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© 2012 McGraw-Hill
Ryerson Limited
© 2009 McGraw-Hill Ryerson Limited
1
Lind
Marchal
Wathen
Waite
© 2012 McGraw-Hill Ryerson Limited
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Learning Objectives
LO 1 Explain the difference between discrete and
LO 2
LO 3
LO 4
LO 5
continuous probability distributions.
Compute the mean and the standard deviation for a
uniform probability distribution.
Compute probabilities by using the uniform probability
distribution.
List the characteristics of the normal probability
distribution.
Define and calculate z-values.
© 2012 McGraw-Hill Ryerson Limited
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Learning Objectives
LO 6 Determine the probability an observation is between
two points on a normal probability distribution.
LO 7 Determine the probability an observation is above or
below a point on a normal probability distribution.
LO 8 Determine the value of a normally distributed random
variable for a given probability.
LO 9 Use the normal probability distribution to approximate
the binomial probability distribution.
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LO
1
Difference Between Discrete and
Continuous Probability
Distribution
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Types of Continuous Probability Distributions
We consider two families of continuous probability
distributions:
1.The uniform probability distribution
2.The normal probability distribution
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© 2012 McGraw-Hill Ryerson Limited
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Continuous Probability Distributions
Usually result from measuring something.
Usually interested in information such as the percentage of
data above or below a certain point; or the percentage of
data in a certain range.
Important continuous probability distribution is the normal
probability distribution.
• Describes the likelihood that a continuous random
variable with an infinite number of possible values lies
within a specified range.
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LO
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The Mean and the Standard
Deviation for a Uniform
Probability Distribution
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MEAN OF THE UNIFORM DISTRIBUTION
The mean of a uniform distribution is located in the middle
of the interval between the minimum and maximum values.
It is computed as:
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Standard Deviation of The Uniform Distribution
The standard deviation describes the dispersion of a
distribution. In the uniform distribution, the standard
deviation is also related to the interval between the
maximum and minimum values.
The equation for the uniform probability distribution is:
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LO
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Compute Probabilities by Using
the Uniform Probability
Distribution
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Example – Uniform Distribution
The time Sue takes to travel from her home to evening
class is uniformly distributed between 25 to 35 minutes.
1. Draw a graph of this distribution.
2. Show that the area of this uniform distribution is 1.00.
3. For how long will Sue “typically” have to travel? In other
words what is the mean travelling time? What is the
standard deviation of the travelling time?
4. What is the probability Sue will travel for more than 33
minutes?
5. What is the probability Sue will travel between 28 to 32
minutes?
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Solution – Uniform Distribution
In this case, the random variable is the length of time Sue
travels. Time is measured on a continuous scale, and the
travel times may range from 25 minutes up to 35 minutes.
1. The graph of the uniform distribution is shown below.
The horizontal line is drawn at a height of .1, found by
1
. The range of this distribution is 10 minutes.
35  25
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Solution – Uniform Distribution
Uniform Distribution
Probability
0.2
0.1
0
25
30
35
Length of Travel (Minutes)
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Solution – Uniform Distribution
Continued
2. The times Sue must travel is uniform over the interval
from 25 minutes to 35 minutes, so in this case
a is 25 and b is 35.
æ 1 ö
Area = height base = ç
35 - 25 = 1.00
÷
è 35 - 25 ø
(
)(
)
(
)
3. To find the mean, we use formula (6–1).
a  b 25  35


