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Chapter Seven
Normal Curves
and Sampling
Distributions
Understanding Basic Statistics
Fourth Edition
By Brase and Brase
Prepared by: Lynn Smith
Gloucester County College
Properties of The Normal Distribution
The curve is bell-shaped with the
highest point over the mean, μ.
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Properties of The Normal Distribution
The curve is symmetrical about a
vertical line through μ.
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Properties of The Normal Distribution
The curve approaches the horizontal
axis but never touches or crosses it.
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Properties of The Normal Distribution
The transition points between
cupping upward and downward occur
above μ + σ and μ – σ .
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The Empirical Rule
• Applies to any symmetrical and bellshaped distribution
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The Empirical Rule
Approximately 68% of the data values lie within
one standard deviation of the mean.
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The Empirical Rule
Approximately 95% of the data values lie within
two standard deviations of the mean.
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The Empirical Rule
Almost all (approximately 99.7%) of the data
values will be within three standard deviations
of the mean.
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Application of the Empirical Rule
The life of a particular type of light bulb is
normally distributed with a mean of
1100 hours and a standard deviation of
100 hours.
What is the probability that a light bulb of
this type will last between 1000 and
1200 hours?
Approximately 68%
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Standard Score
• The z value or z score tells the number
of standard deviations between the
original measurement and the mean.
• The z value is in standard units.
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Formula for z score
x
z

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x Values and Corresponding z Values
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Calculating z scores
The amount of time it takes for a pizza
delivery is approximately normally distributed
with a mean of 25 minutes and a standard
deviation of 2 minutes. Convert 21 minutes
to a z score.
x   21  25
z

 2.00

2
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Calculating z scores
Mean delivery time = 25 minutes
Standard deviation = 2 minutes
Convert 29.7 minutes to a z score.
x   29.7  25
z

 2.35

2
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Raw Score
• A raw score is the result of converting
from standard units (z scores) back to
original measurements, x values.
• Formula:
x=z+
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Interpreting z-scores
Mean delivery time = 25 minutes
Standard deviation = 2 minutes
Interpret a z score of 1.60.
x  z     1 .6( 2 )  25  28 .2
The delivery time is 28.2 minutes.
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Standard Normal Distribution:

=0

=1
Any x values are converted to z
scores.
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Importance of the
Standard Normal Distribution:
Any Normal Curve:
Areas will be equal.
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Areas of a Standard Normal Distribution
• Appendix
• Table 3
• Pages A6 - A7
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Use of the Normal Probability Table
Appendix Table 3 is a left-tail style table.
Entries give the cumulative areas to the
left of a specified z.
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To Find the area to the
Left of a Given z score
• Find the row associated with the sign,
units and tenths portion of z in the left
column of Table 3.
• Move across the selected row to the
column headed by the hundredths digit
of the given z.
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Find the area to the left of
z = – 2.84
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To find the area to the left of
z = – 2.84
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To find the area to the left of
z = – 2.84
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The area to the left of
z = – 2.84 is .0023
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To Find the Area to the Left of a Given
Negative z Value:
Use Table 3 of the Appendix directly.
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To Find the Area to the Left
of a Given Positive z Value:
Use Table 3 of the Appendix directly.
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To Find the Area to the Right
of a Given z Value:
Subtract the area to the left of z from
1.0000.
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Alternate Way To Find the Area to the Right
of a Given Positive z Value:
Use the symmetry of the normal distribution.
Area to the right of z
= area to left of –z.
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To Find the Area Between Two z Values
Subtract area to left of z1 from area to left
of z2 . (When z2 > z1.)
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Convention for Using Table 3
• Treat any area to the left of a z value
smaller than 3.49 as 0.000
• Treat any area to the left of a z value
greater than 3.49 as 1.000
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Use of the Normal Probability Table
a.
.9495
P( z < 1.64 ) = __________
b.
.0034
P( z < - 2.71 ) = __________
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Use of the Normal Probability Table
c.
.3925
P(0 < z < 1.24) = ______
d.
.4452
P(0 < z < 1.60) = _______
e.
.4911
P( 2.37 < z < 0) = ______
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Use of the Normal Probability Table
f.
.9974
P( 3 < z < 3 ) = ________
g.
.9322
P( 2.34 < z < 1.57 ) = _____
h.
.0774
P( 1.24 < z < 1.88 ) = _______
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Use of the Normal Probability Table
i.
P( 2.44 < z < 0.73 ) = _______
.2254
j.
P( z > 2.39 ) = ________
.0084
k.
P( z > 1.43 ) = __________
.9236
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To Work with Any Normal Distributions
• Convert x values to z values using the
formula:
x
z

