Download 6 - Cerritos College

Chapter 6
Modeling Random
Events: The Normal
and Binomial Models
Copyright © 2014 Pearson Education, Inc. All rights reserved
Learning Objectives
Be able to distinguish between discrete and
continuous-valued variables.
 Know when a Normal model is appropriate
and be able to apply the model to find
probabilities.
 Know when the binomial model is
appropriate and be able to apply the model to
find probabilities.

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6.1
Probability
Distributions Are
Models of Random
Experiments
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Probability Models and Distributions

A Probability Model is a description of how a
statistician thinks data are produced.




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Uniform
Linear
Normal
Other
A Probability Distribution or Probability
Distribution Function (pdf) is a table graph or
formula that gives all the outcomes of an
experiment and their probabilities.
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Discrete vs. Continuous

A random variable is called Discrete if the outcomes
are values that can be listed or counted.



A random variable is called Continuous if the
outcomes cannot be listed because they occur over a
range.


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Number of classes taken
The roll of a die
Time to finish the exam
Exact weight
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Discrete or Continuous

Classify the following as
discrete or continuous:
of the left thumb →Continuous
 Number of children in a
→Discrete
the family
 Number of devices in the
house that connect to the →Discrete
Internet
 Sodium concentration in
→Continuous
the bloodstream
 Length
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Discrete Probability Distributions
The most common way to display a pdf for
discrete data is with a table.
 The probability distribution table always has
two columns (or rows).

 The
first, x, displays all the possible outcomes
 The second, P(x), displays the probabilities for
these outcomes.
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Examples of Probability Distribution
Tables
Die Roll
Raffle
x
P(x)
Prize
1
1/6
2
1/6
3
1/6
4
1/6
5
1/6
6
1/6
x
P(x)
95
0.01
995
0.005
-5
0.985
The sum of all the probabilities must equal 1.
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Examples of Probability Distribution
Graphs
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Examples of Probability Distribution
Functions

If 1/3 of the population is Hispanic and 20
randomly selected people from the
population are chosen, then the probability
distribution function for the number of
x
20  x
Hispanics selected is
20!
1  2
P( x) 
Probability
distribution functions can be quite
complicated!
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   
x !(20  x)!  3   3 
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Continuous Data and Probability
Distribution Functions

Often represented as a curve.
 The
area under the curve between two values of x
represents the probability of x being between
these two values.
 The total area under the curve must equal 1.
 The curve cannot lie below the x-axis.
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Finding Probabilities for Uniform
Distributions



The curve above shows the probability distribution
function for time to wait for a bus that comes every
12 minutes. Find the probability that you will wait
between 5 and 10 minutes.
Shade in the area.
Find the area of the rectangle

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P(5 < x <10) = Base x Height = 5 x 0.8333 = 0.41655
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6.2
The Normal Model
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The Normal Model

The Normal Model is a good model if:
 The
distribution is unimodal.
 The distribution is approximately symmetric.
 The distribution is approximately bell shaped.

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The Normal Distribution is also called
Gaussian.
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Center and Spread of the Normal
Distribution



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m stands for the center or mean of a distribution.
s stands for the standard deviation of a distribution
Note that the Greek letters m and s are used for
distributions and x and s are used for sample data.
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Notation and Area
N(6,2) means the normal distribution with
mean m = 6 and standard deviation s = 2.
 The area under the normal curve, above the
x-axis, and to the left of x = 4 represents
P(x < 4).
 P(x < 4) = P(x ≤ 4) for a continuous variable.

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Probability and StatCrunch
1.
2.
3.
4.
5.
6.
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Draw a rough sketch of the normal curve
along with the mean and area to be found.
Go to Stat→ Calculators→ Normal.
Type in the mean and
Choose <= or >=, and the Std. Dev.
Type in the value of x.
Click on “Compute”.
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Using StatCrunch for Probability
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Example: Baby Seals
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
Research has shown that the
mean length of a newborn
Pacific harbor seal is 29.5 in.
and that s = 1.2 in. Suppose that
the lengths follow the Normal model. Find
the probability that a randomly selected pup
will be more than 32 in.