 30
2
2
To find the standard deviation of the travel times, we
use formula (6–2).
 b  a    35  25   2.89
12
12
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 
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2
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Solution – Uniform Distribution
Continued
4. From the area formula:
æ 1 ö
P ( 33 £ travel time £ 35) = ( height ) ( base ) = ç
2) = 0.2
(
÷
è 35 - 25 ø
This conclusion is illustrated by the following graph.
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Solution – Uniform Distribution
Continued
5. The area within the distribution for the interval 28 to 32
represents the probability.
æ 1 ö
P ( 28 £ travel time £ 32 ) = ( height ) ( base ) = ç
4 ) = 0.4
(
÷
è 35 - 25 ø
We can illustrate this probability as follows.
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© 2012 McGraw-Hill Ryerson Limited
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You Try It Out!
The life of a light bulb follows a uniform distribution between
10 and 16 months.
(a) Draw this uniform distribution. What are the height and
base values?
(b) Show the total area under the curve is 1.00.
(c) Calculate the mean and the standard deviation of this
distribution.
(d) What is the probability that a light bulb will work for
between 12 to 14 months?
(e) What is the probability that it will work for less than 11
months?
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© 2012 McGraw-Hill Ryerson Limited
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LO
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Characteristics of the Normal
Probability Distribution
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Normal Probability Distribution Characteristics
1. It is bell-shaped and has a single peak at the centre of
the distribution.
a) The arithmetic mean, median, and mode are equal.
b) The total area under the curve is 1.00; half the area
under the normal curve is to the right of this centre
point and the other half to the left of it
2. It is symmetrical about the mean.
3. It is asymptotic: The curve gets closer and closer to the
X-axis but never actually touches it.
a) To put it another way, the tails of the curve extend
indefinitely in both directions.
4. The location of a normal distribution is determined by
the mean (µ), and the dispersion or spread of the
distribution is determined by the standard deviation (σ).
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The Normal Distribution Graphically
Characteristics of a normal distribution are shown
graphically below:
Characteristics of a Normal Distribution
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Equal Means but Different Standard Deviations
Normal Probability Distributions with
Equal Means but Different Standard Deviations
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Different Means but Equal Standard Deviations
Normal Probability Distributions with Different Means
but Equal Standard Deviations
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Different Means and Standard Deviations
Normal Probability Distributions with Different Means
and Standard Deviations
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LO
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The Standard Normal
Probability Distribution
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The Standard Normal Distribution
The standard normal distribution is a normal distribution
with a mean of 0 and a standard deviation of 1.
It is also called the z distribution.
A z-value is the signed distance between a selected value,
designated X, and the population mean µ, divided by the
population standard deviation, σ.
where:
X is the value of any particular observation or measurement.
 is the mean of the distribution.

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is the standard deviation of the distribution.
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LO
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Applications of the Standard
Normal Distribution
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The Standard Normal Value & Tables
Suppose we wish to compute the probability that boxes of
Sugar Yummies have a weight between 283 g and 285.4 g.
From Chart 6–3, we know that the box weight of Sugar
Yummies follows the normal distribution with a mean of
283 g and a standard deviation of 1.6 g. We want to know
the probability or area under the curve between the mean,
283 g, and 285.4 g.
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The Standard Normal Value & Tables
We can also express this problem using probability notation,
similar to the style used in the previous chapter:
P(283 < mass < 285.4).
P  283  weight  285.4 
285.4  283 
 283  283
 P
z
  P  0  z  1.50   0.4332
1.6
 1.6

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The Standard Normal Value & Tables
Z
0.00
0.01
0.02
0.03
0.04
0.05
1.3
0.4032
0.4049
0.4066
0.4082
0.4099
0.4115
1.4
0.4192
0.4207
0.4222
0.4236
0.4251
0.4265
1.5
0.4332
0.4345
0.4357
0.4370
0.4382
0.4394
1.6
0.4452
0.4463
0.4474
0.4484
0.4495
0.4505
1.7
0.4554
0.4564
0.4573
0.4582
0.4591
0.4599
1.8
0.4641
0.4649
0.4656
0.4664
0.4671
0.4678
1.9
0.4713
0.4719
0.4726
0.4732
0.4738
0.4744
Areas under the Normal Curve
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The Standard Normal Value & Tables
The values of 285.4 g and a z score of 1.50 are the same
distance from µ = 283 and z = 0, respectively.
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© 2012 McGraw-Hill Ryerson Limited
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Example – Calculating a z-value
The monthly returns of certain mutual funds are normally
distributed with a mean of $3000 and a standard deviation
of $300. What is the z-value for return of $3200 on a
mutual fund? For return of $2800 on a mutual fund?
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Solution – Calculating a z-value
The z-values for the two X values ($3200 and $2800) are:
For X  $3200 :
X 
z

$3200  $3000

$300
 0.6667
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For X  $2800 :
X 
z

