Download Chapter 1 Looking at Data – Distributions

Chapter 1
Looking at Data –
Distributions
What is statistics?
• The science of collecting, organizing, and
interpreting numerical facts (data) with the
goal of gaining understanding about a
problem
• Always relate calculations back to the
problem at hand as numbers alone are not
meaningful
• Requires thinking and judgment about data
Variables
• A variable is a characteristic of an
individual, or object of interest (ie. Person,
plant, animal)
– Variables can take on different values for
different individuals
– Ex. Individual
Variable
Person
Flower
Bird
Age or Height
Color
Wingspan
Distributions
• The distribution of a variable tells us what
values the variable takes on (for the group
of individuals under consideration) and how
often it takes them
• Ex. Consider 10 rose bushes in a garden
– What colors are represented?
– How many of each color?
Variables
Categorial
- Value falls into one of
two or more groups, or
categories.
Ex. Blood type, hair color
Quantitative
-takes on numerical values
-Mathematical operations (such as
averaging) make sense
Ex. Height, age, number of credit
cards owned
It makes sense to talk about the average height of the
students in the class, but not the average blood type.
1.1 Displaying Distributions with Graphs
• For a categorical variable, the distribution
lists the categories and the count or percent
of individuals who fall into each one.
• How can we visually display this data?
– Bar graphs
• each category is represented by a bar
– Pie charts
• The slices must represent parts of one whole
Example: Top 10 causes of death in the United
States 2001
Rank Causes of death
Counts
% of top
10s
% of total
deaths
1 Heart disease
700,142
37%
29%
2 Cancer
553,768
29%
23%
3 Cerebrovascular
163,538
9%
7%
4 Chronic respiratory
123,013
6%
5%
5 Accidents
101,537
5%
4%
6 Diabetes mellitus
71,372
4%
3%
7 Flu and pneumonia
62,034
3%
3%
8 Alzheimer’s disease
53,852
3%
2%
9 Kidney disorders
39,480
2%
2%
32,238
2%
1%
10 Septicemia
All other causes
629,967
26%
For each individual who died in the United States in 2001, we record what was
the cause of death. The table above is a summary of that information.
Bar graphs
Top 10 causes of deaths in the United States 2001
The number of individuals
who died of an accident in
2001 is approximately
100,000.
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Counts (x1000)
Each category is represented by one bar. The bar’s height shows the count (or
sometimes the percentage) for that particular category.
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Counts (x1000)
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0
Top 10 causes of deaths in the United States 2001
Bar graph sorted by rank
 Easy to analyze
Sorted alphabetically
 Much less useful
Pie charts
Each slice represents a piece of one whole. The size of a slice depends on what
percent of the whole this category represents.
Percent of people dying from
top 10 causes of death in the United States in 2000
Make sure your
labels match
the data.
Make sure
all percents
add up to 100.
Percent of deaths from top 10 causes
Percent of
deaths from
all causes
How to Chart Quantitative Variables?
• Histograms – Numerical analog of bar graph
– The range of values a variable can take on is
divided into equal size intervals (bins)
– Histogram shows number of data points
(observations) that fall into each interval (bin)
– Choosing the correct bin size is judgment call
Histogram
Student Score
1
75
2
99
3
79
4
71
5
66
6
82
7
89
8
0
9
53
10
73
number of students
• Ex. Test 1 scores for 10 statistics students
10 bins
test score
number of students
What if we change the bin size?
4 bins
test score
Interpreting Histograms
• Look for overall pattern of data, and for any
striking departures from the pattern
• Look for outliers, individual values which fall
outside the overall pattern of a distributions
– Always watch out for outliers, and try to identify and
explain them
– Ex. Was the statistics test really hard, or were there
unusual circumstances for student 8? Did he not show
up for class, or did he cheat on his exam? Should he be
included in the distribution?
