Transcript Chapter 5

Chapter
5
Normal Probability
Distributions
© 2012 Pearson Education, Inc.
All rights reserved.
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Chapter Outline
• 5.1 Introduction to Normal Distributions and the
Standard Normal Distribution
• 5.2 Normal Distributions: Finding Probabilities
• 5.3 Normal Distributions: Finding Values
• 5.4 Sampling Distributions and the Central Limit
Theorem
• 5.5 Normal Approximations to Binomial
Distributions
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Section 5.1
Introduction to Normal Distributions
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Section 5.1 Objectives
• Interpret graphs of normal probability distributions
• Find areas under the standard normal curve
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Properties of a Normal Distribution
Remember: These are measured, not counted (discrete)
Continuous random variable
• Has an infinite number of possible values that can be
represented by an interval on the number line.
Hours spent studying in a day
0
3
6
9
12
15
18
21
24
The time spent
studying can be any
number between 0
and 24.
Continuous probability distribution
• The probability distribution of a continuous random
variable.
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Properties of Normal Distributions
Normal distribution
• A continuous probability distribution for a random
variable, x.
• The most important continuous probability
distribution in statistics.
• The graph of a normal distribution is called the
normal curve.
x
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Properties of Normal Distributions
1. The mean, median, and mode are equal.
2. The normal curve is bell-shaped and is symmetric
about the mean.
3. The total area under the normal curve is equal to 1.
4. The normal curve approaches, but never touches, the
x-axis as it extends farther and farther away from the
mean.
Total area = 1
This means the area
actually approaches
1, but never quite
reaches it
μ
x
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Properties of Normal Distributions
5. Between μ – σ and μ + σ (in the center of the curve),
the graph curves downward. The graph curves
upward to the left of μ – σ and to the right of μ + σ.
The points at which the curve changes from curving
upward to curving downward are called the
inflection points.
μ – 3σ
μ – 2σ
μ–σ
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μ
μ+σ
μ + 2σ
μ + 3σ
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Means and Standard Deviations
• A normal distribution can have any mean and any
positive standard deviation.
• The mean gives the location of the line of symmetry.
• The standard deviation describes the spread of the
data.
μ = 3.5
σ = 1.5
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μ = 3.5
σ = 0.7
μ = 1.5
σ = 0.7
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Normal Distribution Curve
Mean/Median/Mode
Point of
Inflection –
begins to
curve
upward
Total Area
under the
Curve is 1
Point of
Inflection –
begins to
curve
upward
Curves
Downward
Approaches,
but never
touches x
axis
Example: Understanding Mean and
Standard Deviation
1. Which normal curve has the greater mean?
Solution:
Curve A has the greater mean (The line of symmetry
of curve A occurs at x = 15. The line of symmetry of
curve B occurs at x = 12.)
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Example: Understanding Mean and
Standard Deviation
2. Which curve has the greater standard deviation?
Solution:
Curve B has the greater standard deviation (Curve
B is more spread out than curve A.)
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Example: Interpreting Graphs
The scaled test scores for the New York State Grade 8
Mathematics Test are normally distributed. The normal
curve shown below represents this distribution. What is
the mean test score? Estimate the standard deviation.
Solution:
Because a normal curve is
symmetric about the mean,
you can estimate that μ ≈ 675.
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Because the inflection points are
one standard deviation from the
mean, you can estimate that σ ≈
35.
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Standard Normal Distribution
• There are infinitely many “normal” distributions –each
with it’s own mean/median/mode
• A normal distribution with a mean of 0 and a standard
deviation of 1 is called the Standard Normal Distribution
Standardized Normal
Distribution
sZ = 1
s
m
x
m =0 z
The Standard Normal Distribution
Standard normal distribution
• A normal distribution with a mean of 0 and a standard
deviation of 1.
Why convert to a
Area = 1
Z Score?
–3
–2
–1
z
0
1
2
3
• Any x-value can be transformed into a z-score by
using the formula
Value  Mean
xm
z