$2800  $3000

$300
 0.6667
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You Try It Out!
From a certain source it has been found that mean for
creativity is 12.85 and the standard deviation is 3.66.
Convert:
a) The raw creativity score of 7 to a z-value.
b) The raw creativity score of 14 to a z-value.
LO
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© 2012 McGraw-Hill Ryerson Limited
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The Empirical Rule
For a symmetrical, bell-shaped frequency
distribution:
• Approximately 68 percent of the observations will lie
within ±1 standard deviation of the mean.
• About 95 percent of the observations will lie within
±2 standard deviations of the mean.
• Practically all (99.7 percent) will lie within ±3
standard deviations of the mean.
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The Empirical Rule
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The Empirical Rule
This information is summarized in the following graph.
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Example – The Empirical Rule
A training department has conducted a refresher test for
trainee employees. The mean of the score of employees is
34.5 and the standard deviation is 4.8.
1. µ ± 1σ of the employees’ score between what two
values?
2. µ ± 2σ of the employees’ score between what two
values?
3. µ ± 3σ of the employees’ score between what two
values?
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Solution – The Empirical Rule
We can use the results of the Empirical Rule to answer
these questions.
1. µ ± 1σ of the employees’ score between 39.3 and 29.7
found by 34.5 ± 1(4.8).
2. µ ± 2σ of the employees’ score between 44.1 and 24.9,
found by 34.5 ± 2(4.8).
3. µ ± 3σ of the employees’ score between 48.9 and 20.1,
found by 34.5 ± 3(4.8).
This information is summarized on the following chart.
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20.1
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µ +3σ
µ +2σ
µ +1 σ
µ
µ - 1σ
µ -2σ
µ - 3σ
Solution – The Empirical Rule
24.9 29.7 34.5 39.3 44.1 48.9
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You Try It Out!
Adult women’s heights are normally distributed with
μ = 65.5 inches and σ = 2.5 inches.
a)μ ± 1σ of the women’s heights lie between what two
values?
b)μ ± 2σ of the women’s heights lie between what two
values?
c)μ ± 3σ of the women’s heights lie between what two
values?
d)What are the median and the modal heights?
e)Is the distribution of height symmetrical?
LO
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© 2012 McGraw-Hill Ryerson Limited
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LO
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Finding Areas Under the
Normal Curve
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Finding Areas Under the Normal Curve
The applications of the standard normal distribution
involves finding the area in a normal distribution:
1. Between the mean and a selected value, which we
identify as x.
2. Beyond x.
3. Between two points on different sides of the mean.
4. Between two points on the same side of the mean.
5. OR finding the value of the observation X when the
percent above or below the observation is given.
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© 2012 McGraw-Hill Ryerson Limited
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Example – Between the mean and X
In a gift store, there are different gift items with different
prices. The mean value of these items is $1500 and the
standard deviation is $155. What is the likelihood of
selecting at random an item priced between $1500 and
$1800?
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Solution – Between the mean and X
For X  $1500 :
X 
$1500  $1500
z

 0.00

$155
For X  $1800 :
X 
$1800  $1500
z

 1.93

$155
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Solution – Between the mean and XContinued
The area under the normal curve between $1500 and
$1800 is 0.4732. We estimate that 47.32 percent of the
items range between $1500 and $1800. Or, the likelihood
of selecting an item priced between $1500 and $1800 is
0.4732.
LO
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LO
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Value of a Normally Distributed
Random Variable for a Given
Probability
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Normally distributed random variable for a given
probability
We find the probability of selecting an item costing between
the mean of $1500 and $1800. This probability is 0.4732.
Next, recall that half the area, or probability, is above the
mean and half is below. So, the probability of selecting an
item costing less than $1500 is 0.5000. Then add the two
probabilities: 0.4732 + 0.5000 = 0.9732. So 97.32 percent
of the items in the store cost less than $1500.
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Between the mean and X In Excel
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Example – Beyond X
We reported that the mean price of the gift items in the
store is normally distributed with a mean of $1500 and a
standard deviation of $155. What is the probability of
selecting an item that costs less than $1290?
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Solution – Beyond X
z
X 


$1290  $1500
 1.35
$155
To find the area below –1.35,
subtract from 0.50 the area from –1.35 to 0
= 0.50 – 0.4115 = 0.0885
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Solution – Beyond X
Continued
This means that 41.15 percent of the items cost between
$1290 and $1500. Further, we can anticipate that 0.885
percent of the items are less than $1290. This information
is summarized in the following diagram.
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Beyond X In Excel
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© 2012 McGraw-Hill Ryerson Limited
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You Try It Out!
A telephone company has found that the lengths of its
long distance telephone calls are normally distributed,
with a mean of 230 seconds and a standard deviation of
60 seconds.
a)What is the area under the normal curve between 230
and 255 seconds? Write this area in probability
notation.
b)What is the area under the normal curve for greater
than 255 seconds? Write this area in probability
notation.
c)Show the details of this problem in a chart.
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© 2012 McGraw-Hill Ryerson Limited
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Example – Between Two Points on Different Sides
We reported that the mean price of the items in the store is
$1200 and the standard deviation is $155. What is the
likelihood of selecting an item that costs between $1340
and $1700?
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Solution – Between Two Points on Different Sides
For the area between $1240 and the mean of $1500:
$1340  $1500 $160
z