Stem Plots
• Separate each observation into a stem (all
but the final digit) and a leaf (final digit)
• Write the stems in a vertical column with
the smallest value at the top and draw
vertical line to right of column
• Write each leaf in row to right of its stem, in
increasing order
• Note: Some stems may have no leaves
Creating a Stem Plot:
Test scores of 10 students
Student Score
1
75
2
99
3
79
4
71
5
66
6
82
7
89
8
0
9
53
10
73
Score
0
53
66
71
73
75
79
82
89
99
Stemplot
0|0
1|
2|
3|
4|
5|
6|6
7|1359
8|29
9|9
More on Stem Plots
• Back-to-back stem plots with a common stem may
be useful for comparing two related distributions
• Stem plots don’t work too well for large data sets
– If each stem holds a large number of leaves, you can
split each stem into two:
• One for leaves 0-4
• One for leaves 5-9
• If observed values have too many digits, trim
numbers before making stemplot
– Ex. Trim 1234 to 123, then 12 is stem and 3 is leaf.
Indicate leaf unit is 10.
– See example 1.8 in text
Describing Distributions
• Can describe the overall pattern of a distribution by its shape,
center, spread and outliers
• Center – For now, consider the center the midpoint
– Value with approximately half the observations above it and half the
observations below it
• Spread – For now, describe by indicating smallest and largest
values
• Shape
– How many peaks does the distribution have?
• If one, unimodal
• If several, multimodal
– Is the distribution symmetric? Or skewed?
• Outliers – any points that fall far outside the other points
– You can use Tukey’s Rule to determine outliers of data
Most common distribution shapes
Symmetric
distribution
• A distribution is symmetric if the right and left
sides of the histogram are approximately mirror
images of each other.
• A distribution is skewed to the right if the right
side of the histogram (side with larger values)
extends much farther out than the left side. It is
skewed to the left if the left side of the histogram
extends much farther out than the right side.
Skewed
distribution
Complex,
multimodal
distribution

Not all distributions have a simple overall shape,
especially when there are few observations.
Time Plots
• A time plot of a variable plots each
observation against the time at which it was
measured
– Time always on horizontal axis!
• Look for patterns over time
– A trend is a rise or fall that persists over time,
despite small irregularities
– A pattern that repeats itself at regular intervals
of time is called seasonal variation
Ex. Retail price of fresh
oranges over time
Time is on the horizontal, x axis.
The variable of interest—here
“retail price of fresh oranges”—
goes on the vertical, y axis.
This time plot shows a regular pattern of yearly variations. These are seasonal
variations in fresh orange pricing most likely due to similar seasonal variations in
the production of fresh oranges.
There is also an overall upward trend in pricing over time. It could simply be
reflecting inflation trends or a more fundamental change in this industry.
Describing Distributions with Numbers
• Recall: Distributions of variables are
described by shape, center, spread and
outliers
• We now extend beyond inspecting stemplots
and histograms to more precise definitions of
center and spread
• Measures of center: the mean and the median
The Mean (x-bar)
• To find the mean of a set of n observations,
x1, x2, x3, … , xn, add their values and
divide by the number of observations:
x1  x2  x3  ...  xn
x
n
1
x   xi
or
n
S (Sigma) means sum
Example: Test scores on 2nd
exam for 10 statistics students
Exam scores: 80, 73, 92, 85, 75, 98, 93, 55, 80, 90
n = 10
x1  x2  x3  ...  xn
x
n
80  73  92  85  75  98  93  55  80  90
x
10
821
x
 82.1
10
• Note: The mean is sensitive to a few
extreme observations
– NOT a resistant measure of center
– What if there were an 1lth student in the class
who didn’t show up and received a 0 on the 2nd
exam?
• How would this affect the mean?
