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The Standard Deviation as a Ruler and the
Normal Model
Chapter 6
Performance Scales:
1. Use the mean and standard deviation of a data set
to fit it to a normal distribution .
2. Use the mean and standard deviation of a data set
to fit it to a normal distribution and to estimate
population percentages.
3. Use the mean and standard deviation of a data set
to fit it to a normal distribution and to estimate
population percentages. Recognize that there are
data sets for which such a procedure is not
appropriate. Use calculators, spreadsheets, and
tables to estimate areas under the normal curve.
MAFS.912.S-ID.1.4.
4. Adapts and uses the mean and standard deviation
of a data set to fit it to a normal distribution and to
estimate population percentages in different and
more complex ways.
Learning Goals
1. Understand how adding (subtracting) a constant or
multiplying (dividing) by a constant changes the
center and/or spread of a variable.
2. Understand that standardizing uses the standard
deviation as a ruler.
3. Know how to calculate the z-score of an
observation and what it means.
4. Know how to compare values of two different
variables using their z-scores.
5. Recognize when a Normal model is appropriate.
Learning Goals
6. Recognize when standardization can be used to
compare values.
7. Be able to use Normal models and the 68-95-99.7
Rule to estimate the percentage of observations
falling within 1, 2, or 3 standard deviations of the
mean.
8. Know how to find the percentage of observations
falling below any value in a Normal model using a
Normal table or appropriate technology.
9. Know how to check whether a variable satisfies the
Nearly Normal Condition by making a Normal
Probability plot or histogram.
Learning Goal 1
Understand how adding (subtracting) a constant or multiplying
(dividing) by a constant changes the center and/or spread of a variable.
Learning Goal 1:
Linear Transformation of Data
• Linear transformation
– Shifting (moving left or right) the data
or Rescaling (making the size larger or
smaller) the data.
– Changes the original variable x into the
new variable xnew given by
xnew = a + bx
• Adding the constant a shifts all values of x
upward (right) or downward (left) by the
same amount.
• Multiplying by the positive constant b
changes the size of the values or rescales
the data.
Learning Goal 1:
Shifting Data
• Shifting data:
– Adding (or subtracting) a
constant amount to each value
just adds (or subtracts) the same
constant to (from) the mean. This
is true for the median and other
measures of position too.
– In general, adding a constant to
every data value adds the same
constant to measures of center
and percentiles, but leaves
measures of spread unchanged.
Learning Goal 1:
Shifting Data Example: Adding a Constant
• Given the data: 2, 4, 6, 8, 10
– Center: mean = 6, median = 6
– Spread: s = 3.2, IQR = 6
• Add a constant 5 to each value, new data
7, 9, 11, 13, 15
– New center: mean = 11, median = 11
– New spread: s = 3.2, IQR = 6
• Effects of adding a constant to each data
value
– Center increases by the constant 5
– Spread does not change
– Shape of the distribution does not
change
Learning Goal 1: Example - Subtracting a Constant
• The following histograms show a shift from men’s actual
weights to kilograms above recommended weight.
• Weights of 80 men age
19 to 24 of average height
(5'8" to 5'10") 𝑥 = 82.36 kg.
• NIH recommends maximum healthy
weight of 74 kg. To compare their
weights to the recommended
maximum, subtract 74 kg from each
weight; 𝑥 new = 𝑥 – 74.
So, 𝑥 new = 𝑥 – 74 = 8.36 kg.
1.
No change in shape
2.
No change in spread
3.
Shift by 74
Shift Down
Learning Goal 1:
Rescaling Data
• Rescaling data:
– When we divide or multiply all
the data values by any constant
value, all measures of position
(such as the mean, median and
percentiles) and measures of
spread (such as the range, IQR,
and standard deviation) are
divided and multiplied by that
same constant value.
Learning Goal 1:
Rescaling Example: Multiplying by a Constant
• Given the data: 2, 4, 6, 8, 10
– Center: mean = 6, median = 6
– Spread: s = 3.2, IQR = 6
• Multiple a constant 3 to each value, new data:
6, 12, 18, 24, 30
– New center: mean = 18, median = 18
– New spread: s = 9.6, IQR = 18
• Effects of multiplying each value by a constant
– Center increases by a factor of the
constant (times 3)
– Spread increases by a factor of the
constant (times 3)
– Shape of the distribution does not change
Learning Goal 1: Example - Rescaling Data
• The men’s weight data set measured weights in kilograms. If
we want to think about these weights in pounds, we would
• Change from kilograms to pounds
rescale the data:
•
•
•
•
•
•
Weights of 80 men age 19 to 24, of
average height (5'8" to 5'10")
𝑥 = 82.36 kg
min=54.30 kg
max=161.50 kg
range=107.20 kg
s = 18.35 kg
•
•
•
•
•
(multiple each observation by 2.2 lb/kg):
𝑥 new = 2.2(82.36)=181.19 lb
minnew = 2.2(54.30)=119.46 lb
maxnew = 2.2(161.50)=355.3 lb
rangenew= 2.2(107.20)=235.84 lb
snew = 2.2(18.35) = 40.37 lb
1.
No change in shape
2.
Increase in spread
3.
Shift by to the right
Shifts up
Size (spread) increases
Learning Goal 1:
Summary of Linear Transformations
• Multiplying each observation by a
positive number b multiples both
measures of center (mean and median)
and measures of spread (IQR and
standard deviation) by b.
• Adding the same number a (either
positive or negative) to each
observation adds a to measures of
center and to quartiles, but does not
change measures of spread.
• Linear transformations do not change
the shape of a distribution.
Learning Goal 1:
Summary of Linear Transformations
• Linear transformations do not affect the
shape of the distribution of the data.
-for example, if the original data is rightskewed, the transformed data is rightskewed.
• Example, changing the units of data from
minutes to seconds (multiplying by 60
sec/min).
Assembly Time (seconds)
Assembly Time (minutes)
30
20
10
0
Shape remains
the same
25
Frequency
Frequency
30
20
15
10
5
0
Learning Goal 1: Example
Los Angeles Laker’s Salaries (2000)
𝑥 = $4.14 mil, Med. = $2.6 mil, s = $4.76 mil, IQR = $3.6 mil
(a)
Suppose that each member of the team receives a
$100,000 bonus for winning the NBA championship.
How will this affect the center and spread?
$100,000 = $0.1 mil
Learning Goal 1: Example - Solution
a) The mean and median will all increase
by $0.1 mil and the standard deviation
and IQR will not change.
Original: 𝑥 = $4.14 mil, Med. = $2.6 mil,
s = $4.76 mil, IQR = $3.6 mil
𝑥 new = 0.1 + 4.14 = $4.24 mil
Med.new = 0.1 + 2.6 = $2.7 mil
snew = no change = $4.76 mil
IQRnew = no change = $3.6 mil
Learning Goal 1: Example
(b)
Each player is offered a 10%
increase base salary. How will
this affect the center and
spread?
𝑥 = $4.14 mil, Med. = $2.6 mil,
s = $4.76 mil, IQR = $3.6 mil
Learning Goal 1: Example - Solution
(b)
The mean, median, IQR and
standard deviation will all increase
by a factor of 1.1 (100% + 10% = 110%,
as decimal 1.1).
Original: 𝑥 = $4.14 mil, Med. = $2.6 mil,
s = $4.76 mil, IOR = $3.6 mil
𝑥 new = (1.1)(4.14) = $4.55 mil
Med.new = (1.1)(2.6) = $2.86 mil
snew = (1.1)(4.76) = $5.24 mil
IQRnew = (1.1)(3.6) = $3.96 mil
Learning Goal 1: Your Turn
•
Maria measures the lengths of 5
cockroaches that she finds at school. Here
are her results (in inches):
1.4
a.
b.
2.2
1.1
1.6
1.2
Find the mean and standard deviation of
Maria’s measurements (use calc).
Maria’s science teacher is furious to
discover that she has measured the
cockroach lengths in inches rather than
centimeters (There are 2.54 cm in 1 inch).
She gives Maria two minutes to report the
mean and standard deviation of the 5
cockroaches in centimeters. Find the mean
and standard deviation in centimeters.
Learning Goal 1: Solution
a. Mean 𝑥 = 1.5 in
Standard Deviation s = .436 in
b. To change from cm to in, multiple
by a factor of 2.54 cm/in.
𝑥 new = (2.54)(1.5) = 3.81 cm
snew = (2.54)(.436) = 1.107 cm
Learning Goal 1: Class Problem
We have a company with employees with the following
salaries:
1200 900 1400 2100 1800 1000 1300 700
1700 2300 1200
1. What is the mean and standard deviation of the company
salaries?
1418.18
503.62
mean = _________
st dev = _________
2. Suppose we give everyone a $500 raise. What is the new
mean and standard deviation?
1918.18
503.62
mean = _________
st dev = _________
3. Suppose we have to cut everyone’s pay by $500 due to the
economy. What is the mean and standard deviation now?
918.18
503.62
mean = _________
st dev = _________
Learning Goal 1: Class Problem (continued)
1. What was the mean and standard deviation of the
company salaries?
1418.18
503.62
mean = _________
st dev = _________
4. Suppose we give everyone a 30% raise. What is the new
mean and standard deviation?
1843.63
654.71
mean = _________
st dev = _________
5. Suppose we cut everyone’s pay by 7%. What is the new
mean and standard deviation?
1318.91
468.37
mean = _________
st dev = _________
Learning Goal 2
Understand that standardizing uses the standard deviation as a
ruler.
Learning Goal 2:
Comparing Apples to Oranges
• In order to compare data values
with different units (apples and
oranges), we need to make sure we
are using the same scale.
• The trick is to look at how the values
deviate from the mean. Look at
whether the data point is above or
below the mean, and by how much.
• Standard deviation measures that,
the deviation of the data values
from the mean.
Learning Goal 2:
The Standard Deviation as a Ruler
• As the most common measure of
variation, the standard deviation
plays a crucial role in how we look
at data.
• The trick in comparing very
different-looking values is to use
standard deviations as our ruler.
• The standard deviation tells us how
the whole collection of values
varies, so it’s a natural ruler for
comparing an individual to a group.
Learning Goal 2:
The Standard Deviation as a Ruler
• We compare individual data values
to their mean, relative to their
standard deviation using the
following formula:
x  x

