, are
used instead.
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Slide 14
Probability Density Curve: Example
This probability density curve shows the probability of
wait times before service at a coffee shop. The shaded
area represents the probability of a wait time between
0 and 2 minutes. The total area under the curve = 1.
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Slide 15
Finding Probabilities: Uniform Distributions
A bus arrives at a certain stop every 12 minutes. The
graph shows the probability distribution for wait times
before the bus arrives. Use the distribution to find the
probability of a wait time between 4 and 10 minutes.
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Slide 16
Finding Probabilities: Uniform Distributions
The shaded area represents the probability of a wait
time between 4 and 10 minutes. We can find the area
of this rectangle: (length x height).
6 x 0.08333 = 0.49998 ≈ 0.5000
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Slide 17
Finding Areas Under Curves
In general, finding areas under curves that are
not geometric shapes like rectangles is difficult
and requires using some type of computerbased technology.
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Slide 18
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Section 6.2
THE NORMAL MODEL
• Finding Probabilities Using the Normal Model
• Finding Values for a Random Variable Given a
Normal Probability
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Slide 19
The Normal Model
• One of the most widely used probability
models for continuous numerical random
variables
• Also called the “bell curve” because of its
shape
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Slide 20
Visualizing the Normal Model
This histogram shows heights for a sample of
males. Since it is symmetric and unimodal, it
can be modeled using a Normal distribution.
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Slide 21
The Normal Distribution: Same Mean,
Different Standard Deviation
These two Normal distributions have the same mean
but different standard deviations.
Note: The area under both curves = 1.
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Slide 22
The Normal Distribution: Different
Means, Same Standard Deviation
These normal curves show the distribution of heights for
adult men and women in the US. Notice that the spread
of each curve (the standard deviation) is about the same,
but that the center of each curve (the mean) is different.
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Slide 23
Notation
• We use the symbol μ (“mu”) to represent the
mean of a probability distribution.
• We use the symbol σ (“sigma”) to represent
the standard deviation of a probability
distribution.
• The notation N(μ,σ) represents a Normal
distribution with mean μ and standard
deviation σ.
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Slide 24
Finding Probability with Technology
• It is always helpful to sketch a picture of the
area you are interested in finding before using
technology to find the area.
• Most technologies require you to enter the
mean and standard deviation of the
distribution, as well as the beginning and
ending x-values of the area you are finding in
the Normal model.
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Slide 25
Using the TI-84 Calculator
To find a probability for a Normal distribution on the
TI-84 calculator:
1. Push 2ND Dist then select option 2: normalcdf.
2. Enter the left boundary of your shaded area, the
right boundary of your shaded area, the mean,
and the standard deviation and press ENTER.
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Slide 26
Example: Women’s Heights
The Normal model N(64, 3) gives a good
approximation of adult women’s heights (in
inches) in the US. Find the probability that a
randomly selected adult woman from the US is
between 62 and 67 inches tall.
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Slide 27
Example: Women’s Heights
To find this area using a TI-84 calculator:
normalcdf(62, 67, 64, 3) = 0.589
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Slide 28
Using StatCrunch
Research has shown that the distribution of
newborn Pacific harbor seals’ birth lengths is
approximately Normal with a mean of 29.5
inches and a standard deviation of 2.5 inches.
Find the probability that a randomly selected
harbor seal pup is longer than 31 inches.
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Slide 29
Using StatCrunch
1.
2.
3.
4.
Select STAT > Calculators > Normal.
Enter the mean and standard deviation.
Enter x ≥ 31.
Press COMPUTE.
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Slide 30
Seal Pups
P(x ≥ 31) = 0.274
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Slide 31
Without Technology
• To find Normal probabilities without
technology, we convert the information in the
problem to standard units and find the
probability using a standard Normal table.
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Slide 32
Standard Normal Model
The Standard Normal Model is a normal
distribution with a mean = 0 and a standard
deviation = 1.