Use Table 3 of the Appendix to find
corresponding areas and probabilities.
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Rounding
• Round z values to the hundredths
positions before using Table 3.
• Leave area results with four digits to
the right of the decimal point.
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Application of the Normal Curve
The amount of time it takes for a pizza delivery is
approximately normally distributed with a mean of 25
minutes and a standard deviation of 2 minutes. If you
order a pizza, find the probability that the delivery time
will be:
a. between 25 and 27 minutes.
.3413
a. __________
b. less than 30 minutes.
.9938
b. __________
c. less than 22.7 minutes.
.1251
c. __________
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Inverse Normal Probability Distribution
• Finding z or x values that correspond to
a given area under the normal curve
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Inverse Normal Left Tail Case
Look up area A in body of Table 3 and use
corresponding z value.
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Inverse Normal Right Tail Case:
Look up the number 1 – A in body of
Table 3 and use corresponding z value.
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Inverse Normal Center Case:
Look up the number (1 – A)/2 in body of
Table 3 and use corresponding ± z value.
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Using Table 3 for
Inverse Normal Distribution
• Use the nearest area value rather than
interpolating.
• When the area is exactly halfway between
two area values, use the z value exactly
halfway between the z values of the
corresponding table areas.
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When the area is exactly
halfway between two area values
• When the z value corresponding to an
area is smaller than 2, use the z value
corresponding to the smaller area.
• When the z value corresponding to an
area is larger than 2, use the z value
corresponding to the larger area.
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Find the indicated z score:
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Find the indicated z score:
– 2.57
z = _______
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Find the indicated z score:
2.33
z = _______
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Find the indicated z scores:
–1.23
–z = _____
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1.23
z = ____
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Find the indicated z scores:
± 2.58
± z =__________
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Find the indicated z scores:
± 1.96
± z = ________
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Application of Determining z Scores
The Verbal SAT test has a mean score
of 500 and a standard deviation of 100.
Scores are normally distributed. A
major university determines that it will
accept only students whose Verbal SAT
scores are in the top 4%. What is the
minimum score that a student must
earn to be accepted?
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Application of Determining z Scores
The cut-off score is 1.75 standard deviations
above the mean.
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Application of Determining z Scores
Mean = 500
standard
deviation = 100
The cut-off score is 500 + 1.75(100) = 675.
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Introduction to Sampling Distributions
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Review of Statistical Terms
• Population
• Sample
• Parameter
• Statistic
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Population
• the set of all measurements or counts
• (either existing or conceptual)
• under consideration
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Sample
• a subset of measurements from a
population
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Parameter
• a numerical descriptive measure of a
population
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Statistic
• a numerical descriptive measure of a
sample
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We use a statistic to make
inferences about a population parameter.
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Some Common Statistics
and Corresponding Parameters
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Principal Types of Inferences
• Estimation: estimate the value of a
population parameter
• Testing: formulate a decision about the
value of a population parameter
• Regression: Make predictions or
forecasts about the value of a statistical
variable
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Sampling Distribution
• a probability distribution for the sample
statistic
• based on all possible random samples
of the same size from the same
population
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Example of a Sampling Distribution
• Select samples with two elements each
(in sequence with replacement) from
the set
• {1, 2, 3, 4, 5, 6}.
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Constructing a Sampling Distribution of the
Mean for Samples of Size n = 2
List all samples (with 2 items in each) and compute
the mean of each sample.
sample:
mean:
sample:
mean
{1,1}
{1,2}
{1,3}
{1,4}
{1,5}
1.0
1.5
2.0
2.5
3.0
{1,6}
{2,1}
{2,2}
…
3.5
1.5
4
...
How many different samples are there? 36
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Sampling Distribution of the Mean
x
p
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
1/36
2/36
3/36
4/36
5/36
6/36
5/36
4/36
3/36
2/36
1/36
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Let x be a random variable with a
normal distribution with mean  and
standard deviation . Let x be the
sample mean corresponding to
random samples of size n taken from
the distribution .
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Theorem 7.1:
• The x distribution is a normal distribution.
• The mean of the x distribution is  (the
same mean as the original distribution).
• The standard deviation of the x distribution
is / n (the standard deviation of the
original distribution, divided by the square
root of the sample size).
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We can use this theorem to draw
conclusions about means of
samples taken from normal
distributions.
If the original distribution is
normal, then the sampling
distribution will be normal.
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The Mean of the Sampling Distribution
x
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The mean of the sampling distribution is
equal to the mean of the original
distribution.
x  
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The Standard Deviation
of the Sampling Distribution
x
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The standard deviation of the sampling
distribution is equal to the standard
deviation of the original distribution divided
by the square root of the sample size.