z
s
• We call the resulting values
standardized values, denoted as z.
They can also be called z-scores.
Learning Goal 3
Know how to calculate the z-score of an observation
and what it means.
Learning Goal 3: Standardizing with z-scores
A z-score measures the number of standard deviations that a data
value 𝑥 is from the mean 𝑥.
When 𝑥 is 1 standard deviation larger
than the mean, then z = 1.
(x  x)
z
s
for x  x  s, z 
( x  s)  x s
 1
s
s
When 𝑥 is 2 standard deviations larger
than the mean, then z = 2.
for x  x  2 s,
z
( x  2s)  x 2s

2
s
s
When x is larger than the mean, z is positive.
When x is smaller than the mean, z is negative.
Learning Goal 3:
Standardizing with z-scores
• A z-score puts values on a common
scale. Standardized values have no
units.
• z-scores measure the distance of each
data value is from the mean in
standard deviations.
• z-scores farther from 0 are more
extreme. z-scores beyond -2 or 2 are
considered unusual.
• A negative z-score tells us that the data
value is below the mean, while a
positive z-score tells us that the data
value is above the mean.
Learning Goal 3: Standardizing with z-scores
• Gives a common scale.
2.15 SD
Z=-2.15
This Z-Score tells
us it is 2.15
Standard
Deviations from
the mean
– We can compare two
different distributions with
different means and standard
deviations.
• Z-Score tells us how many
standard deviations the
observation falls away from
the mean.
• Observations greater than
the mean are positive when
standardized and
observations less than the
mean are negative.
Learning Goal 3:
Standardizing with z-scores - Example
•
If 𝑥 is from a unimodal symmetric
distribution with mean of 100 and
standard deviation of 50, the z-score
for 𝑥 = 200 is
x  x 200  100
z