821  0 821
x

 74.6
10  1
11
The Median (M)
•
The median is the midpoint of a distribution
–
•
Half the observations are smaller and half the
observations are larger than M
To find the median:
1. Arrange data from smallest to largest
2. If the number of observations (n) is odd, M is
the center observation in the ordered list, located
(n+1)/2 observations up from the bottom
3. If the number of observations (n) is even, M is
the mean of the two center observations in the
ordered list. M is still located at the (n+1)/2
position
Finding the Median
• Consider again exam scores for 10 students:
Exam scores: 80, 73, 92, 85, 75, 98, 93, 55, 80, 90
• Arrange data from smallest to largest:
55, 73, 75, 80, 80, 85, 90, 92, 93, 98
• n = 10, so n is even and M is the mean of the
5th and 6th observations in the ordered list.
• M is located at (10+1)/2, or 5.5th position in
ordered list
• M = (80+85)/2 = 82.5
• What happens to M if we include the 11th
student who received a 0 in the data set?
Exam scores (in order): 0, 55, 73, 75, 80, 80, 85, 90, 92, 93, 98
• There are now 11 data points, so n = 11 and
is odd
• M is therefore center observation in
ordered list, located in position (12+1)/2,
or 6th position
• M = 80
The median is a more resistant measure of
center than the mean.
Comparing the mean and the median
The mean and the median are the same only if the distribution is
symmetrical. The median is a measure of center that is resistant to skew
and outliers. The mean is not.
Mean and median for
a symmetric
distribution
Mean
Median
Left skew
Mean and median
for skewed
distributions
Mean
Median
Mean
Median
Right skew
Impact of skewed data
Symmetric distribution…
Disease X:
x  3.4
M  3.4
Mean and median are the same.
… and a right-skewed distribution
Multiple myeloma:
x  3.4
M  2.5
The mean is pulled toward
the skew.
Measure of spread: the quartiles
The first quartile, Q1, is the value in the
sample that has 25% of the data at or
below it ( it is the median of the lower
half of the sorted data, excluding M).
M = median = 3.4
The third quartile, Q3, is the value in the
sample that has 75% of the data at or
below it ( it is the median of the upper
half of the sorted data, excluding M).
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25
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2
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0.6
1.2
1.6
1.9
1.5
2.1
2.3
2.3
2.5
2.8
2.9
3.3
3.4
3.6
3.7
3.8
3.9
4.1
4.2
4.5
4.7
4.9
5.3
5.6
6.1
Q1= first quartile = 2.2
Q3= third quartile = 4.35
Five-number summary and boxplot
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2
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6.1
5.6
5.3
4.9
4.7
4.5
4.2
4.1
3.9
3.8
3.7
3.6
3.4
3.3
2.9
2.8
2.5
2.3
2.3
2.1
1.5
1.9
1.6
1.2
0.6
Largest = max = 6.1
BOXPLOT
7
Q3= third quartile
= 4.35
M = median = 3.4
6
Years until death
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20
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5
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3
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1
5
4
3
2
1
Q1= first quartile
= 2.2
Smallest = min = 0.6
0
Disease X
Five-number summary:
min Q1 M Q3 max
Boxplots for skewed data
Years until death
Comparing box plots for a normal
and a right-skewed distribution
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10
9
8
7
6
5
4
3
2
1
0
Boxplots remain
true to the data and
depict clearly
symmetry or skew.
Disease X
Multiple Myeloma
Suspected Outliers
• Outliers are troublesome data points, and it is
important to be able to identify them.
• One way to raise the flag for a suspected outlier is
to compare the distance from the suspicious data
point to the nearest quartile (Q1 or Q3). We then
compare this distance to the interquartile range
(distance between Q1 and Q3).
• We call an observation a suspected outlier if it
falls more than 1.5 times the size of the
interquartile range (IQR) above the first quartile or
below the third quartile. This is called the
“1.5 * IQR rule for outliers.”
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7.9
6.1
5.3
4.9
4.7
4.5
4.2
4.1
3.9
3.8
3.7
3.6
3.4
3.3
2.9
2.8
2.5
2.3
2.3
2.1
1.5
1.9
1.6
1.2
0.6
8
7
Q3 = 4.35
Distance to Q3
7.9 − 4.35 = 3.55
6
Years until death
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24
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6
5
4
3
2
1
5
Interquartile range
Q3 – Q 1
4.35 − 2.2 = 2.15
4
3
2
1
Q1 = 2.2
0
Disease X
Individual #25 has a value of 7.9 years, which is 3.55 years above
the third quartile. This is more than 3.225 years, 1.5 * IQR. Thus,
individual #25 is a suspected outlier.