x 
n
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To Calculate z Scores
x  x x  
z

x
 n
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The time it takes to drive between cities A and
B is normally distributed with a mean of 14
minutes and a standard deviation of 2.2
minutes.
1. Find the probability that a trip between
the cities takes more than 15 minutes.
2. Find the probability that mean time of
nine trips between the cities is more
than 15 minutes.
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Mean = 14 minutes,
standard deviation = 2.2 minutes
• Find the probability that a trip between
the cities takes more than 15 minutes.
15  14
z
 0.45
2.2
P( z  0.45)  1  0.6736  0.3264
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Mean = 14 minutes,
standard deviation = 2.2 minutes
• Find the probability that the mean time
of nine trips between the cities is more
than 15 minutes.
 x    14

2 .2
x 

 0.73
n
9
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Mean = 14 minutes, standard deviation =
2.2 minutes
• Find the probability that mean time of nine
trips between the cities is more than 15
minutes.
15  14
z
 1.37
0.73
P ( z  1.37 )  1  0.9147  0.0853
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Standard Error of the Mean
• The standard error of the mean is the
standard deviation of the sampling
distribution
• For the x sampling distributi on, the standard error 
of the mean
x  
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n
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What if the Original
Distribution Is Not Normal?
• Use the Central Limit Theorem.
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Central Limit Theorem
If x has any distribution with mean  and
standard deviation , then the sample
mean based on a random sample of
size n will have a distribution that
approaches the normal distribution
(with mean  and standard deviation 
divided by the square root of n) as n
increases without limit.
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How large should the sample size be to
permit the application of the Central Limit
Theorem?
• In most cases a sample size of n = 30
or more assures that the distribution
will be approximately normal and the
theorem will apply.
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Central Limit Theorem
• For most x distributions, if we use a
sample size of 30 or larger, the x
distribution will be approximately
normal.
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Central Limit Theorem
• The mean of the sampling distribution
is the same as the mean of the original
distribution.
• The standard deviation of the sampling
distribution is equal to the standard
deviation of the original distribution
divided by the square root of the
sample size.
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Central Limit Theorem Formula
x  
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Central Limit Theorem Formula

x 
n
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Central Limit Theorem Formula
x x x
z