 2.0
s
50
•
This says that 𝑥 = 200 is two
standard deviations (2 increments of
50 units) above the mean of 100.
Learning Goal 3:
Standardizing with z-scores – Your Turn
Bob is 64 inches tall. The heights of
men are unimodal symmetric with a
mean of 69 inches and standard
deviation of 2.5 inches. How does
Bob’s height compare to other men.
Learning Goal 3:
Standardizing with z-scores – Solution
Bob is 64 inches tall. The heights of men are unimodal
symmetric with a mean of 69 inches and standard
deviation of 2.5 inches. How does Bob’s height compare
to other men.
Solution:
𝑥 − 𝑥 64 − 69
𝑧=
=
= −2
𝑠
2.5
Bob’s height is 2 standard deviations
below the mean men’s height.
Learning Goal 3:
Problem
Which one of the following is a FALSE
statement about a standardized value
(z-score)?
a) It represents how many standard
deviations an observation lies from the
mean.
b) It represents in which direction an
observation lies from the mean.
c) It is measured in the same units as the
variable.
Learning Goal 3:
Problem (answer)
Which one of the following is a FALSE
statement about a standardized value
(z-score)?
a) It represents how many standard
deviations an observation lies from the
mean.
b) It represents in which direction an
observation lies from the mean.
c) It is measured in the same units as
the variable.
Learning Goal 3:
Problem
Rachael got a 670 on the analytical
portion of the Graduate Record Exam
(GRE). If GRE scores are unimodal
symmetric and have mean 𝑥 = 600 and
standard deviation s = 30, what is her
standardized score?
a)
670  600
 2.33
30
b)
600  670
 2.33
30
Learning Goal 3:
Problem (answer)
Rachael got a 670 on the analytical
portion of the Graduate Record Exam
(GRE). If GRE scores are unimodal
symmetric and have mean 𝑥 = 600 and
standard deviation s = 30, what is her
standardized score?
a)
670  600
 2.33
30
b)
600  670
 2.33
30
Learning Goal 4
Know how to compare values of two different variables
using their z-scores.
Learning Goal 4:
Benefits of Standardizing
• Standardized values have been
converted from their original units
to the standard statistical unit of
standard deviations from the mean
(z-score).
• Thus, we can compare values that
are measured on different scales,
with different units, or from
different populations.
Learning Goal 4:
Standardizing – Comparing Distributions
• The men’s combined skiing event in the in
the winter Olympics consists of two races:
a downhill and a slalom. In the 2006
Winter Olympics, the mean slalom time
was 94.2714 seconds with a standard
deviation of 5.2844 seconds. The mean
downhill time was 101.807 seconds with a
standard deviation of 1.8356 seconds. Ted
Ligety of the U.S., who won the gold
medal with a combined time of 189.35
seconds, skied the slalom in 87.93 seconds
and the downhill in 101.42 seconds.
• On which race did he do better compared
with the competition?
Learning Goal 4:
Standardizing – Comparing Distributions
•
Slalom time (𝑥): 87.93 sec.
Slalom mean 𝑥 : 94.2714 sec.
Slalom standard deviation (s): 5.2844 sec.
x  x

z
s
•
zSlalom 
87.93  94.2714
 1.2
5.2844
Downhill time (𝑥): 101.42 sec.
Downhill mean 𝑥 : 101.807 sec.
Downhill standard deviation (s): 1.8356 sec.
zDownhill 
•
101.42  101.807
 0.21
1.8356
The z-scores show that Ligety’s time in the slalom is
farther below the mean than his time in the downhill.
Therefore, his performance in the slalom was better.
Learning Goal 4:
Standardizing – Your Turn
• Timmy gets a 680 on the math of
the SAT. The SAT score distribution is
Unimodal symmetric with a mean of
500 and a standard deviation of
100. Little Jimmy scores a 27 on the
math of the ACT. The ACT score
distribution is unimodal symmetric
with a mean of 18 and a standard
deviation of 6.
• Who does better? (Hint: standardize
both scores then compare z-scores)
Learning Goal 4: Standardizing – Solution
• Timmy:
680  500
z
 1.8
100
• Timmy’s z score is further
away from the mean so he
does better than Little
Jimmy who’s only 1.5 SD’s
from the mean
• Little Jimmy:
27  18
z
 1.5
6
• Little Jimmy does better
than average and is 1.5 SD’s
from the mean but Timmy
beats him because he is .3
SD further.
Learning Goal 4: Standardizing – Your Turn
Which is better, an ACT score of 28 or a combined
SAT score of 2100?
• ACT: 𝑥 = 21, s = 5
• SAT: 𝑥 = 1500, s = 325
Assume ACT and SAT scores have unimodal
symmetric distributions.
28  21 7
z
  1.4
5
5
a)
ACT score of 28
2100  1500 600
z

 1.85
b)
SAT score of 2100
325
325
c)
I don’t know
Learning Goal 4: Standardizing – Class Problem
A town’s January high temp averages 36 ̊F with a
standard deviation of 10, while in July, the mean
high temp is 74 ̊F with a standard deviation of 8. In
which month is it more unusual to have a day with
a high temp of 55 ̊F?
Solution:
55−36
zJanuary =
10
55−74
zJuly =
=
8
= 1.9
−2.375
It is more unusual to
have a high temp. of
55 F in July, because
July’s z-score is further
from the mean.
Learning Goal 4:
Standardizing – Combining z-scores
Because z-scores are standardized
values, measure the distance of each
data value from the mean in standard
deviations and have no units, we can
also combine z-scores of different
variables.
Learning Goal 4:
Standardizing – Example: Combining z-scores
• In the 2006 Winter Olympics men’s
combined event, Ted Ligety of the
U.S. won the gold medal with a
combined time of 189.35 seconds.
Ivica Kostelic of Croatia skied the
slalom in 89.44 seconds and the
downhill in 100.44 seconds, for a
combined time of 189.88 seconds.
• Considered in terms of combined zscores, who should have won the
gold medal?
Learning Goal 4: Solution
• Ted Ligety:
zSlalom 
zDownhill
• Combined z-score: -1.41
• Ivica Kostelic:
87.93  94.2714
 1.2
5.2844
101.42  101.807