Measure of Spread: Standard Deviation
• The most common numerical description of a
distribution is given by the mean to measure
center and the standard deviation (s) to measure
spread
– Looks at how far observations are from their mean
• The variance of a set of observations (s2) is the
average of the squares of the deviations of the
observations from their mean:
• The standard deviation (s) is then given by the
square root of the variance:
1 n
2
s
(
x

x
)
 i
n 1 1
• The deviations xi – x are large in magnitude if
observations lie far from the mean
• Some deviations will be positive and some will be
negative depending on if the observations are
smaller or larger than the mean
• The sum of the deviations of the observations
from the mean will always be zero
• s and s2 will be large for widely spread distributions
and small if observations do not lie far from the mean
Steps for finding variance and
standard deviation:
1. Find the mean
2. subtract each value from the mean
3. Square each of the results
4. Add them together
5. Divide by n-1 (where n is the number of observations)
*** This value is the variance
6. take the square root to get the standard deviation
• Why divide by n-1?
– Since the sum of the deviations are zero, the
last observation/deviation can be calculated
once the other n-1 are known
– Thus we say there are only n-1 degrees of
freedom
• Why emphasize s over s2?
– s has the same unit of measurement as the
original observations
– Natural measure of spread for Normal
distribution (section 1.3)
Calculations …
s
1
df
n
 ( xi  x ) 2
1
Mean = 63.4
Sum of squared deviations from mean
= 85.2
Degrees freedom (df) = (n − 1) = 13
s2 = variance = 85.2/13 = 6.55 inches
squared
s = standard deviation = √6.55 = 2.56
inches
Women’s height (inches)
i
xi
x
(xi-x)
(xi-x)2
1
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63.4
-4.4
19.0
2
60
63.4
-3.4
11.3
3
61
63.4
-2.4
5.6
4
62
63.4
-1.4
1.8
5
62
63.4
-1.4
1.8
6
63
63.4
-0.4
0.1
7
63
63.4
-0.4
0.1
8
63
63.4
-0.4
0.1
9
64
63.4
0.6
0.4
10
64
63.4
0.6
0.4
11
65
63.4
1.6
2.7
12
66
63.4
2.6
7.0
13
67
63.4
3.6
13.3
14
68
63.4
4.6
21.6
Sum
0.0
Sum
85.2
Mean
63.4
Mean = 63.4 inches
x
s = 2.56 inches
Mean
± 1 s.d.
Standard Deviation in the
calculator:
Input the values in L1 (under STAT enter)
STAT-CALC-enter-enter
The Sx value is the sample
standard deviation
Another Standard Deviation Example
Find the SD for 3, 5, 6, 6, 7, 9, 10, 10, 14
Step 1: Find the mean:
(3 + 5 + 6 + 6 + 7 + 9 + 10 + 10 +14) / 9 = 7.8
Step 2: Subtract each value
from the mean:
(3-7.8) = -4.8
(5-7.8) = -2.8
(6-7.8) = -1.8
(6-7.8) = -1.8
(7-7.8) = -.8
(9-7.8) = 1.2
(10-7.8) = 2.2
(10-7.8) = 2.2
(14-7.8) = 6.2
Step 3: Square each value (be
sure to use parenthesis!)
(-4.8)²= 23.04
(-2.8)²= 7.84
(-1.8)²= 3.24
(-1.8)²= 3.24
(-.8)²= .64
(1.2)²= 1.44
(2.2)²= 4.84
(2.2)²= 4.84
(6.2)²= 38.44
Step 4: Add them all together
23.04 + 7.84 + 3.24 + 3.24 +
.64 + 1.44 + 4.84 + 4.84 +
38.44 = 87.56
Step 5: Divide by n-1 (n is the
number of observations)
84.32 / 8 = 10.945 (this is the
variance)
Step 6: Take the square root
sqrt(10.54) = 3.31
Properties of the Standard Deviation
• s measures spread about the mean
– Only use when mean is measure of center
• s = 0 only when there is NO spread
– Occurs when all observations have same value
– Otherwise, s > 0
• Like the mean, s is not resistant
– A few outliers can make s very large
– Remember, the deviation is squared!