 x / n
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Application of the Central Limit Theorem
Records indicate that the packages shipped by
a certain trucking company have a mean weight
of 510 pounds and a standard deviation of 90
pounds. One hundred packages are being
shipped today. What is the probability that their
mean weight will be:
a.
b.
c.
more than 530 pounds?
less than 500 pounds?
between 495 and 515 pounds?
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Are we authorized to
use the Normal Distribution?
• Yes, we are attempting to draw
conclusions about means of large
samples.
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Applying the Central Limit Theorem
What is the probability that the mean weight will
be more than 530 pounds?
Consider the distribution of sample means:
 x  510 ,  x  90 / 100  9
P( x > 530): z = 530 – 510 = 20 = 2.22
9
9
.0132
P(z > 2.22) = _______
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7 | 91
Applying the Central Limit Theorem
What is the probability that their mean weight
will be less than 500 pounds?
P( x < 500): z = 500 – 510 = –10 = – 1.11
9
9
P(z < – 1.11) = _______
.1335
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7 | 92
Applying the Central Limit Theorem
What is the probability that their mean weight
will be between 495 and 515 pounds?
P(495 < x < 515) :
for 495: z = 495 – 510 = - 15 = - 1.67
9
9
for 515: z = 515 – 510 = 5 = 0.56
9
9
.6648
P(–1.67 < z < 0.56) = ______
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Normal Approximation of
The Binomial Distribution:
• If n (the number of trials) is sufficiently
large, a binomial random variable has a
distribution that is approximately
normal.
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Define “sufficiently large”
The sample size, n, is considered to be
"sufficiently large" if
np and nq
are both greater than 5.
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Mean and Standard
Deviation: Binomial Distribution
  np and   npq
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Experiment: tossing a coin 20 times
Problem: Find the probability of getting
exactly 10 heads.
Distribution of the number of heads appearing should look
like:
0
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10
20
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Using the Binomial Probability Formula
n=
20
x=
10
P(10) = 0.176197052
p=
0.5
q = 1 – p = 0.5
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Normal Approximation of the Binomial
Distribution
First calculate the mean
and standard deviation:
 = np = 20 (.5) = 10

np(1  p ) 
20(. 5 )(. 5 ) 
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5  2.24
7 | 99
The Continuity Correction
Continuity Correction allows us to approximate
a discrete probability distribution with a
continuous distribution.
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The Continuity Correction
• We are using the area under the curve
to approximate the area of the
rectangle.
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The Continuity Correction
• If r is the left-point of an interval,
subtract 0.5 to obtain the
corresponding normal variable.
• x = r  0.5
• If r is the right-point of an interval, add
0.5 to obtain the corresponding normal
variable.
• x = r + 0.5
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The Continuity Correction
• Continuity Correction: to compute the
probability of getting exactly 10 heads,
find the probability of getting between
9.5 and 10.5 heads.
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Using the Normal Distribution
P(9.5 < x < 10.5 ) = ?
For x = 9.5: z = – 0.22
For x = 10.5: z = 0.22
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Using the Normal Distribution
P(9.5 < x < 10.5 ) =
P( – 0.22 < z < 0.22 ) =
.5871 – .4129 = .1742
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7 | 105
Application of Normal Distribution
If 22% of all patients with high blood pressure
have side effects from a certain medication,
and 100 patients are treated, find the
probability that at least 30 of them will
have side effects.
Using the Binomial Probability Formula we would
need to compute:
P(30) + P(31) + ... + P(100) or 1 - P( x < 29)
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Using the Normal Approximation to the
Binomial Distribution
Is n sufficiently large?
Check: np =
nq =
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Using the Normal Approximation to the
Binomial Distribution
Is n sufficiently large?
np = 22
nq = 78
Both are greater than five.
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Find the mean and standard deviation
 = 100(.22) = 22
and  = 100(.22)(.78) 
17.16  4.14
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7 | 109
Applying the Normal Distribution
To find the probability that at least 30 of
them will have side effects, find P(x  29.5)
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Applying the Normal Distribution
z = 29.5 – 22 = 1.81
4.14
Find P(z  1.81)
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The probability that
at least 30 of the
patients will have
side effects is
0.0351.
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Reminders:
• Use the normal distribution to
approximate the binomial only if both
np and nq are greater than 5.
• Always use the continuity correction
when using the normal distribution to
approximate the binomial distribution.
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7 | 112