 0.21
1.8356
zSlalom 
89.44  94.2714
 0.91
5.2844
zDownhill 
100.44  101.807
 0.74
1.8356
• Combined z-score: -1.65
• Using standardized scores, overall Kostelic did better
and should have won the gold.
Learning Goal 4:
Combining z-scores - Your Turn
• The distribution of SAT scores has a
mean of 500 and a standard
deviation of 100. The distribution of
ACT scores has a mean of 18 and a
standard deviation of 6. Jill scored a
680 on the math part of the SAT and
a 30 on the ACT math test. Jack
scored a 740 on the math SAT and a
27 on the math ACT.
• Who had the better combined
SAT/ACT math score?
Learning Goal 4: Combining z-scores - Solution
• Jill
zSAT 
680  500
 1.8
100
z ACT 
30  18
 2.0
6
• Combined math score: 3.8
• Jack
zSAT 
740  500
 2.4
100
z ACT 
27  18
 1.5
6
• Combined math score: 3.9
• Jack did better with a combined math score of 3.9, to
Jill’s combined math score of 3.8.
Learning Goal 5
Recognize when a Normal model is appropriate.
Learning Goal 5:
Smooth Curve
• Sometimes the overall pattern of a
histogram is so regular that it can be
described by a Smooth Curve.
• This can help describe the location
of individual observations within the
distribution.
Learning Goal 5:
Smooth Curve
• The distribution of a histogram depends
on the choice of classes, while with a
smooth curve it does not.
• Smooth curve is a mathematical model of
the distribution.
– How?
• The smooth curve describes what
proportion of the observations fall in each
range of values, not the frequency of
observations like a histogram.
• Area under the curve represents the
proportion of observations in an interval.
• The total area under the curve is 1.
Learning Goal 5: Mathematical Model
A density curve is a mathematical model of a distribution.
The total area under the curve, by definition, is equal to 1, or 100%.
The area under the curve for a range of values is the proportion of all observations for
that range.
A mathematical model more represents a population then a sample like a histogram.
Therefore, when calculating z-scores for a mathematical model we use 𝜇 and 𝜎 , the
population mean and standard deviation, instead of 𝑥 and s, sample mean and
standard deviation.
Histogram of a sample with the
smoothed, density curve
describing theoretically the
population.
z
x

Learning Goal 5:
The Normal Model
• There is no universal standard for zscores, but there is a model that
shows up over and over in Statistics.
• This model is called the Normal
Model (You may have heard of “bellshaped curves.”).
• Normal models are appropriate for
distributions whose shapes are
unimodal and roughly symmetric.
• These distributions provide a
measure of how extreme a z-score
is.
Learning Goal 5:
The Normal Model
• Normal Model: One Particular class
of distributions or model.
1. Symmetric
2. Single Peaked
3. Bell Shaped
• All have the same overall shape.
Learning Goal 5:
The Normal Model or Normal Distribution
• The normal distribution is
considered the most important
distribution in all of statistics.
• It is used to describe the
distribution of many natural
phenomena, such as the height of a
person, IQ scores, weight, blood
pressure etc.
9-57
Learning Goal 5:
The Normal Distribution
• The mathematical equation for the
normal distribution is given below:
 x    /2 2
2
y
e
Not required
to know.
 2
where e  2.718,   3.14,  = population mean, and  =
population standard deviation.
Learning Goal 5:
Properties of the Normal Distribution
• When this equation is graphed for a
given  and , a continuous, bellshaped, symmetric graph will result.
• Thus, we can display an infinite
number of graphs for this equation,
depending on the value of  and .
• In such a case, we say we have a
family of normal curves.
• Some representations of the
normal curve are displayed in the
following slides.
9-59
Learning Goal 5:Properties of the Normal Dist.
Here, means are the same ( = 15)
while standard deviations are
different ( = 2, 4, and 6).
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
Here, means are different
( = 10, 15, and 20) while standard
deviations are the same ( = 3)
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
Learning Goal 5:
Properties of the Normal Distribution
Normal distributions with the same mean but with
different standard deviations.
9-61
Learning Goal 5:
Properties of the Normal Distribution
Normal distributions with different means but with
the same standard deviation.
9-62
Learning Goal 5:
Properties of the Normal Distribution
Normal distributions with different means and
different standard deviations.
9-63
Learning Goal 5: Describing a Normal Dist.
The exact curve for a particular normal distribution is
described by its Mean (μ) and Standard Deviation (σ).
μ located at the center of
the symmetrical curve
σ controls
the spread
Normal Distribution
Notation: N(μ,σ)
Learning Goal 5:
Properties of the Normal Distribution
• These normal curves have similar
shapes, but are located at different
points along the x-axis.
• Also, the larger the standard
deviation, the more spread out the
distribution, and the curves are
symmetrical about the mean.
• A normal distribution is a
continuous, symmetrical, bellshaped distribution.
9-65
Learning Goal 5:
Properties of the Normal Distribution
• Summary of the Properties of the
normal Distribution:
• The curve is continuous.
• The curve is bell-shaped.
• The curve is symmetrical about the
mean.
• The mean, median, and mode are
located at the center of the
distribution and are equal to each
other.
• The curve is unimodal (single mode)
• The curve never touches the x-axis.
• The total area under the normal curve
is equal to 1.
9-66
Learning Goal 5: Not Normal Curves
• Why
a)
b)
c)
d)
Normal curve gets closer and closer to the horizontal axis, but
never touches it.
Normal curve is symmetrical.
Normal curve has a single peak.
Normal curve tails do not curve away from the horizontal axis.
Learning Goal 5:
More Normal Distribution
• For a normal distribution the Mean
(μ) is located at the center of the
single peak and controls location of
the curve on the horizontal axis.
• The standard deviation (σ) is located
at the inflection points of the curve
and controls the spread of the
curve.
Learning Goal 5: Inflection Points
• The point on the curve where the curve changes from
falling more steeply to falling less steeply (change in
curvature – concave down to concave up).
Inflection point
Inflection point
• Located one standard deviation (σ) from the mean (μ).
• Allows us to visualize on any normal curve the width of one
standard deviation.
Learning Goal 5:
More Normal Model
• There is a Normal model for every
possible combination of mean and
standard deviation.
– We write N(μ,σ) to represent a Normal
model with a mean of μ and a standard
deviation of σ.
• We use Greek letters because this mean
and standard deviation are not numerical
summaries of the data. They are part of
the model and a model is more like a
population. They don’t come from the
data. They are numbers that we choose to
help specify the model.
• Such numbers are called parameters of
the model.
Learning Goal 5:
More Normal Model
• Summaries of data, like the sample
mean and standard deviation, are
written with Latin letters. Such
summaries of data are called
statistics.
• When we standardize Normal data,
we still call the standardized value a
z-score also, and we write
z
x

Learning Goal 5:
Standardizing the Normal Distribution
• All normal distributions are the
same general shape and share many
common properties.
• Normal distribution notation:
N(μ,σ).
• We can make all normal
distributions the same by measuring
them in units of standard deviation
(σ) about the mean (μ), z-scores.
• This is called standardizing and gives
us the Standard Normal Curve.
Learning Goal 5:
Standardizing the Normal Distribution
• How do linear transformations apply to zscores?
• When we convert a data value to a zscores, we are shifting it by the mean (to
set the scale at 0) and then rescaling by
the standard deviation (to reset the
standard deviation to 1).
– Standardizing into z-scores does not
change the shape of the distribution.
– Standardizing into z-scores changes the
center by making the mean 0.
– Standardizing into z-scores changes the
spread by making the standard
deviation 1.
Learning Goal 5: Standardizing the Normal Dist.
• We can standardize a variable that has a normal
distribution to a new variable that has the standard
normal distribution using the z-score formula:
Substitute your
variable as x
z
BAM! Pops out your
z-score
x