Standard deviation
s
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The Standard Normal Distribution
• If each data value of a normally distributed random
variable x is transformed into a z-score, the result will
be the standard normal distribution.
Normal Distribution
σ
z
m
x
xm
Standard Normal
Distribution
s
σ1
m0
z
• Then you can use the Standard Normal Table to find
the cumulative area under the standard normal curve.
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Properties of the Standard Normal
Distribution
1. The cumulative area is close to 0 for z-scores close
to z = –3.49. KNOW THIS NUMBER
2. The cumulative area increases as the z-scores
increase.
Area is
close to 0
z = –3.49
z
–3
–2
–1
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0
1
2
3
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Properties of the Standard Normal
Distribution
3. The cumulative area for z = 0 is 0.5000.
4. The cumulative area is close to 1 for z-scores close
to z = 3.49. KNOW THIS TOO
Area
is close to 1
z
–3
–2
–1
0
1
z=0
Area is 0.5000
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2
3
z = 3.49
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Example: Using The Standard Normal Table
Find the cumulative area that corresponds to a z-score of
1.15.
Solution:
Find 1.1 in the left hand column.
Move across the row to the column under 0.05
The area to the left of z = 1.15 is 0.8749.
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Example: Using The Standard Normal Table
Find the cumulative area that corresponds to a z-score of
–0.24.
Solution:
Find –0.2 in the left hand column.
Move across the row to the column under 0.04
The area to the left of z = –0.24 is 0.4052.
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On the Calculator
• You can calculate the cumulative area under the curve
using a left limit, a right limit, a mean and a standard
deviation
• If they tell you it has a normal distribution, then you
know that the mean is zero, and the standard
deviation is 1. So all you need is the left limit, and the
right limit
• Since this is a normal curve, we use “NormalCDF”
• Go to 2nd Vars Normal CDF
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NormalCDF
•
•
•
•
•
•
•
NormalCDF requires 4 pieces of information:
NormalCDF(ll,rl,µ,ơ)
ll= Left Limit
rl= Right Limit
µ= Mean
ơ= Standard Deviation
If you are calculating a Standard Normal Deviation,
then you don’t need to type in µ and ơ
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NormalCDF
•
•
•
•
•
•
•
•
•
•
note “EE” is the
2nd comma
button
In our previous example, we had z = 1.15
Go to 2nd Vars NormalCDF(
The first number we have to type in is the left limit
In this case, we want to start at the bottom of the curve and
extend all the way to the z-score
Our Left Limit therefore is theoretically zero
The calculator will not let us enter zero
So we approximate zero by typing this:
-EE99 which is 10 to the -99th power
What do we use for our Right Limit?
The Z-Score  NormalCDF(-E99,1.15 = .8749
Larson/Farber 5th ed
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Normal CDF
•
•
•
•
In our second example, Z = -.24
Calculate the area under the curve for Z = -.24
What does this mean?
This is also the probability for an event occurring
with a corresponding z-score of -.24
Larson/Farber 5th ed
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Finding Areas Under the Standard
Normal Curve
1. Sketch the standard normal curve and shade the
appropriate area under the curve.
2. Find the area by following the directions for each
case shown.
a. To find the area to the left of z, find the area that
corresponds to z in the Standard Normal Table.
2.
The area to the left
of z = 1.23 is 0.8907
1. Use the table to find the
area for the z-score
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We use
NormalCDF to
calculate this area
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Finding Areas Under the Standard
Normal Curve
b. To find the area to the right of z, use the Standard
Normal Table to find the area that corresponds to
z. Then subtract the area from 1.
2. The area to the
left of z = 1.23
is 0.8907.
3. Subtract to find the area
to the right of z = 1.23:
1 – 0.8907 = 0.1093.
1. Use the table to find the
area for the z-score.
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There are 2 ways
to solve this
problem on the
calculator
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Finding area ABOVE a Z-Score on the
Calculator
• Remember we use NormalCDF(ll,rl,µ,ơ)
• If we wanted to solve ABOVE the Z-Score, we can either:
 Solve the same way as previous, and then subtract our
answer from 1
 Use the Z-SCORE as the LEFT Limit
 What do we use for the RIGHT Limit?
 E99  We get as close to infinity as we can (remember the
curve goes to infinity)
 Find the area above a Z-Score of 1.23
 NormalCDF(1.23,E99
 .1093
Larson/Farber 5th ed
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Finding Areas Under the Standard
Normal Curve
c. To find the area between two z-scores, find the
area corresponding to each z-score in the
Standard Normal Table. Then subtract the
smaller area from the larger area.
2. The area to the
left of z = 1.23
is 0.8907.
3. The area to the
left of z = –0.75
is 0.2266.
1. Use the table to find the
area for the z-scores.
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4. Subtract to find the area of
the region between the two
z-scores:
0.8907 – 0.2266 = 0.6641.
How would we
do this on the
calculator?
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Example: Finding Area Under the
Standard Normal Curve
Find the area under the standard normal curve to the left
of z = –0.99.
Solution:
0.1611
–0.99
z
0
From the Standard Normal Table, the area is
equal to 0.1611.
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Example: Finding Area Under the
Standard Normal Curve
Find the area under the standard normal curve to the
right of z = 1.06.
Solution:
1 – 0.8554 = 0.1446
0.8554
z
0
1.06
From the Standard Normal Table, the area is equal to
0.1446.
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Example: Finding Area Under the
Standard Normal Curve
Find the area under the standard normal curve between
z = –1.5 and z = 1.25.
Solution:
0.8944 – 0.0668 = 0.8276
0.8944
0.0668
–1.50
0
1.25
z
From the Standard Normal Table, the area is equal to
0.8276.
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Using the Calculator to Draw the Curve
• You can calculate and draw the area under the curve
simultaneously
• 2nd Vars Draw ShadeNorm(ll,rl,µ,ơ)
• Again, you do not need the mean and the S.D. if it is
a normal distribution
• For you to get a good picture, you will have to
manually set the window:
 Xmin: -4
Xmax: 4
 Ymin: -1 Ymax: 1
 X and Y scale: 1
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Section 5.1 Summary
• Interpreted graphs of normal probability distributions
• Found areas under the standard normal curve
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Assignment
• Page 244 11-60
Larson/Farber 5th ed
34
Chapter 5 Quiz 1 (on your own)
•
•
•
•
Do “Try it Yourself 1” on page 249
1) Do part B
2) Do part C
3) If 100 people traveled on this stretch of highway in
a given time, how many would you expect to see
violating the 70 mile per hour speed limit?
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Section 5.2
Normal Distributions: Finding
Probabilities
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Section 5.2 Objectives
• Find probabilities for normally distributed variables
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Probability and Normal Distributions
• If a random variable x is normally distributed, you
can find the probability that x will fall in a given
interval by calculating the area under the normal
curve for that interval.
μ = 500
σ = 100
P(x < 600) = Area
x
μ = 500 600
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Probability and Normal Distributions
Normal Distribution
Standard Normal Distribution
μ = 500 σ = 100
μ=0 σ=1
P(x < 600)
x  m 600  500
z