 1.03
$155
$155
For the area between the mean of $1500 and $1700:
z
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$1700  $1500 $200

 1.29
$155
$155
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Solution – Between Two Points on Different Sides
Continued
The area under the curve for a z of –1.03 is 0.0517. The
area under the curve for a z of 1.29 is 0.4015. Adding the
two areas: 0.3485 + 0.4015 = 0.7500. Thus, the probability
of selecting an article whose price is between $1340 and
$1700 is 0.7500.
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© 2012 McGraw-Hill Ryerson Limited
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Example – Between Two Points on The Same
Side
We reported that the mean price of the gift article in a store
is normally distributed with a mean of $1500 and a
standard deviation of $155. What is the likelihood of
selecting a supervisor whose weekly income is between
$1650 and $1750?
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Solution – Between Two Points on The Same Side
For x = 1650
$1650  $1500
z
$155
z  0.9677
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For x = 1750
$1750  $1500
z
$155
 1.61
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Summary of Normal Curve Applications
1. To find the area between 0 and z (or –z), look up the
probability directly in the table.
2. To find the area beyond z or (–z), locate the probability
of z in the table and subtract that probability from
0.5000.
3. To find the area between two points on different sides
of the mean, determine the z-values and add the
corresponding probabilities.
4. To find the area between two points on the same side
of the mean, determine the z-values and subtract the
smaller probability from the larger.
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© 2012 McGraw-Hill Ryerson Limited
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You Try It Out!
A psychologist finds that the intelligence quotients of a
group of patients are normally distributed, with a mean
of 104 and a standard deviation of 26.
a) What percent of the patients have intelligence
quotients between 89 and 120? Draw a normal
curve and shade the desired area on your diagram.
b) What percent of the patients have intelligence
quotients between 110 and 120? Draw a normal
curve and shade the desired area on your diagram.
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Example – Finding X When the Percentage is
Given
The Flash Bulbs Company wishes to set a minimum
duration guarantee on its new range of bulbs. Tests reveal
the mean number of minutes is 107 000 with a standard
deviation of 3100 minutes and that the distribution of
minutes follows the normal distribution. Flash Bulbs
Company wants to set the minimum guaranteed number of
minutes so that no more than 3 percent of the bulbs will
have to be replaced. What minimum guaranteed minutes
should Flash Bulbs announce?
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Solution – Finding X When the Percentage is
Given
The details of the problem are shown in the below diagram,
where X represents the minimum guaranteed number of
minutes.
Inserting the mean and the standard deviation for z:
z
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X 

X  107000

3100
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Solution – Finding X When the Percentage is
Given
Continued
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Solution – Finding X When the Percentage is
Given
Continued
Knowing that the distance between µ and X is –1.88σ or
z = –1.88, we can now solve for X:
X -107000
z=
3100
X -107000
-1.88 =
3100
-1.88 ( 3100 ) = X -107000
X = 107000 -1.88 ( 3100 ) = 101172
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Finding X When the Percentage is Given In Excel
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You Try It Out!
An analysis of performance of the employees in a
month follows normal distribution. The mean of the
distribution is 95 and the standard deviation is 9. The
team leader decided to award an A category to the
employees whose performance are in the highest 20
percent. What is the dividing point for those employees
who earn an A and those earning a B?
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LO
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The Normal Approximation to the
Binomial Distribution
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The Normal Approximation to the Binomial
As n gets large, the binomial distribution gets timeconsuming to use.
As n increases, a binomial distribution gets closer and
closer to a normal distribution.
n=1
0.50
0.40
0.2
0.40
p(x)
n = 20
n=3
0.30
0.15
0.30
0.20
0.1
0.20
0.10
0.10
x
0.00
0
1
Number of
occurrences
0.05
x
0.00
0
1
2
3
Number of occurrences
x
0
0
2
4
6 8 10 12 14 16 18 20
Number of occurrences
Binomial Distributions for an n of 1, 3, and 20,
where p = .50
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The Normal Approximation to the Binomial – When
To Use
When np and n(1 – p) are both at least 5.
All binomial criteria are met:
1. There are only two mutually exclusive outcomes to an
experiment: a “success” and a “failure”
2. The distribution results from counting the number of
successes in a fixed number of trials.
3. Each trial is independent.
4. The probability, p, remains the same from trial to trial.
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The Continuity Correction Factor
The value 0.5 subtracted or added, depending on the
question, to a selected value when a discrete probability
distribution is approximated by a continuous probability
distribution.
Only four cases may arise. These cases are:
1. For the probability that at least X occur, use the area
above (X – 0.5).
2. For the probability that more than X occur, use the area
above (X + 0.5).
3. For the probability that X or fewer occur, use the area
below (X + 0.5).
4. For the probability that fewer than X occur, use the area
below (X – 0.5).
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Example – Normal Approximation of the Binomial
Suppose the management of the drama theatre found that
70 percent of its new audience return for another drama.
For a week in which 80 new (first-time) audiences visited
theatre, what is the probability that 60 or more will return for
another drama?
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Solution – Normal Approximation of the Binomial
Step 1. Find the z corresponding to an X of 59.5 using
formula (6–4), and formulas (5–4) and (5–5) for the
mean and the variance of a binomial distribution:
  np  80  0.70   56.00
 2  np 1  p   80  0.70  (0.30)  16.8
  16.8  4.10
X   59.5  56
z