Then divide by your
Standard Deviation
Subtract the mean
from your variable
Learning Goal 5: Standardizing the Normal Dist.
x
x

Learning Goal 5: Standardizing the Normal Dist.
Standardizing Data into z-scores
The Standard Normal Distribution
Learning Goal 5: Standardizing the Normal Dist.
Distribution – any 𝝁 and 𝝈


X
Learning Goal 5: Standardizing the Normal Dist.
z 
Distribution – any 𝝁 and 𝝈
x

Standard Normal Distribution
𝝁 = 𝟎 𝒂𝒏𝒅 𝝈 = 𝟏

= 1

X
= 0
Results in a Standardized Normal Distribution (curve)
One Distribution → One set of areas under the curve → One Table
Z
Learning Goal 5:
Standardizing the Normal Distribution
• Subtracting Mu from each value X
just moves the curve around, so
values are centered on 0 instead of
on Mu.
• Once the curve is centered, dividing
each value by sigma>1 moves all
values toward 0, smushing the
curve.
Learning Goal 5: Standard Normal Dist. - Example
Learning Goal 5: Standard Normal Dist. - Example
Distribution 𝝁 = 𝟓, 𝝈 = 𝟏𝟎 𝒂𝒏𝒅 𝒙 = 𝟔. 𝟐
 = 10
= 5 6.2 X
Learning Goal 5: Standard Normal Dist. - Example
x   6.2  5
z