1
s
100
P(z < 1)
z
x
μ =500 600
μ=0 1
Same Area
P(x < 600) = P(z < 1)
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Example: Finding Probabilities for
Normal Distributions
A survey indicates that people use their cellular phones
an average of 1.5 years before buying a new one. The
standard deviation is 0.25 year. A cellular phone user is
selected at random. Find the probability that the user
will use their current phone for less than 1 year before
buying a new one. Assume that the variable x is
normally distributed. (Source: Fonebak)
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Solution: Finding Probabilities for
Normal Distributions
Normal Distribution
μ = 1.5 σ = 0.25
z
P(x < 1)
Standard Normal Distribution
μ=0 σ=1
xm
s
1  1.5

 2
0.25
P(z < –2)
0.0228
z
x
1
1.5
–2
0
P(x < 1) = 0.0228
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On the Calculator
• You may recall that we can calculate area as:
• NormalCDF(ll,rl,µ,ơ)
• We DO NOT have to convert the data to a Z-Score to
get the same result
• We would type in:
• NormalCDF(-E99,1,1.5,.25) = 0.0228
Larson/Farber 5th ed
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Example: Finding Probabilities for
Normal Distributions
A survey indicates that for each trip to the supermarket,
a shopper spends an average of 45 minutes with a
standard deviation of 12 minutes in the store. The length
of time spent in the store is normally distributed and is
represented by the variable x. A shopper enters the store.
Find the probability that the shopper will be in the store
for between 24 and 54 minutes.
On the calculator: NormalCDF(24,54,45,12)
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Solution: Finding Probabilities for
Normal Distributions
Normal Distribution
μ = 45 σ = 12
xm
Standard Normal Distribution
μ=0 σ=1
24  45
 1.75
s
12
x  m 54  45
z2 

 0.75
s
12
z1 
P(24 < x < 54)

P(–1.75 < z < 0.75)
0.7734
0.0401
x
24
45 54
z
–1.75
0 0.75
P(24 < x < 54) = P(–1.75 < z < 0.75)
= 0.7734 – 0.0401 = 0.7333
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Example: Finding Probabilities for
Normal Distributions
If 200 shoppers enter the store, how many shoppers
would you expect to be in the store between 24 and 54
minutes?
Solution:
Recall P(24 < x < 54) = 0.7333
200(0.7333) =146.66 (or about 147) shoppers
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Example: Finding Probabilities for
Normal Distributions
Find the probability that the shopper will be in the store
more than 39 minutes. (Recall μ = 45 minutes and
σ = 12 minutes)
On the calculator: NormalCDF(39,EE99,45,12)
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Solution: Finding Probabilities for
Normal Distributions
Normal Distribution
μ = 45 σ = 12
z
P(x > 39)
Standard Normal Distribution
μ=0 σ=1
xm
s