P(x > 32) ≈ 0.019
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Finding the Probability Between Two
Values
Since StatCrunch can only
handle probabilities
involving “≤” and “≥”,
use geometry to find the probability of x
falling between two values.
 P(3 < x < 6) = P(x < 6) – P(x < 3)

≈ 0.9772 – 0.1587

= 0.8185

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Finding Probabilities from Percentiles

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Newborn seals in the
bottom 10th percentile
will probably not survive.
Given that new born
seals are N(29.5,1.2),
find the 10th percentile.
With StatCrunch put
in 0.1 for the probability and click on Compute.
The 10th percentile is about 27.96 in.
Copyright © 2014 Pearson Education, Inc. All rights reserved
The Normal Model and the Empirical
Rule
The Empirical Rule told us that if a
distribution is approximately normal, then
68% of the data will fall within 1 standard
deviation of the mean, 95% within 2, and
99.7% within 3.
 If the distribution is exactly normal, then
these numbers are just the corresponding
areas under the normal curve.

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6.3
The Binomial Model
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The Binomial Model

The Binomial Model applies if:
1.
2.
3.
4.
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There are a fixed number of trials.
Only two outcomes are possible for each trial:
Yes or No, Success or Failure, Heads or Tails,
etc.
The probability of success, p, is the same for
each trial.
The trials are independent.
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Binomial or Not?

40 randomly selected college students were asked if
they selected their major in order to get a good job.


35 randomly selected Americans were asked what
country their mothers were born.


Not Binomial, more than two possible answers per trial.
To estimate the probability that students will pass an
exam, the professor records a study group’s success
on the exam.

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Binomial
Not Binomial, since the outcomes are not independent.
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Surveys and Independence
The Binomial Model may be used if the
respondents of a survey are selected with
replacement.
 If the selection is done without replacement
and the population size is at least 10 times
larger than the sample size, then we may still
use the Binomial Model as an approximation.

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Visualizing the Binomial Distribution
 If n is large and p is
close to 0.5, then the
binomial distribution is
approximately normal
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Words and Inequalities
Exactly
 Less Than
 At Least
 More Than
 At Most

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→
→
→
→
→
=
<
=>
>
<=
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 Notice that
“Less Than” and
“At Least” are
complements and
“More Than” and
“At Most” are
Complements.
Finding a Binomial Probability

12% of all US women will eventually develop
breast cancer. If 30 women are randomly
selected, what is the probability that exactly 4 of
them will eventually develop breast cancer?
 b(30,0.12,4)
=?
 StatCrunch: Stat→Calculators→Binomial
 n = 30, p = 0.12, P(x = 4)
 The probability that exactly 4 of the 30 women
will develop breast cancer is about 0.20.
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Finding a Binomial Probability

14% of all clothing bought online is returned. If
an online retailer sells 35 items of clothing, what
is the probability that:
At least 5 will be returned?
 n = 35, p = .14, P(x => 5) ≈ 0.55
 Fewer than 7 will be returned?
 n = 35, p = .14, P(x < 7) ≈ 0.79
 More than 6 will be returned?
 n = 35, p = .14, P(x > 6) ≈ 0.21
 At most 4 will be returned?
 n = 35, p = .14, P(x <= 4) ≈ 0.45

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The Expected Value
If we roll a six sided die 30 times then we
would expect to roll a two 30 x 1/6 = 5 times.
 For a Binomial Distribution, m = np is called
the mean or the expected value.
 A Binomial Distribution with n trials and
probability of success p has standard
deviation

s  np(1  p)
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The Standard Deviation
On any particular day, there is a 6% chance
of a fatal accident in the city. Find the mean
and standard deviation for the number of
accidents in a (365 day) year.
m

= np = 365 x 0.06 = 21.9
s  365  0.06  0.94  4.5
 We
expect about 22 fatal accidents per year give
or take four or five accidents.
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Chapter 6
Case Study
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Too Heavy or Not?
McDonald’s claims that its ice cream cones
weigh 3.18 ounces. However, one of
the authors bought five cones and found that
all five weighed more than that. Is this
surprising?
 Assume the ice cream weights are normally
distributed or at least symmetric about the
mean.