N(0, 1)
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Slide 33
Standard Units or “Z-Scores”
• Tell us how many standard deviations from
the mean an observation lies.
xx
z
s
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Slide 34
Seal Pups: Using a Table
Small newborn seal pups have a lower chance of
survival than larger newborn pups. Suppose the
length of newborn seal pups follow a Normal
distribution with a mean of 29.5 inches and a
standard deviation of 1.2 inches. What is the
probability that a randomly selected newborn
seal pup is shorter than 28.0 inches?
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Slide 35
Seal Pups: Using a Table
Convert the information in the problem into a
standard score or z-score:
28 - 29.5 -1.5
z=
=
= -1.25
1.2
1.2
Now look up –1.25 on the Normal table.
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Slide 36
Going “Backwards”:
Finding x-values Given a Probability
Sometimes we are interested in finding the
value for the random variable associated with a
given probability.
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Slide 37
Example: Seal Pups
The birth lengths of newborn seal pups follow
the model N(29.5, 1.2). Suppose pups with
lengths at the 15th percentile and below are
unlikely to survive. What birth lengths would
make it unlikely that a pup would survive?
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Slide 38
Using StatCrunch
The shaded area = 0.15.
The x-value that
corresponds with the
15th percentile is
x = 28.3 inches.
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Slide 39
Using the TI-84 Calculator
To find an x-value for a Normal distribution on the
TI-84 calculator given an area:
1. Push 2nd DIST then select option 3: invNorm.
2. Enter the area to the left of the x-value of
interest, the mean, and the standard deviation.
To find the 15th percentile for the seal pup lengths,
invNorm(0.15, 29.5, 1.2) = 28.3 inches.
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Slide 40
Normal or Inverse Normal?
It is known that the heights of adult men follow a
Normal model. For each of the following situations,
decide if it is a Normal or Inverse Normal problem.
1. A clothing store manager wonders what
percentage of her customers are taller than six
feet.
2. A clothing store manager wants to cater to the
tallest 20% of men and wants to know what
heights she should accommodate.
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Slide 41
Normal or Inverse Normal?
1. A clothing store manager wonders what
percentage of her customers are taller than six
feet. This is a Normal problem.
2. A clothing store manager wants to cater to the
tallest 20% of men and wants to know what
heights she should accommodate. This is an
Inverse Normal problem.
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Slide 42
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Section 6.3
THE BINOMIAL MODEL
• Finding Probabilities Using the Binomial Model
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Slide 43
Binomial Probability Model
• Useful for many situations with discrete
numerical variables
Look for:
1. A fixed number of trials (n).
2. Only two outcomes possible at each trial.
3. The probability (p) of a success is the same at
each trial.
4. The trials are independent.
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Slide 44
Example: Tossing a Coin
Suppose you toss a coin eight times and count
the number of heads. Explain why this is an
example of a binomial experiment.
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Slide 45
Binomial Experiment:
Tossing a Coin
1. There is a fixed number of trials (8).
2. Each trial has only two outcomes (heads or
tails).
3. The probability of a landing on heads is the
same at each trial (1/2).
4. Trials are independent (outcome of one toss
does not affect the outcome of any other
toss).
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Slide 46
Binomial or Not Binomial
For each of the following situations, identify
which are binomial. For the ones that are not
binomial, explain why it is not binomial.
1. Record the ages of a group of 20 randomly
selected statistics students.
2. Ask each partner of a married couple
whether or not they exercise regularly.
3. Ask a random sample of workers whether
their annual salary is greater than $50,000.
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Slide 47
Binomial or Not Binomial
1. Record the ages of a group of 20 randomly
selected statistics students. Not binomial (each
trial does not have only two possible outcomes).
2. Ask each partner of a married couple whether or
not they exercise regularly. Not binomial (since
the couple is married, the trials are not
independent).
3. Ask a random sample of workers whether their
annual salary is greater than $50,000. Binomial
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Slide 48
Shape of a Binomial Distribution
• Depends on both n and p
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Shape of a Binomial Distribution
• Symmetric if p = 0.5, but also when n is large
even if p is close to 0 or 1.