39  45
 0.50
12
P(z > –0.50)
0.3085
z
x
39 45
–0.50 0
P(x > 39) = P(z > –0.50) = 1– 0.3085 = 0.6915
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Example: Finding Probabilities for
Normal Distributions
If 200 shoppers enter the store, how many shoppers
would you expect to be in the store more than 39
minutes?
Solution:
Recall P(x > 39) = 0.6915
200(0.6915) =138.3 (or about 138) shoppers
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Example: Using Technology to find
Normal Probabilities
Triglycerides are a type of fat in the bloodstream. The
mean triglyceride level in the United States is
134 milligrams per deciliter. Assume the triglyceride
levels of the population of the United States are
normally distributed with a standard deviation of
35 milligrams per deciliter. You randomly select a
person from the United States. What is the probability
that the person’s triglyceride level is less than 80? Use a
technology tool to find the probability.
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Solution: Using Technology to find
Normal Probabilities
Must specify the mean and standard deviation of the
population, and the x-value(s) that determine the
interval.
Note: They use
-10,000 instead of
-EE99
–it accomplishes the
same result, but our
way is more accurate
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Section 5.2 Summary
• Found probabilities for normally distributed variables
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Assignment
• Page 252 1-30
Larson/Farber 5th ed
52
Section 5.3
Normal Distributions: Finding
Values
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Section 5.3 Objectives
• Find a z-score given the area under the normal curve
• Transform a z-score to an x-value
• Find a specific data value of a normal distribution
given the probability
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Finding values Given a Probability
• In section 5.2 we were given a normally distributed
random variable x and we were asked to find a
probability.
• In this section, we will be given a probability and we
will be asked to find the value of the random variable
x.
5.2
x
z
probability
5.3
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Example: Finding a z-Score Given an
Area
Find the z-score that corresponds to a cumulative area of
0.3632.
Solution:
0.3632
z
z 0
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Solution: Finding a z-Score Given an
Area
• Locate 0.3632 in the body of the Standard Normal
Table.
The z-score
is –0.35.
• The values at the beginning of the corresponding row
and at the top of the column give the z-score.
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Example: Finding a z-Score Given an
Area
Find the z-score that has 10.75% of the distribution’s
area to its right.
Solution:
1 – 0.1075
= 0.8925
0.1075
z
0
z
Because the area to the right is 0.1075, the
cumulative area is 0.8925.
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Solution: Finding a z-Score Given an
Area
• Locate 0.8925 in the body of the Standard Normal
Table.
The z-score
is 1.24.
• The values at the beginning of the corresponding row
and at the top of the column give the z-score.
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Using a Calculator to determine Z-Scores
•
•
•
•
•
•
•
•
•
You can use 2nd Vars InvNorm(a, µ, σ) on the calculator
“a” is the area you are given
Fine the z-score that corresponds to a cumulative area of .3632
InvNorm(.3632,0,1) = -.3499
Find the z-score that corresponds to a cumulative area of .8925
in a standard normal distribution
InvNorm(.8925) = 1.2399
NOTE: If you are using a standard Normal distribution, you do
not have to enter the mean (0), and the Standard Deviation (1)
Find the z-score that has 10.75% of the distribution’s area to its
right
What do we have to do to correct this score?
Larson/Farber 5th ed
60
Percentiles
• Percentiles divide a data set into 100 equal parts
• To find a z-score that corresponds to a percentile, you
can use the standard-normal table
• Remember that if a value of x represents the 83rd
percentile (for example), then 83% of the data are at
or below x, and 17% of the data are above x
Example: Finding a z-Score Given a
Percentile
Find the z-score that corresponds to P5.
Solution:
The z-score that corresponds to P5 is the same z-score that
corresponds to an area of 0.05.
0.05
z
0
z
The areas closest to 0.05 in the table are 0.0495 (z = –1.65)
and 0.0505 (z = –1.64). Because 0.05 is halfway between the
two areas in the table, use the z-score that is halfway
between –1.64 and –1.65. The z-score is –1.645.
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Converting Percentiles to Z-Score
• On a calculator:
• 2nd Vars  InvNorm(perc)
• Note: type the percentile as a decimal
a) Convert P5 to a z-score
b) Convert P50 to a z-score
c) Convert P90 to a z-score
• P5 = InvNorm(.05) = -1.6445
• P50 = InvNorm(.50) = 0
• P90 = InvNorm(.90) = 1.282
Larson/Farber 5th ed
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Transforming a z-Score to an x-Score
To transform a standard z-score to a data value x in a
given population, we take this formula:
xm
z
s
And re-write it to solve for x. So we use this formula:
x = μ + zσ
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Example: Finding an x-Value
A veterinarian records the weights of cats treated at a
clinic. The weights are normally distributed, with a
mean of 9 pounds and a standard deviation of 2 pounds.
Find the weights x corresponding to z-scores of 1.96,
–0.44, and 0.
Solution: Use the formula x = μ + zσ
•z = 1.96:
x = 9 + 1.96(2) = 12.92 pounds
•z = –0.44: x = 9 + (–0.44)(2) = 8.12 pounds
•z = 0:
x = 9 + (0)(2) = 9 pounds
Notice 12.92 pounds is above the mean, 8.12 pounds is
below the mean, and 9 pounds is equal to the mean.
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Example: Finding a Specific Data Value
Scores for the California Peace Officer Standards and
Training test are normally distributed, with a mean of 50
and a standard deviation of 10. An agency will only hire
applicants with scores in the top 10%. What is the
lowest score you can earn and still be eligible to be
hired by the agency?
Solution:
An exam score in the top
10% is any score above the
90th percentile. Find the zscore that corresponds to a
cumulative area of 0.9.
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Solution: Finding a Specific Data Value
From the Standard Normal Table, the area closest to 0.9
is 0.8997. So the z-score that corresponds to an area of
0.9 is z = 1.28.
Use: InvNorm(.90)
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Solution: Finding a Specific Data Value
Using the equation x = μ + zσ
x = 50 + 1.28(10) = 62.8
The lowest score you can earn and still be eligible
to be hired by the agency is about 63.
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Finding A Specific Data Value
• Scores for a civil service exam are normally
distributed with a mean of 75 and a standard
deviation of 6.5. To be eligible for civil service
employment, you must score in the top 5 percent.
What is the lowest score you can earn and still be
eligible for employment?
• Note: You are looking for the top 5 %, which means
you want the area to the left of the 95% point
Example
• Referring to the previous slide, we need the top 5%.
Therefore, a score in the 95th percentile (or higher)
will suffice
• Looking at the standard z –score for 95th percentile 
InvNorm(.95) = 1.644
• X = µ + z*σ
• X = 75 + 1.644*(6.5) = 85.69
• Therefore, the lowest score you can achieve and still
be in the top 5 percent is an 86
Example
• The braking distance of a sample of Ford F150s is
normally distributed. The mean braking distance is
158 feet, with an S.D. of 6.51 feet. What is the
longest braking distance on one of these pickups
could have and still be in the best (top) 1%?
• First, find the z-score for top (first) 1%
• InvNorm(.01) = -2.32
• X = 158 + -2.32(6.51) = 142.89 feet
• To be in the top 1%, the pickup would have to stop
within 143 feet
Example
• In a randomly selected sample of 1169 men aged 40-49,
the mean total cholesterol level was 211 mg with a
standard deviation of 39.2 mg. Assume these are normally
distributed. Find the highest total cholesterol level a man
in this age group can have and be in the lowest 1%
• Again, this is the 1st 1 percent in the shaded region
• We find the z-value from InvNorm(.01) = -2.33
• X = 211 + (-2.33)(39.2) = 119.66
• If a man is to be in the top 1%, his cholesterol level
cannot exceed 119.66
Section 5.3 Summary
• Found a z-score given the area under the normal
curve
• Transformed a z-score to an x-value
• Found a specific data value of a normal distribution
given the probability
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Assignment
• Page 262 1-35
Larson/Farber 5th ed
74
Chapter 5 Quiz 2
• On your own:
• Do question 36 on page 264 –both parts
• 20 points…
Larson/Farber 5th ed
75
Section 5.4
Sampling Distributions and the
Central Limit Theorem
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Section 5.4 Objectives
• Find sampling distributions and verify their properties
• Interpret the Central Limit Theorem
• Apply the Central Limit Theorem to find the
probability of a sample mean
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Sampling Distributions
Sampling distribution
• The probability distribution of a sample statistic.
• Formed when samples of size n are repeatedly taken
from a population.
• If the sample statistic is the sample mean, then the
distribution is the sampling distribution of sample
means
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Sampling Distribution of Sample Means
Population with μ, σ
Sample 5
Sample 3
x3
Sample 1
x1
Sample 2
x2
Sample 4
x5
x4
The sampling distribution consists of the values of the
sample means, x1, x2 , x3 , x4 , x5 ,...
If you have a large population, you may have samples which are not the
same. Over time (with replacement) you should see those samples look like
the population when you put them together. This is how you do it…
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Properties of Sampling Distributions of
Sample Means
1. The mean of the sample means, m x , is equal to the
population mean μ.
m x  m This is known as “Central Tendency”
2. The standard deviation of the sample means, s x , is
equal to the population standard deviation, σ,
divided by the square root of the sample size, n.
sx 
s
n
Remember: this is also known as
the sample Standard Deviation
• Called the standard error of the mean.
Remember on the calculator, the population mean is ơx, and the
sample mean is Sx now you know why they are different. And
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this difference is called the standard error of the mean
Example: Sampling Distribution of
Sample Means
The population values {1, 3, 5, 7} are written on slips of
paper and put in a box. Two slips of paper are randomly
selected, with replacement.
a. Find the mean, variance, and standard deviation of
the population.
Solution:
Mean:
x
m
4
N
2