Choosing among summary statistics
• Because the mean is not resistant to
Height of 30 Women
outliers or skew, use it to describe
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symmetrical and don’t have outliers.
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 Plot the mean and use the
standard deviation for error bars.
• Otherwise use the median in the five
Height in Inches
distributions that are fairly
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number summary which can be
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plotted as a boxplot.
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Box Plot
Boxplot
Mean +/- SD
Mean ± SD
What should you use, when, and why?
Arithmetic mean or median?
• Middletown is considering imposing an income tax on citizens. City
hall wants a numerical summary of its citizens’ income to estimate the
total tax base.
– Mean: Although income is likely to be right-skewed, the city
government wants to know about the total tax base.
• In a study of standard of living of typical families in Middletown, a
sociologist makes a numerical summary of family income in that city.
– Median: The sociologist is interested in a “typical” family and wants to
lessen the impact of extreme incomes.
Changing the unit of measurement
Variables can be recorded in different units of measurement. Most
often, one measurement unit is a linear transformation of another
measurement unit: xnew = a + bx.
Temperatures can be expressed in degrees Fahrenheit or degrees Celsius.
TemperatureFahrenheit = 32 + (9/5)* TemperatureCelsius  a + bx.
Linear transformations do not change the basic shape of a
distribution (skew, symmetry, multimodal). But they do change the
measures of center and spread:
– Multiplying each observation by a positive number b multiplies both
measures of center (mean, median) and spread (IQR, s) by b.
– Adding the same number a (positive or negative) to each observation adds a
to measures of center and to quartiles but it does not change measures of
spread (IQR, s).
Density Curves and Normal Distributions
• A density curve is a mathematical
idealization of a distribution of data,
picturing the overall pattern of the data and
ignoring minor irregularities as well as any
outliers
• A smooth approximation to the irregular
bars of a histogram
• A density curve is always on or above the
horizontal axis, and has area exactly 1
beneath it
• Recall, in a histogram, the areas of bars represent either counts or
proportions of observations (differ in scale on y-axis)
• If proportion, then total area of all bars is 1, and area of
shaded bars gives proportion of test scores 6.0 or lower
• Similarly, the total area under a density curve is 1, and the area
under the density curve for a range of values is the proportion of all
observations for that range.
Histogram of a sample with the
smoothed, density curve
describing theoretically the
population.
Density curves come in any
imaginable shape.
Some are well known mathematically
and others aren’t.
Median and mean of a density curve
The median of a density curve is the equal-areas point: the point that divides
the area under the curve in half.
The mean of a density curve is the balance point, at which the curve would
balance if it were made of solid material.
The median and mean are the same for a symmetric density curve.
The mean of a skewed curve is pulled in the direction of the long tail.
Notation
• We use x and s to denote the mean and
standard deviation, respectively, as
computed from a set of actual observations
• To distinguish an idealized distribution
from a sampled distribution, we denote the
mean of a density curve by m (the Greek
letter mu) and the standard deviation of a
density curve by s (the Greek letter sigma)
Normal Distributions
• Mean at center of symmetric distribution
• Standard deviation natural measure of spread
– Points of inflection of density curve are located
distance s on either side of m (ms, ms)
• Density curve notation: N(m,s)
Larger s, more spread out
Smaller s, less spread out
Why is the Normal distribution so important?
• Good description of data sets such as test
scores, characteristics of biological
populations, and repeated measurements of
the same quantity
• Good approximation to results of chance
outcomes such as tossing a coin many times
• Basis for many statistical inference
procedures
A family of density curves
Here, means are the same (m = 15)
while standard deviations are different
(s = 2, 4, and 6).