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McDonald’s claims that its ice cream cones
weigh 3.18 ounces. However, one of the authors
bought five cones and found that all five weighed
more than that. Is this surprising?
n = 5, p = 0.5
 P(x = 5) ≈ 0.031
 If the mean is 3.18 ounces as McDonald’s
claims, then there is only a 3.1% chance that out
of 5 randomly chosen cones, they will all weigh
above 3.18 ounces.
 With such a small probability, suspicion is
raised.

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Chapter 6
Guided Exercise 1
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SAT Scores

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According to data from the College Board,
the mean quantitative SAT score for female
college-bound high school seniors in 2009
was 500. SAT scores are approximately
Normally distributed with a population
standard deviation of 100. What percentage
of the female college-bound high school
seniors had scores above 675?
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N(500,100)

To find the z-score for 675, subtract the mean
and divide by the standard deviation. Report
the z-score.
z
xm
s
675  500

100
 1.75
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N(500,100)


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Refer to the Normal curve. Explain why the SAT
score of 500 is right below the z-score of 0. The
dots on the axis mark the location of z-scores that
are integers from -3 to 3.
The mean is always at the center of the Normal
curve. The mean SAT score is 500 and the mean
z-score is 0 by definition.
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N(500,100)


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Carefully sketch a copy of the curve. Pencil in the
SAT scores of 200, 300, 400, 600, and 700 in the
correct places.
Notice that the mean is 500 and the standard
deviation is 100. The z-scores are -3, -2, -1, and 1.
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N(500,100)

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Draw a vertical line through the curve at 675. Just
above the 675 (indicated on the graph with “???”).
Put in the corresponding z-score. We want to find
what percentage of students had scores above 675.
Shade the area to the right of this boundary, because
numbers to the right are larger.
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N(500,100)


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Use StatCrunch to find the area to the right of this
z-score
P(z > 1.75) ≈ 0.04
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N(500,100)
P(z > 1.75) ≈ 0.04
 Finally, write a sentence telling what you
found.
 About 4% of the female college-bound high
school seniors had scores above 675.

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Chapter 6
Guided Exercise 2
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SAT Scores

6 - 45
According to data from the College Board, the mean
quantitative SAT score for female college-bound
high school seniors in 2009 was 500. SAT scores
are approximately Normally distributed with a
population standard deviation of 100. A scholarship
committee wants to give awards to college-bound
women who score at the 96th percentile or above on
the SAT. What score does an applicant need?
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SAT Scores: N(500,100)



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A scholarship committee wants to give awards to
college-bound women who score at the 96th
percentile or above on the SAT. What score does an
applicant need?
Will the SAT test score be above the mean or below
it? Explain.
The 96th percentile is the score such that 96% of all
scores are at or below this score. The 50th percentile
is the mean, so the 96th percentile is above the
mean.
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SAT Scores: N(500,100)

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Label the curve with integer z-scores. The dots
represent the position of integer z-scores from -3 to
3.
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SAT Scores: N(500,100)

Use StatCrunch to find the z-score that has area to
the left 0.96.

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z ≈ 1.75
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SAT Scores: N(500,100) , z ≈ 1.75

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Add that z-score to the sketch and draw a vertical
line above it through the curve. Shade the left side
because the area to the left is what is given.
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SAT Scores: N(500,100) , z ≈ 1.75

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Find the SAT score that corresponds to the z-score.
The score should be z standard deviations above the
mean, so
 x = m + zs
 = 500 + (1.75)(100)
  675
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SAT Scores: N(500,100) ,
z ≈ 1.75, x = 675
6 - 51
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SAT Scores: N(500,100) ,
z ≈ 1.75, x = 675
Finally, write a sentence stating what you
found.
 The applicant needs a score of at least 675 to
receive the scholarship.

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