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Slide 50
Finding Binomial Probabilities
• Binomial probabilities can be found by hand
using the Binomial formula if n and x are
relatively small.
• In most applications, technology that has the
binomial distribution built in is used to find
binomial probabilities.
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Slide 51
Example: Finding Binomial
Probabilities for 1 Value of x
According to a recent Pew poll, 47% of American
say it is important to control gun ownership. In
a random sample of 200 Americans, find the
probability that exactly 90 Americans say it is
important to control gun ownership.
Source: www.Pewresearch.org
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Slide 52
Example: Gun Control
Check that this is a binomial problem:
1. Fixed number of trials (n = 200)
2. Only two outcomes for each trial (Is it
important to control gun ownership? Yes or no)
3. Probability of a “success” remains the same
for each trial (p = 0.47)
4. Trials are independent
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Slide 53
Notation: Binomial Probability
In this problem we are looking for the binomial
probability where n = 200, p = 0.47, and x = 90.
We use the notation b(200, 0.47, 90) to denote
this probability.
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Slide 54
Finding Binomial Probabilities:
Single Value of x Using StatCrunch
Use STAT > Calculators > Binomial
Enter n, p, and
x=90.
P(X=90) = 0.048
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Slide 55
Using the TI-84 Calculator
To find a binomial probability for a single value of x on
the TI-84 calculator:
1. Push 2ND Dist then select option binompdf.
2. Enter binompdf(n, p, x).
binompdf(200, 0.47, 90) = 0.048
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Slide 56
Finding Binomial Probabilities for
More Than One Value of x
According to the Bureau of Labor Statistics, the
unemployment rate in Detroit in June 2014 was
10.2%. If a random sample of 15 people living in
Detroit at that time was taken, find the
probability that three or fewer were
unemployed.
Source: www.bls.gov
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Slide 57
Example: Unemployment in Detroit
This problem is binomial because:
1. There is a fixed number of trials (n = 15).
2. There are only two outcomes for each trial
(Unemployed? Yes/no).
3. The probability of a “success” is the same for
each trial (p = 0.102).
4. The trials are independent.
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Slide 58
Example: Unemployment in Detroit
n = 15, p = 0.102
“Three or fewer” means x = 0, 1, 2, or 3
We need to compute:
b(15, 0.102, 0)+ b(15, 0.102,1)+ b(15, 0.102, 2)+ b(15, 0.102,3)
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Slide 59
Probability Cumulative
Density Functions
Instead of computing four binomial probabilities
separately and adding them together, we can
take advantage of the probability cumulative
density functions built into certain technologies.
These functions add together probabilities for a
range of x values.
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Slide 60
Binomial Cumulative Density Function:
StatCrunch
In the unemployment problem for a random
sample of 15 people from Detroit in June 2014,
n = 15, p = 0.102, and x = 0, 1, 2, 3.
In StatCrunch, use
STAT > Calculators > Binomial
Enter n, p, and set x ≤ 3.
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Slide 61
Cumulative Probabilities Using
StatCrunch
The sum of the red bars
represents P(X≤3).
P(X≤3) = 0.941
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Slide 62
Using the TI-84 Calculator
To find a binomial probability involving more than 1
value of x on the TI-84 calculator:
1. Push 2ND Dist then select option binomcdf.
2. Enter binomcdf(n, p, x).
NOTE: The binomcdf command always adds probabilities
starting with x = 0. Finding the sum of probabilities for x
values not starting at 0 requires subtracting several binomcdf
calculations.
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Slide 63
Binomial Cumulative Density Function:
TI-84 Calculator
In the unemployment problem for a random
sample of 15 people from Detroit in June 2014,
n = 15, p = 0.102, and x = 0, 1, 2, 3.
NOTE: We want to add probabilities starting
with x = 0, so we can use the binomcdf
command.
binomcdf(15, 0.102, 3) = 0.941
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Slide 64
Being Careful with Language
Because the binomial probability distribution
models the probability of discrete random
variables, we have to pay close attention to
language.