(
x

m
)
Variance: s 2 
5
N
Standard Deviation: s  5  2.236
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Example: Sampling Distribution of
Sample Means
b. Graph the probability histogram for the population
values.
Solution:
Probability Histogram of
Population of x
P(x)
0.25
Probability
All values have the
same probability of
being selected (uniform
distribution)
x
1
3
5
7
Population values
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Example: Sampling Distribution of
Sample Means
c. List all the possible samples of size n = 2 and
calculate the mean of each sample.
Solution:
Sample x
Sample x
1, 1
1, 3
1, 5
1, 7
3, 1
3, 3
3, 5
3, 7
1
2
3
4
2
3
4
5
© 2012 Pearson Education, Inc. All rights reserved.
5, 1
5, 3
5, 5
5, 7
7, 1
7, 3
7, 5
7, 7
3
4
5
6
4
5
6
7
These means
form the
sampling
distribution of
sample means.
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Example: Sampling Distribution of
Sample Means
d. Construct the probability distribution of the sample
means.
Solution:
f
Probability
xx f Probability
Enter xbar as
list 1, and f as
list 2
Run 1 var
stats on both
lists
1
1
0.0625
2
3
4
5
2
3
4
3
0.1250
0.1875
0.2500
0.1875
6
7
2
1
0.1250
0.0625
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Example: Sampling Distribution of
Sample Means
e. Find the mean, variance, and standard deviation of
the sampling distribution of the sample means.
Solution:
The mean, variance, and standard deviation of the
16 sample means are:
mx  4
5
s   2.5
2
s x  2.5  1.581
2
x
These results satisfy the properties of sampling
distributions of sample means.
mx  m  4
sx 
© 2012 Pearson Education, Inc. All rights reserved.
s
n