 0.85

4.10
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Solution – Normal Approximation of the Binomial
Continued
Step 2. Determine the area under the normal curve
between a µ of 56 and an X of 59.5. From step 1,
we know that the z-value corresponding to 59.5 is
0.85. So we go to Appendix A.1 and read down the
left margin to 0.8, and then we go horizontally to the
area under the column headed by 0.05. That area
is 0.3023.
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Solution – Normal Approximation of the Binomial
Continued
Step 3. Calculate the area beyond 59.5 by subtracting
0.3023 from 0.5000, that is, 0.5000 – 0.3023 =
0.1977. Thus, 0.1977 is the probability that 60 or
more first-time drama audience out of 80 will return
for another drama. In probability notation:
P(audience > 59.5) = 0.5000 – 0.3023 = 0.1977
The details of this problem are shown graphically:
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You Try It Out!
A survey revealed that 85 percent of customers in a
certain store bargain while shopping.
a) During a period in which 300 customers visited the
store and 155 items were sold, what is the
probability that the customers bargained?
b) During a period in which 300 customers visited the
store and 180 items were sold, what is the
probability that the customers bargained?
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Chapter Summary
I. The uniform distribution is a continuous probability
distribution with the following characteristics.
A. It is rectangular in shape.
B. The mean and the median are equal.
C. It is completely described by its minimum value a
and its maximum value b.
D. It is also described by the following equation for the
region from a to b:
1
[6–3]
P  x 
ba
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Chapter Summary
E. The mean and standard deviation of a uniform
distribution are computed as follows:
ab
[6–1]

2
 
b  a 
2
12
[6–2]
II. The normal probability distribution is a continuous
probability distribution with the following characteristics.
A. It is bell-shaped and has a single peak at the centre
of the distribution.
B. The distribution is symmetrical.
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Chapter Summary
C. It is asymptotic, meaning the curve approaches but
never touches the X-axis.
D. It is completely described by the mean and standard
deviation.
E. There is a family of normal distributions.
1. Another normal probability distribution is created
when either the mean or the standard deviation
changes.
2. The area under a normal curve expresses the
probability of an outcome.
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Chapter Summary
III. The standard normal probability distribution is a
particular normal distribution.
A. It has a mean of 0 and a standard deviation of 1.
B. Any normal distribution can be converted to the
standard normal distribution by the following
X 
formula.
z

[6–4]
C. By standardizing a normal distribution, we can report
the distance from the mean in units of the standard
deviation.
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Chapter Summary
IV. The normal distribution can approximate a binomial
distribution under certain conditions.
A. np and n(1 – p) must both be at least 5.
1. n is the number of observations.
2. p is the probability of a success.
B. The four conditions for a binomial distribution are:
1. There are only two possible outcomes.
2. p remains the same from trial to trial.
3. The trials are independent.
4. The distribution results from a count of the
number of successes in a fixed number of trials.
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Chapter Summary
C. The mean and variance of a binomial distribution are
computed as follows:
  np
 2  np 1  p 
D. The continuity correction factor of .5 is used to
extend the continuous value of X one-half unit in
either direction. This correction compensates for
estimating a discrete distribution by a continuous
distribution.
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