 .12

10
Distribution 𝝁 = 𝟓, 𝝈 = 𝟏𝟎 𝒂𝒏𝒅 𝒙 = 𝟔. 𝟐
 = 10
= 5 6.2 X
Standard Normal Distribution
𝝁 = 𝟎, 𝝈 = 𝟏 𝒂𝒏𝒅 𝒛 =. 𝟏𝟐
=1
= 0 .12
Z
The area under the original curve and the standard normal curve are the same.
Learning Goal 5:
The Standard Normal Curve
• Let x be a normally distributed variable with
mean μ and standard deviation σ, and let a
and b be real numbers with a < b. The
percentage of all possible observations of x
that lie between a and b is the same as the
percentage of all possible observations of z
that lie between (a −μ)/σ and (b−μ)/σ. This
latter percentage equals the area under the
standard normal curve between (a −μ)/σ and
(b−μ)/σ.
Learning Goal 5:
Standard Normal Curve and z-scores
• Same as with any Normal
Distribution.
• A z-score gives us an indication of
how unusual a value is because it tells
us how far it is from the mean.
• A data value that sits right at the
mean, has a z-score equal to 0.
• A z-score of 1 means the data value is
1 standard deviation above the mean.
• A z-score of –1 means the data value
is 1 standard deviation below the
mean.
Learning Goal 5:
Standard Normal Curve and z-scores
• How far from 0 does a z-score have
to be to be interesting or unusual?
• z-scores beyond -2 or 2 are
considered unusual.
• Remember that a negative z-score
tells us that the data value is below
the mean, while a positive z-score
tells us that the data value is above
the mean.
Learning Goal 5:
The Standard Normal Model
• Once we have standardized, we
need only one model:
– The N(0,1) model is called the
Standard Normal model (or the
Standard Normal distribution).
• Be careful—don’t use a Normal
model for just any data set, since
standardizing does not change the
shape of the distribution.
Learning Goal 5:
Properties of the Standard Normal Dist.
•
•
•
•
•
•
Shape – normal curve
Mean (μ) = 0
Standard Deviation (σ) = 1
Horizontal axis scale – Z score
No vertical axis
Notation: N(0, 1)
Learning Goal 5:
Standard Normal Dist. Problem
Which one of the following is a FALSE
statement about the standard
normal distribution?
a) The mean is greater than the
median.
b) It is symmetric.
c) It is bell-shaped.
d) It has one peak.
Learning Goal 5:
Standard Normal Dist. Problem (answer)
Which one of the following is a FALSE
statement about the standard
normal distribution?
a) The mean is greater than the
median.
b) It is symmetric.
c) It is bell-shaped.
d) It has one peak.
Learning Goal 5:
Standard Normal Dist. Problem
If you knew that the  = 0 and  = 3, which
normal curve would match the data?
a) Dataset 1
b) Dataset 2
Learning Goal 5:
Standard Normal Dist. Problem (answer)
If you knew that the  = 0 and  = 3, which
normal curve would match the data?
a) Dataset 1
b) Dataset 2
Learning Goal 5:
Standard Normal Dist. Problem
Which one of the following is a FALSE
statement about the standard normal
curve?
a) Its standard deviation  can vary with
different datasets.
b) It is bell-shaped.
c) It is symmetric around 0.
d) Its mean  = 0.
Learning Goal 5:
Standard Normal Dist. Problem (answer)
Which one of the following is a FALSE
statement about the standard normal
curve?
a) Its standard deviation  can vary with
different datasets.
b) It is bell-shaped.
c) It is symmetric around 0.
d) Its mean  = 0.
Learning Goal 5:
Standard Normal Dist. Problem
Suppose the lengths of sport-utility vehicles (SUV) are
normally distributed with mean  = 190 inches
and standard deviation  = 5 inches. Marshall just
bought a brand-new SUV that is 194.5 inches long
and he is interested in knowing what percentage
of SUVs is longer than his. Using his statistical
knowledge, he drew a normal curve and labeled
the appropriate area of interest. Which picture
best represents what Marshall drew?
a) Plot A
b) Plot B
Learning Goal 5:
Standard Normal Dist. Problem (answer)
Suppose the lengths of sport-utility vehicles (SUV) are
normally distributed with mean  = 190 inches
and standard deviation  = 5 inches. Marshall just
bought a brand-new SUV that is 194.5 inches long
and he is interested in knowing what percentage
of SUVs is longer than his. Using his statistical
knowledge, he drew a normal curve and labeled
the appropriate area of interest. Which picture
best represents what Marshall drew?
a) Plot A
b) Plot B
Learning Goal 6
Recognize when standardization can be used to
compare values.
Learning Goal 6:
Why We Standardize
• Standardizing allows us to compare
distributions by giving them a
common scale.
• If the distribution is Normal, then it
can be standardized.
• ALWAYS check to make sure the
Normal Model is appropriate before
standardizing data or using z-scores.
Learning Goal 6:
Nearly Normal Condition
• When we use the Normal model, we
are assuming the distribution is
Normal.
• We cannot check this assumption in
practice, so we check the following
condition:
– Nearly Normal Condition: The shape
of the data’s distribution is
unimodal and symmetric.
– This condition can be checked with
a histogram or a Normal probability
plot (to be explained later).
Learning Goal 6:
Nearly Normal Condition
• Standardization (z-scores)
can only be used when the
Nearly Normal Condition is
met.
Learning Goal 7
Be able to use Normal models and the 68-95-99.7 Rule to estimate the percentage
of observations falling within 1, 2, or 3 standard deviations of the mean.
Learning Goal 7:
The 68-95-99.7 Rule
• Normal models give us an idea of
how extreme a value is by telling us
how likely it is to find one that far
from the mean.
• We can find these numbers
precisely, but until then we will use
a simple rule that tells us a lot about
the Normal model…
• The 68-95-99.7 Rule or
Empirical Rule
Learning Goal 7:
The 68-95-99.7 Rule
• A very important property of any
normal distribution is that within a
fixed number of standard deviations
from the mean, all normal
distributions have the same fraction
of their probabilities.
• We will illustrate for for 1, 2,
and 3 from the mean .
9-102
Learning Goal 7:
The 68-95-99.7 Rule
• One-sigma rule: Approximately 68%
of the data values should lie within
one standard deviation of the mean.
• That is, regardless of the shape of
the normal distribution, the
probability that a normal random
variable will be within one standard
deviation of the mean is
approximately equal to 0.68.
• The next slide illustrates this.
9-103
Learning Goal 7: The 68-95-99.7 Rule
One sigma rule.
9-104
Learning Goal 7:
The 68-95-99.7 Rule
• Two-sigma rule: Approximately 95%
of the data values should lie within
two standard deviations of the
mean.
• That is, regardless of the shape of
the normal distribution, the
probability that a normal random
variable will be within two standard
deviations of the mean is
approximately equal to 0.95.
• The next slide illustrates this.
9-105
Learning Goal 7: The 68-95-99.7 Rule
Two sigma rule.
9-106
Learning Goal 7:
The 68-95-99.7 Rule
• Three-sigma rule: Approximately
99.7% of the data values should lie
within three standard deviations of
the mean.
• That is, regardless of the shape of
the normal distribution, the