“More than 5” mean x = 6, 7, 8, …, n
“5 or more” mean x = 5, 6, 7, 8, …, n
NOTE: The second case includes b(n, p, 5) in the sum while the
first case does not.
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Slide 65
A Difference between the
Normal and Binomial Distributions
We did not need to worry about this distinction
with Normal probabilities because for
continuous numerical variables, the probability
of getting five or more is the same as the
probability of getting more than five.
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Slide 66
Practice Identifying X-Values
What x value(s) are implied in each of these
phrases?
1. More than 6
2. Fewer than 5
3. At least 7
4. Between 3 and 7 (inclusive)
5. 10 or more
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Slide 67
Practice Identifying X-Values
1.
2.
3.
4.
5.
More than 6
x = 7, 8, 9, …, n
Fewer than 5
x = 0, 1, 2, 3, 4
At least 7
x = 7, 8, 9, …, n
Between 3 and 7 (inclusive) x = 3, 4, 5, 6, 7
10 or more
x = 10, 11, 12, …, n
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Slide 68
Example: Medical Coverage
According to a report in the Sacramento Bee, only 11% of
Californians lacked medical insurance after the passage of
the Affordable Care Act (down from a previous rate of 17%).
Suppose a sample of 80 Californians was taken after the
passage of the Affordable Care Act. Find the probability
that:
a. More than 15 lack medical insurance.
b. Between 10 and 20 (inclusive) lack medical insurance.
Source: www.sacbee.com
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Slide 69
Example: Medical Coverage
a. More than 15 lack medical insurance
n = 80, p = 0.11, x =16, 17, …,
80
NOTE: x > 15
P(X>15) = 0.013
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Slide 70
Example: Medical Coverage
b. Between 10 and 20 (inclusive) lack medical
insurance.
n = 80, p = 0.11, x = 10, 11,
…, 20
P(10≤x≤20) = 0.384
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Slide 71
The Mean of a Binomial Distribution
• Sometimes called the “expected value”
• Can be thought of as the number of
“successes” we would expect if the
experiment were repeated n times
• As with all distributions, tells us where the
distribution “balances”
• Formula:
m = np
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Slide 72
Example: Mean of a Binomial
Experiment
Suppose you toss a fair coin 15 times and count
the number of heads. Find and interpret the
mean of the probability distribution.
Mean = 15(0.5) = 7.5
We expect between 7 and 8 heads in 15 tosses.
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Slide 73
Example: Mean of a Binomial
Experiment
Note: The mean represents the balancing point of the distribution.
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Slide 74
The Standard Deviation of a Binomial
Distribution
• Measures the spread of a distribution
• Formula: s = np(1- p)
• Can help us approximate a range of values we
would expect and a range of values we would
find surprising in a binomial distribution
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Slide 75
Example: Standard Deviation of a
Binomial Distribution
During his career, major league baseball player
Buster Posey of the San Francisco Giants has a
batting average of 0.308. This means that he
gets a hit about 30.8% of the time when he bats.
If he bats 480 times in a season, we would
expect him to get, give or take, how many hits?
Would it be unusual if he got 120 hits?
Source: www.baseball-reference.com
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Slide 76
Example: Buster Posey
Expected value: μ = 480(0.308) = 147.8
Standard deviation:
  480(0.308)(1  0.308)  10.1
We expect Posey to get 147.8 ± 10.1 hits.
147.8 – 10.1 = 137.7
147.8 + 10.1 = 157.9
We expect Posey to get between 137.7 – 157.9
hits.
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2017,2014
2014Pearson
PearsonEducation,
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Inc.
Slide 77
Example: Buster Posey
Would it be unusual if he got 120 hits?
We know 120 is beyond one standard deviation
below the mean. To determine if this would be
surprising we can compute a z-score:
120 -147.8
z=
= -2.75
10.1
Yes, this would be surprising since is it over two
standard deviations below the mean.
Copyright
Copyright©©2017,
2017,2014
2014Pearson
PearsonEducation,
Education,Inc.
Inc.
Slide 78