5 2.236

 1.581
2
2
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Example: Sampling Distribution of
Sample Means
f. Graph the probability histogram for the sampling
distribution of the sample means.
Solution:
P(x)
Probability
0.25
Probability Histogram of
Sampling Distribution of x
0.20
0.15
0.10
0.05
x
2
3
4
5
6
The shape of the
graph is symmetric
and bell shaped. It
approximates a
normal distribution.
7
Sample mean
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The Central Limit Theorem
• The Central Limit Theorem is possibly the most
important and useful theorem in Statistics
• Forms the foundation for the inferential branch of
statistics (where we make predictions)
• Describes the relationship between the sampling
distribution of sample means and the population that
the samples are taken from
• The previous exercise was used to show how many
samples can approximate the population
• It also shows that many samples are approximately
normally distributed, even if the population is NOT
The Central Limit Theorem
1. If samples of size n ≥ 30 are drawn from any
population with mean = µ and standard deviation = σ,
x
m
then the sampling distribution of sample means
approximates a normal distribution. The greater the
sample size, the better the approximation.
xx
x x
x x x
x x x x x
m
© 2012 Pearson Education, Inc. All rights reserved.
x
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The Central Limit Theorem
2. If the population itself is normally distributed,
x
m
then the sampling distribution of sample means is
normally distribution for any sample size n.
xx
x x
x x x
x x x x x
m
© 2012 Pearson Education, Inc. All rights reserved.
x
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The Central Limit Theorem
• In either case, the sampling distribution of sample
means has a mean equal to the population mean.
m x  m Mean
• The sampling distribution of sample means has a
variance equal to 1/n times the variance of the
population and a standard deviation equal to the
population standard deviation divided by the square
root of n.
2
s x2 
sx 
s
n
s
n
© 2012 Pearson Education, Inc. All rights reserved.
Sample Variance
Sample Standard deviation
(standard error of the mean)
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The Central Limit Theorem
1.
Any Population Distribution
Distribution of Sample Means,
n ≥ 30
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2.
Normal Population Distribution
Distribution of Sample Means,
(any n)
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Central Limit Theorem
As
Sample
Size Gets
Large
Enough
Sampling
Distribution
Becomes
Almost
Normal
regardless of
shape of
population
X
X
Example: Interpreting the Central Limit
Theorem
Cellular phone bills for residents of a city have a mean
of $63 and a standard deviation of $11. Random
samples of 100 cellular phone bills are drawn from this
population and the mean of each sample is determined.
Find the mean and standard error of the mean of the
sampling distribution. Then sketch a graph of the
sampling distribution of sample means.
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Solution: Interpreting the Central Limit
Theorem
• The mean of the sampling distribution is equal to the
population mean
mx  m  63
• The standard error of the mean is equal to the
population standard deviation divided by the
square root of n.
s x  s  11  1.1
n
100
Remember: This is another way of saying the Sample
Standard Deviation
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Solution: Interpreting the Central Limit
Theorem
• Since the sample size is greater than 30, the
sampling distribution can be approximated by a
normal distribution with
s x  $1.10
mx  $63
Because the standard
deviation is small, we
know the curve will
be narrow
© 2012 Pearson Education, Inc. All rights reserved.
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Example: Interpreting the Central Limit
Theorem
Suppose the training heart rates of all 20-year-old
athletes are normally distributed, with a mean of 135
beats per minute and standard deviation of 18 beats per
minute. Random samples of size 4 are drawn from this
population, and the mean of each sample is determined.
Find the mean and standard error of the mean of the
sampling distribution. Then sketch a graph of the
sampling distribution of sample means.
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Solution: Interpreting the Central Limit
Theorem
• The mean of the sampling distribution is equal to the
population mean
mx  m  135
• The standard error of the mean is equal to the
population standard deviation divided by the
square root of n.
s x  s  18  9
n
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4
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Solution: Interpreting the Central Limit
Theorem
• Since the population is normally distributed, the
sampling distribution of the sample means is also
normally distributed.
s x  9 Since the Sample
mx  135
Standard
Deviation is larger,
we know the curve
will be wider
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Probability and the Central Limit
Theorem
• To transform x to a z-score
Value  Mean x  m x x  m
z


s
Standard error
sx
n
Remember, this is
the Sample
Standard
Deviation
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Example: Probabilities for Sampling
Distributions
The graph shows the length of
time people spend driving each
day. You randomly select 50
drivers ages 15 to 19. What is
the probability that the mean
time they spend driving each
day is between 24.7 and 25.5
minutes? Assume that σ = 1.5
minutes.
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Solution: Probabilities for Sampling
Distributions
From the Central Limit Theorem (sample size is greater
than 30), the sampling distribution of sample means is
approximately normal with
s
1.5
sx 

 0.21213
mx  m  25
n
50
We’re going to
convert these
into Z-scores
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Solution: Probabilities for Sampling
Distributions On a Calculator: NormalCDF(-1.41,2.36)
Normal Distribution
Standard Normal Distribution
μ = 25 σ = 0.21213 x  m 24.7 - 25
μ=0 σ=1
z1 

 1.41
s
1.5
n
50
P(–1.41 < z < 2.36)
P(24.7 < x < 25.5)
z2 
x m
s
n

25.5  25
 2.36
1.5
50
0.9909
0.0793
x
24.7
25
25.5
z
–1.41
0
2.36
P(24 < x < 54) = P(–1.41 < z < 2.36)
= 0.9909 – 0.0793 = 0.9116
© 2012 Pearson Education, Inc. All rights reserved.
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Example: Probabilities for x and x
An education finance corporation claims that the
average credit card debts carried by undergraduates are
normally distributed, with a mean of $3173 and a
standard deviation of $1120. (Adapted from Sallie Mae)
1. What is the probability that a randomly selected
undergraduate, who is a credit card holder, has a
credit card balance less than $2700?
Solution:
You are asked to find the probability associated with
a certain value of the random variable x.
NOTE: This does not have a sample size, so we use
our original Z-Score Equation
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Solution: Probabilities for x and x
Normal Distribution
μ = 3173 σ = 1120
P(x < 2700)
z
Standard Normal Distribution
μ=0 σ=1
xm
s