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
Here, means are different
(m = 10, 15, and 20) while standard
deviations are the same (s = 3)
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
The 68-95-99.7% Rule for Normal Distributions
• About 68% of all observations
are within 1 standard deviation
Inflection point
(s) of the mean (m) (for ALL
Normal distributions!).
• About 95% of all observations
are within 2 s of the mean m.
• Almost all (99.7%) observations
are within 3 s of the mean.
mean µ = 64.5
standard deviation s = 2.5
N(µ, s) = N(64.5, 2.5)
Reminder: µ (mu) is the mean of the idealized curve, while x¯ is the mean of a sample.
s (sigma) is the standard deviation of the idealized curve, while s is the s.d. of a sample.
The standard Normal distribution
Because all Normal distributions share the same properties, we can standardize our
data to transform any Normal curve N(m,s) into the standard Normal curve N(0,1).
X
N(64.5, 2.5)
Z
N(0,1)
=>
x
z
Standardized height (no units)
If a variable X has any Normal distribution N(m,s) then the standardized
variable Z = (X – m)/s has the standard normal distribution N(0,1).

For each x we calculate a new value, z (called a z-score).
Standardizing: calculating z-scores
A z-score measures the number of standard deviations that a data
value x is from the mean m.
z
(x  m )
s
When x is 1 standard deviation
larger than the mean, then z = 1.
m s  m s
for x  m  s , z 
 1
s
s
When x is 2 standard deviations
smaller than the mean, then z = -2.
for x  m  2s , z 
m  2s  m  2s

 2
s
s
When x is larger than the mean, z is positive.
When x is smaller than the mean, z is negative.
Ex. Women heights
Women’s heights follow the
N(64.5”,2.5”) distribution. What
percent of women are shorter than
67 inches tall (that’s 5’7”)?
mean µ = 64.5"
standard deviation s = 2.5"
x (height) = 67"
N(µ, s) =
N(64.5, 2.5)
Area= ???
Area = ???
m = 64.5” x = 67”
z=0
z=1
We calculate z, the standardized value of x:
z
(x  m)
s
(67  64.5) 2.5
, z

 1  1 stand. dev. from mean
2.5
2.5
Because of the 68-95-99.7 rule, we can conclude that the
percent of women shorter than 67” should be, approximately,
0.68 + half of (1 - 0.68) = 0.84 or 84%.
Using the standard Normal table
Table A gives the area under the standard Normal curve to the left of any z value.
.0082 is the
area under
N(0,1) left
of z = 2.40
.0080 is the area
under N(0,1) left
of z = -2.41
(…)
0.0069 is the area
under N(0,1) left
of z = -2.46
Percent of women shorter than 67”
For z = 1.00, the area under
the standard Normal curve
to the left of z is 0.8413.
N(µ, s) =
N(64.5”, 2.5”)
Area ≈ 0.84
Conclusion:
Area ≈ 0.16
84.13% of women are shorter than 67”.
By subtraction, 1 - 0.8413, or 15.87% of
women are taller than 67".
m = 64.5” x = 67”
z=1
What percent of women are shorter than 65”?
Height distributed
according to:
N(µ, s) = N(64.5”, 2.5”)
Tips on using Table A
Because the Normal distribution is
symmetrical, there are 2 ways that
Area = 0.9901
you can calculate the area under
the standard Normal curve to the
Area = 0.0099
right of a z value.
z = -2.33
area right of z = area left of -z
area right of z =
1
-
area left of z
More Tips on using Table A
To calculate the area between 2 z-values, first get the area under N(0,1)
to the left for each z-value from Table A.
Then subtract the
smaller area from the
larger area.
A common mistake made
by students is to subtract
both z values. The area
between z1 and z2 is NOT
the same as the area to the
left of z2 – z1 = 0.8
area between z1 and z2 =
area left of z1 – area left of z2
Note: The area under N(0,1) for a single value of z is zero.