probability that a normal random
variable will be within three
standard deviations of the mean is
approximately equal to 0.997.
• The next slide illustrates this.
9-107
Learning Goal 7: The 68-95-99.7 Rule
Three sigma rule.
9-108
Learning Goal 7:
The 68-95-99.7 Rule
• The following shows what the
68-95-99.7 Rule tells us:
Learning Goal 7: The 68-95-99.7 Rule
Because all Normal distributions share the same properties, we can
standardize our data to transform any Normal curve N(,) into the
standard Normal curve N(0,1).
N(64.5, 2.5)
N(0,1)
=>
x
z
Standardized height (no units)
And then use the 68-95-99.7 rule to find areas under the curve.
Learning Goal 7:
More 68-95-99.7% Rule
You can further divide the area
under the normal curve into the
following parts.
Using the 68-95-99.7 Rule
• SOUTH AMERICAN RAINFALL
• The distribution of rainfall in South
American countries is approximately
normal with a (mean) µ = 64.5 cm
and (standard deviation) σ = 2.5 cm.
• The next slide will demonstrate the
empirical rule of this application.
N(64.5,2.5)
• 68% of the countries receive rain fall
between 64.5(μ) – 2.5(σ) cm (62)
and 64.5(μ)+2.5(σ) cm (67).
– 68% = 62 to 67
• 95% of the countries receive rain fall
between 64.5(μ) – 5(2σ) cm (59.5)
and 64.5 (μ) + 5(2σ) cm (69.5).
– 95% = 59.5 to 69.5
• 99.7% of the countries receive rain
fall between 64.5(μ) – 7.5(3σ) cm
(57) and 64.5(μ) + 7.5(3σ) cm (72).
– 99.7% = 57 to 72
The middle 68% of the
countries (µ ± σ) have
rainfall between 62 –
67 cm
The middle 95% of the
countries (µ ± 2σ) have
rainfall between 59.5 –
69.5 cm
Almost all of
the data
(99.7%) is
within 57 – 72
cm (µ ± 3σ)
Example: IQ Test
• The scores of a referenced
population on the IQ Test are
normally distributed with μ=100
and σ=15.
1) Approximately what percent of
scores fall in the range from 70 to
130?
2) A score in what range would
represent the top 16% of the
scores?
Example: IQ Test
1) 70 to 130 is μ±2σ, therefore it
would 95% of the scores.
μ=100
2) The top 16% of the scoresσ=15
is one σ
above the μ, therefore the score
would be 115.
Your Turn:
• Runner’s World reports that the times
of the finishes in the New York City 10km run are normally distributed with a
mean of 61 minutes and a standard
deviation of 9 minutes.
1) Find the percent of runners who take
more than 70 minutes to finish.
16%
2) Find the percent of runners who finish
in less than 43 minutes.
2.5%
The First Three Rules for Working with Normal
Models
• Make a picture.
• Make a picture.
• Make a picture.
• And, when we have data, make a
histogram to check the Nearly
Normal Condition to make sure we
can use the Normal model to model
the distribution.
Finding Normal Percentiles by Hand
• When a data value doesn’t fall
exactly 1, 2, or 3 standard deviations
from the mean, we can look it up in
a table of Normal percentiles.
• Table Z in Appendix D provides us
with normal percentiles, but many
calculators and statistics computer
packages provide these as well.
Finding Normal Percentiles by Hand (cont.)
• Table Z is the standard Normal table. We
have to convert our data to z-scores
before using the table.
• The figure shows us how to find the area
to the left when we have a z-score of
1.80:
Standard Normal Distribution Table
• Gives area under the
curve to the left of a
positive z-score.
• Z-scores are in the
1st column and the
1st row
– 1st column – whole
number and first
decimal place
– 1st row – second
decimal place
Table Z
• The table entry for each
value z is the area under
the curve to the LEFT of z.
USING THE Z TABLE
•
You found your z-score to be 1.40 and
you want to find the area to the left of
1.40.
1.
2.
3.
Find 1.4 in the left-hand column of the Table
Find the remaining digit 0 as .00 in the top
row
The entry opposite 1.4 and under .00 is
0.9192. This is the area we seek: 0.9192
Other Types of Tables
Using Left-Tail Style Table
1. For areas to the left of a specified z
value, use the table entry directly.
2. For areas to the right of a specified z
value, look up the table entry for z
and subtract the area from 1. (can
also use the symmetry of the normal
curve and look up the table entry for
–z).
3. For areas between two z values, z1
and z2 (where z2 > z1), subtract the
table area for z1 from the table area
for z2.
More using Table Z (left tailed table)
Use table directly
Example: Find Area Greater Than a Given
Z-Score
• Find the area from the
standard normal
distribution that is greater
than -2.15
THE ANSWER IS 0.9842
• Find the corresponding Table Z value
using the z-score -2.15.
• The table entry is 0.0158
• However, this is the area to the left of 2.15
• We know the total area of the curve =
1, so simply subtract the table entry
value from 1
– 1 – 0.0158 = 0.9842
– The next slide illustrates these areas
Practice using Table A to find areas under the Standard Normal Curve
1. z<1.58
2. z<-.93
3. z>-1.23
4. z>2.48
5. .5<z<1.89
6. -1.43<z<1.43
1. .9429 (directly from table)
2. .1762 (directly from table)
3. .8907 (1-.1093 z<-1.23 or
use symmetry z<1.23)
4. .0066 (1-.9934 z<2.48 or
use symmetry z<-2.48)
5. .2791 (z<1.89=.9706 –
z<.5=.6915)
6. .8472 (z<1.43=.9236 – z<1.43=>0764)
Using the TI-83/84 to Find the Area Under
the Standard Normal Curve
• Under the DISTR menu, the 2nd entry is
“normalcdf”.
• Calculates the area under the Standard
Normal Curve between two z-scores (1.43<z<.96).
• Syntax normalcdf(lower bound, upper
bound). Upper and lower bounds are zscores.
• If finding the area > or < a single zscore use a large positive value for the
upper bound (ie. 100) and a large
negative value for the lower bound (ie.
-100) respectively.
Practice use the TI-83/84 to find areas under the standard normal curve
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
z>-2.35 and z<1.52
.85<z<1.56
-3.5<z<3.5
0<z<1
z<1.63
z>.85
z>2.86
z<-3.12
z>1.5
z<-.92
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
.9264
.1383
.9995
.3413
.9484
.1977
.0021
.0009
.0668
.1789
Using TI-83/84 to Find Areas Under the
Standard Normal Curve Without Z-Scores
• The TI-83/84 can find areas under the
standard normal curve without first
changing the observation x to a z-score
• normalcdf(lower bound, upper bound,
mean, standard deviation) If finding
area < or > use very large observation
value for the lower and upper bound
receptively.
• Example: N(136,18) 100<x<150
• Answer: .7589
• Example: N(2.5,.42) x>3.21
• Answer: .0455
Procedure for Finding Normal Percentiles
1.
State the problem in terms of the observed
variable y.
– Example : y > 24.8
2. Standardize y to restate the problem in terms
of a z-score.
– Example: z > (24.8 - μ)/σ, therefore z > ?
3. Draw a picture to show the area under the
standard normal curve to be calculated.
4.
Find the required area using Table Z or the TI83/84 calculator.
Example 1:
• The heights of men are
approximately normally distributed
with a mean of 70 and a standard
deviation of 3. What proportion of
men are more than 6 foot tall?
Answer:
1. State the problem in terms of y.
(6’=72”)
y  72
2. Standardize and state in terms of z.
z
y