2700  3173
 0.42
1120
P(z < –0.42)
0.3372
x
z
–0.42 0
2700 3173
P( x < 2700) = P(z < –0.42) = 0.3372
On a Calculator: NormalCDF(-EE99,-.42)
© 2012 Pearson Education, Inc. All rights reserved.
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Example: Probabilities for x and x
2. You randomly select 25 undergraduates who are
credit card holders. What is the probability that
their mean credit card balance is less than $2700?
• NOTE: This uses information from
previous example
Solution:
You are asked to find the probability associated with
a sample mean x.
mx  m  3173
© 2012 Pearson Education, Inc. All rights reserved.
s x  s  1120  224
n
25
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Solution: Probabilities for x and x
Normal Distribution
μ = 3173 σ = 1120
z
Standard Normal Distribution
μ=0 σ=1
xm
s

n
2700  3173 473

 2.11
1120
224
25
P(z < –2.11)
P(x < 2700)
0.0174
x
2700
z
–2.11
3173
0
P( x < 2700) = P(z < –2.11) = 0.0174
On a Calculator: NormalCDF(-EE99,-2.11)
© 2012 Pearson Education, Inc. All rights reserved.
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Solution: Probabilities for x and x
• There is about a 34% chance that an undergraduate
will have a balance less than $2700.
• There is only about a 2% chance that the mean of a
sample of 25 will have a balance less than $2700
(unusual event).
• It is possible that the sample is unusual or it is
possible that the corporation’s claim that the mean is
$3173 is incorrect.
© 2012 Pearson Education, Inc. All rights reserved.
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Section 5.4 Summary
• Found sampling distributions and verified their
properties
• Interpreted the Central Limit Theorem
• Applied the Central Limit Theorem to find the
probability of a sample mean
© 2012 Pearson Education, Inc. All rights reserved.
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Assignment
• Page 275 13-38 (no pictures/graphs are required to be
drawn)
Larson/Farber 5th ed
109
Chapter 5 Quiz 3
• Do problems 13-16 on page 275
• 2 points each
• Do them on your own
Larson/Farber 5th ed
110
Section 5.5
Normal Approximations to Binomial
Distributions
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Section 5.5 Objectives
• Determine when the normal distribution can
approximate the binomial distribution
• Find the continuity correction
• Use the normal distribution to approximate binomial
probabilities
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Normal Approximation to a Binomial
• The normal distribution is used to approximate the
binomial distribution when it would be impractical to
use the binomial distribution to find a probability.
Normal Approximation to a Binomial Distribution
• If np ≥ 5 and nq ≥ 5, then the binomial random
variable x is approximately normally distributed with
 mean μ = np
 standard deviation σ  npq
where n is the number of independent trials, p is the
probability of success in a single trial, and q is the
probability of failure in a single trial.
© 2012 Pearson Education, Inc. All rights reserved.
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Normal Approximation to a Binomial
• Binomial distribution: p = 0.25
np not greater than 5 and
nq not greater than 5
np not greater than 5 and
nq not greater than 5
• As n increases the histogram approaches a normal curve.
© 2012 Pearson Education, Inc. All rights reserved.
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Example: Approximating the Binomial
Decide whether you can use the normal distribution to
approximate x, the number of people who reply yes. If
you can, find the mean and standard deviation.
1. Sixty-two percent of adults in the U.S. have an
HDTV in their home. You randomly select 45
adults in the U.S. and ask them if they have an
HDTV in their home.
© 2012 Pearson Education, Inc. All rights reserved.
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Solution: Approximating the Binomial
• You can use the normal approximation
n = 45, p = 0.62, q = 0.38
np = (45)(0.62) = 27.9
nq = (45)(0.38) = 17.1
• Mean: μ = np = 27.9
• Standard Deviation: σ  npq  45  0.62  0.38  3.26
© 2012 Pearson Education, Inc. All rights reserved.
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Example: Approximating the Binomial
Decide whether you can use the normal distribution to
approximate x, the number of people who reply yes. If
you can, find the mean and standard deviation.
2. Twelve percent of adults in the U.S. who do not
have an HDTV in their home are planning to
purchase one in the next two years. You randomly
select 30 adults in the U.S. who do not have an
HDTV and ask them if they are planning to
purchase one in the next two years.
© 2012 Pearson Education, Inc. All rights reserved.
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Solution: Approximating the Binomial
• You cannot use the normal approximation
n = 30, p = 0.12, q = 0.88
np = (30)(0.12) = 3.6
nq = (30)(0.88) = 26.4
• Because np < 5, you cannot use the normal
distribution to approximate the distribution of x.
© 2012 Pearson Education, Inc. All rights reserved.
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Correction for Continuity
• The binomial distribution is discrete and can be
represented by a probability histogram.
• To calculate exact binomial probabilities, the
binomial formula is used for each value of x and the
results are added.
• Geometrically this corresponds to adding the areas of
bars in the probability histogram.
© 2012 Pearson Education, Inc. All rights reserved.
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Correction for Continuity
• When you use a continuous normal distribution to
approximate a binomial probability, you need to
move 0.5 unit to the left and right of the midpoint to
include all possible x-values in the interval
(continuity correction).
Exact binomial probability
P(x = c)
c
© 2012 Pearson Education, Inc. All rights reserved.
Normal approximation
P(c – 0.5 < x < c + 0.5)
c – 0.5 c c + 0.5
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Example: Using a Correction for
Continuity
Use a continuity correction to convert the binomial
interval to a normal distribution interval.
1. The probability of getting between 270 and 310
successes, inclusive.
Solution:
• The discrete midpoint values are 270, 271, …, 310.
• The corresponding interval for the continuous normal
distribution is
269.5 < x < 310.5
© 2012 Pearson Education, Inc. All rights reserved.
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Example: Using a Correction for
Continuity
Use a continuity correction to convert the binomial
interval to a normal distribution interval.
2. The probability of getting at least 158 successes.
Solution:
• The discrete midpoint values are 158, 159, 160, ….
• The corresponding interval for the continuous normal
distribution is
x > 157.5
© 2012 Pearson Education, Inc. All rights reserved.
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Example: Using a Correction for
Continuity
Use a continuity correction to convert the binomial
interval to a normal distribution interval.
3. The probability of getting fewer than 63 successes.
Solution:
• The discrete midpoint values are …, 60, 61, 62.
• The corresponding interval for the continuous normal
distribution is
x < 62.5
© 2012 Pearson Education, Inc. All rights reserved.
123 of 105
Using the Normal Distribution to
Approximate Binomial Probabilities
In Words
1. Verify that the binomial
distribution applies.
2. Determine if you can use
the normal distribution to
approximate x, the binomial
variable.
3. Find the mean µ and
standard deviation σ for the
distribution.
© 2012 Pearson Education, Inc. All rights reserved.
In Symbols
Specify n, p, and q.
Is np ≥ 5?
Is nq ≥ 5?
m  np
s  npq
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Using the Normal Distribution to
Approximate Binomial Probabilities
In Words
4. Apply the appropriate
continuity correction.
Shade the corresponding
area under the normal
curve.
5. Find the corresponding
z-score(s).
6. Find the probability.
© 2012 Pearson Education, Inc. All rights reserved.
In Symbols
Add or subtract 0.5
from endpoints.
z
xm
s
Use the Standard
Normal Table.
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Example: Approximating a Binomial
Probability
Sixty-two percent of adults in the U.S. have an HDTV
in their home. You randomly select 45 adults in the U.S.
and ask them if they have an HDTV in their home.
What is the probability that fewer than 20 of them
respond yes? (Source: Opinion Research Corporation)
Solution:
• Can use the normal approximation (see slide 91)
μ = 45 (0.62) = 27.9 s  450.620.38  3.26
© 2012 Pearson Education, Inc. All rights reserved.
126 of 105
Solution: Approximating a Binomial
Probability
• Apply the continuity correction:
Fewer than 20 (…17, 18, 19) corresponds to the
continuous normal distribution interval x < 19.5.
Normal Distribution
μ = 27.9 σ ≈ 3.26
Standard Normal
μ=0 σ=1
z
xm
s