Finding the percentage on the TI-84
Women’s heights follow the N(64.5”,2.5”) distribution. What percent of women are
shorter than 67 inches tall (that’s 5’7”)?
In the calculator:
2nd VARS normalcdf(lower bound, upper bound, mean, standard deviation)
OR:
2nd VARS – DRAW ShadeNorm(lower bound, upper bound, mean, standard deviation)
If the image doesn’t
appear, alter your
WINDOW
After viewing the graph, you must do 2nd PRGM ClrDraw
Example 1.27. The National Collegiate Athletic Association (NCAA) requires
Division I athletes to score at least 820 on the combined math and verbal SAT
exam to compete in their first college year. The SAT scores of 2003 were
approximately normal with mean 1026 and standard deviation 209.
What proportion of all students would be NCAA qualifiers (SAT ≥ 820)?
x  820
m  1026
s  209
(x  m)
z
s
(820  1026)
209
 206
z
 0.99
209
Table A : area under
z
N(0,1) to the left of
z  -.99 is 0.1611
or approx. 16%.
area right of 820
=
=
≈ 84%
In the calculator:
total area
1
-
area left of 820
0.1611
Ex. 1.28. The NCAA defines a “partial qualifier” eligible to practice and receive an
athletic scholarship, but not to compete, with a combined SAT score of at least 720.
What proportion of all students who take the SAT would be partial qualifiers?
That is, what proportion have scores between 720 and 820?
x  720
m  1026
s  209
(x  m)
z
s
(720  1026)
z
209
 306
z
 1.46
209
Table A : area under
N(0,1) to the left of
z  - 1.46 is 0.0721
or approx. 7%.
area between
720 and 820
≈ 9%
=
=
area left of 820
0.1611
-
area left of 720
0.0721
About 9% of all students who take the SAT have scores
between 720 and 820.
Inverse normal calculations
We may also want to find the observed range of values that
correspond to a given proportion/ area under the curve.
For that, we use Table A backward:
• we first find the
desired area/ proportion
in the body of the table,
• we then read the
corresponding z-value
from the left column
and top row.
For a left area of 1.25 % (0.0125),
the z-value is -2.24
Inverse Normal Calculations
Scores on the SAT verbal test in recent years follow the N(505,110)
distribution. How high must a student score to place in the top 5% of
all students taking the SAT?
1. To be in the top 5%, must find z value for standard normal
distribution with 95% of area to the left of z – Use Table A
z value closest to 0.95 is between 1.64 and 1.65. Use z = 1.645
2. Unstandardize. Transform
from z back to original x scale.
3. Interpret: This is the x that
lies 1.645 standard deviations above
the mean on the N(505,110) curve.
Scores above 685.95 are in the upper
5% of scores.
z
(x  m)
s
x  sz  m
x  (110)(1.645)  505
x  685.95
Inverse Normal Calculations
in the calculator
Scores on the SAT verbal test in recent years follow the N(505,110)
distribution. How high must a student score to place in the top 5% of
all students taking the SAT?
2nd VARS 3: invNorm (percent to the left, mean, standard deviation)
Since we are looking for the top 5%, the percent to the left is 95%.
Normal probability plots
One way to assess if a distribution is indeed approximately normal is to plot the
data on a normal probability plot.
The data points are ranked and the percentile ranks are converted to z-scores.
The z-scores are then used for the x axis against which the data are plotted on
the y axis of the normal probability plot.

If the distribution is indeed normal the plot will show a straight line,
indicating a good match between the data and a normal distribution.

Systematic deviations from a straight line indicate a non-normal
distribution. Outliers appear as points that are far away from the overall
pattern of the plot.
Good fit to a straight line: the
distribution of rainwater pH
values is close to normal.
Curved pattern: the data are not
normally distributed. Instead, it shows
a right skew: a few individuals have
particularly long survival times.
Normal probability plots are complex to do by hand, but you can create
them on your calculator using 2nd Y=. Choose the last option for Type.