72  70
z
 .67
3
3. Draw a picture of the area under the
curve to be calculated.
4. Calculate the area under the curve.
Example 2:
• Suppose family incomes in a town
are normally distributed with a
mean of $1,200 and a standard
deviation of $600 per month. What
are the percentage of families that
have income between $1,400 and
$2,250 per month?
Answer:
1. State the problem in terms of y.
1400  y  2250
2. Standardize and state in terms of z.
1400  1200
2250  1200
z
600
600
 .33  z  1.75
3. Draw a picture.
4. Calculate the area.
Your Turn:
• The Chapin Social Insight (CSI) Test
evaluates how accurately the subject
appraises other people. In the reference
population used to develop the test,
scores are approximately normally
distributed with mean 25 and standard
deviation 5. The range of possible scores is
0 to 41.
1. What percent of subjects score above a
32 on the CSI Test?
2. What percent of subjects score at or
below a 13 on the CSI Test?
3. What percent of subjects score between
16 and 34 on the CSI Test?
Solution:
1) What percent of subjects score
above a 32 on the CSI Test?
1. y>32
32  25
 1.4
2. z 
5
3. Picture
4. 8.1%
Solution:
2) What percent of subjects score at
or below a 13 on the CSI Test?
1) y≤13
13  25
2) z 
 2.4
5
3) Picture
4) .82%
Solution:
3) What percent of subjects score
between 16 and 34 on the CSI Test?
1) 16<y<34
2) 16  25  z  34  25 ,
5
3) Picture
4) 92.8%
5
 1.8  z  1.8
From Percentiles to Scores: z in Reverse
• Sometimes we start with areas and
need to find the corresponding zscore or even the original data
value.
• Example: What z-score represents
the first quartile in a Normal model?
z in Reverse
• Given a normal distribution proportion
(area under the standard normal
curve), find the corresponding
observation value.
• Table Z – find the area in the table
nearest the given proportion and read
off the corresponding z-score.
• TI-83/84 Calculator – Use the DISTR
menu, 3rd entry invNorm. Syntax for
invNorm(area,[μ,σ]) is the area to the
left of the z-score (or Observation y)
wanted (left-tail area).
From Percentiles to Scores: z in Reverse (cont.)
• Look in Table Z for an area of 0.2500.
• The exact area is not there, but 0.2514
is pretty close.
• This figure is associated with z = –0.67,
so the first quartile is 0.67 standard
deviations below the mean.
Inverse Normal Practice
• Proportion (area
• Z-Score
Using Table Z
1. .3409
2. .7835
3. .9268
4. .0552
Using TI-83/84
1. .3409
2. .7835
3. .9268
4. .0552
Using Table Z
1. Z = -.41
2. Z = .78
3. Z = 1.45
4. Z = -1.60
Using the TI-83/84
1. Z = -.4100
2. Z = .7841
3. Z = 1.4524
4. Z = -1.5964
under curve, left tail)
Procedure for Inverse Normal Proportions
1. Draw a picture showing the given
proportion (area under the curve).
2. Find the z-score corresponding to
the given area under the curve.
3. Unstandardize the z-score.
4. Solve for the observational value y
and answer the question.
Example 1: SAT VERBAL SCORES
• SAT Verbal scores are approximately
normal with a mean of 505 and a
standard deviation of 110
• How high must a student score in
order to place in the top 10% of all
students taking the verbal section of
the SAT.
Analyze the Problem and Picture It.
• The problem wants to know the SAT
score y with the area 0.10 to its
right under the normal curve with a
mean of 505 and a standard
deviation of 110. Well, isn't that the
same as finding the SAT score y with
the area 0.9 to its left? Let's draw
the distribution to get a better look
at it.
1. Draw a picture showing the given
proportion (area under the curve).
y=505
y=?
2.
Find Your Z-Score
1. Using Table Z - Find the entry
closest to 0.90. It is 0.8997. This is
the entry corresponding to z =
1.28. So z = 1.28 is the
standardized value with area 0.90
to its left.
2. Using TI-83/84 –
DISTR/invNorm(.9). It is 1.2816.
3. Unstandardize
• Now, you will need to unstandardize
to transform the solution from the z,
back to the original y scale. We
know that the standardized value of
the unknown y is z = 1.28. So y itself
satisfies:
y  505
 1.28
110
4. Solve for y and Summarize
• Solve the equation for y:
y  505  (1.28)(110)  645.8
• The equation finds the y that
lies 1.28 standard deviations
above the mean on this
particular normal curve. That is
the "unstandardized" meaning
of z = 1.28.
• Answer: A student must score
at least 646 to place in the
highest 10%
Example 2:
• A four-year college will accept any
student ranked in the top 60
percent on a national examination.
If the test score is normally
distributed with a mean of 500 and
a standard deviation of 100, what is
the cutoff score for acceptance?
Answer:
1. Draw picture of given proportion.
2. Find the z-score. From TI-83/84,
invNorm(.4) is z = -.25.
y  500
3. Unstandardize:
0.25 
100
4. Solve for y and answer the question.
y = 475, therefore the minimum score
the college will accept is 475.
Your Turn:
• Intelligence Quotients are normally
distributed with a mean of 100 and
a standard deviation of 16. Find the
90th percentile for IQ’s.
Answer:
1. Draw picture of given proportion.
2. Find the z-score. From TI-83/84,
invNorm(.9) is z = 1.28.
y  100
3. Unstandardize:
1.28 
16
4. Solve for y and answer the question.
y = 120.48, what this means; the 90th
percentile for IQ’s is 120.48. In other
words, 90% of people have IQ’s below
120.48 and 10% have IQ’s above
120.48.
Are You Normal? How Can You Tell?
• When you actually have your own
data, you must check to see
whether a Normal model is
reasonable.
• Looking at a histogram of the data is
a good way to check that the
underlying distribution is roughly
unimodal and symmetric.
Are You Normal? How Can You Tell? (cont.)
• A more specialized graphical display
that can help you decide whether a
Normal model is appropriate is the
Normal probability plot.
• If the distribution of the data is
roughly Normal, the Normal
probability plot approximates a
diagonal straight line. Deviations
from a straight line indicate that the
distribution is not Normal.
The Normal Probability Plot
A normal probability plot for data from a normal
distribution will be approximately linear:
X
90
60
30
-2
-1
0
1
2
Z
The Normal Probability Plot
Left-Skewed
Right-Skewed
X 90
X 90
60
60
30
30
-2 -1 0
1
2 Z
-2 -1 0
Rectangular
X 90
60
30
-2 -1 0
1
2 Z
1
2 Z
Nonlinear
plots
indicate a
deviation
from
normality
Are You Normal? How Can You Tell? (cont.)
• Nearly Normal data have a histogram and a Normal
probability plot that look somewhat like this
example:
Are You Normal? How Can You Tell? (cont.)
• A skewed distribution might have a histogram and
Normal probability plot like this:
Summary Assessing Normality
(Is The Distribution Approximately Normal)
1. Construct a Histogram or Stemplot.
See if the shape of the graph is
approximately normal.
2. Construct a Normal Probability Plot
(TI-83/84). A normal Distribution
will be a straight line. Conversely,
non-normal data will show a
nonlinear trend.
Assess the Normality of the Following Data
• 9.7, 93.1, 33.0, 21.2, 81.4, 51.1,
43.5, 10.6, 12.8, 7.8, 18.1, 12.7
• Histogram – skewed right
• Normal Probability Plot – clearly not
linear
Normal Distribution Problem – Your Turn:
• Suppose a normal model describes
the fuel efficiency of cars currently
registered in your state. The mean is
24 mpg, with a standard deviation
of 6 mpg.
• Sketch the normal model,
illustrating the 68-95-99.7 rule.
Normal Distribution Problem – Your Turn:
• What percent of all cars get less
than 15 mpg?
Normal Distribution Problem – Your Turn:
• What percent of all cars get
between 20 and 30 mpg?
Normal Distribution Problem – Your Turn:
• What percent of cars get more than
40 mpg?
Normal Distribution Problem – Your Turn:
• Describe the fuel efficiency of the
worst 20% of all cars?
Normal Distribution Problem – Your Turn:
• What gas mileage represents the
third quartile?
Normal Distribution Problem – Your Turn:
• Describe the gas mileage of the
most efficient 5% of all cars.
Normal Distribution Problem – Your Turn:
• What gas mileage would you
consider unusual? Why?
Normal Distribution Problem – Your Turn:
• What percent of cars get under 20
mpg?
Normal Distribution Problem – Your Turn:
• An ecology group is lobbying for a
national goal calling for no more
than 10% of all cars to be under 20
mpg. If the standard deviation does
not change what average fuel
efficiency must be attained?
Normal Distribution Problem – Your Turn:
• Car manufacturers argue that they
cannot raise the average that much
– they believe they can only get to
26 mpg. What standard deviation
would allow them to meet the “only
10% under 20 mpg” goal?
Normal Distribution Problem – Your Turn:
• What change in the fuel economy of
cars would achieving that standard
deviation bring about? What are the
advantages and disadvantages?
Assignment
• Chapter 6 Notes Worksheet
• Chapter 6, Exercises pg. 129 – 133: #3‐17 odd, 23,
35‐45 odd.
• Read Ch-7, pg. 146 - 163