19.5  27.9
 2.58
3.26
P(z < –2.58)
0.0049
–2.58
z
μ=0
P(z < –2.58) = 0.0049
On a calculator: NormalCDF(-E99,19.5,27.9,3.26)
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Example: Approximating a Binomial
Probability
A survey reports that 62% of Internet users use
Windows® Internet Explorer® as their browser. You
randomly select 150 Internet users and ask them
whether they use Internet Explorer® as their browser.

What is the probability that exactly
96 will say yes?
(Source: Net Applications)
Solution:
• Can use the normal approximation
np = 150∙0.62 = 93 ≥ 5 nq = 150∙0.38 = 57 ≥ 5
μ = 150∙0.62 = 93
© 2012 Pearson Education, Inc. All rights reserved.
σ  150  0.62  0.38  5.94
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Solution: Approximating a Binomial
Probability
• Apply the continuity correction:
Rewrite the discrete probability P(x=96) as the
continuous probability P(95.5 < x < 96.5).
Normal Distribution
μ = 93 σ = 5.94
xm
Standard Normal
μ=0 σ=1
95.5  93
 0.42
s
5.94
x  m 96.5  93
z2 

 0.59
s
5.94
z1 

P(0.42 < z < 0.59)
0.7224
0.6628
z
μ = 0 0.59
0.42
P(0.42 < z < 0.59) = 0.7224 – 0.6628 = 0.0596
On a calculator: NormalCDF(95.5,96.5,93,5.94)
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Section 5.5 Summary
• Determined when the normal distribution can
approximate the binomial distribution
• Found the continuity correction
• Used the normal distribution to approximate binomial
probabilities
© 2012 Pearson Education, Inc. All rights reserved.
130 of 105
Assignment
• Page 287 7-26
Larson/Farber 5th ed
131