Download The Normal Distribution

The Normal Distribution
Outline
1
Introduction: Continuous Random Variables (3.6)
2
Introduction: Normal Distribution (4.1-4.2)
3
Areas Under a Normal Curve (4.3)
Standard Normal Table: Computing Probabilities
Standard Normal Table: Finding Quantiles
General Normal Distribution
Using the Computer
A Case Study
4
Assessing Normality (4.4)
Outline
1
Introduction: Continuous Random Variables (3.6)
2
Introduction: Normal Distribution (4.1-4.2)
3
Areas Under a Normal Curve (4.3)
Standard Normal Table: Computing Probabilities
Standard Normal Table: Finding Quantiles
General Normal Distribution
Using the Computer
A Case Study
4
Assessing Normality (4.4)
Recall: Discrete Random Variables
We can describe a discrete random variable with a table.
For example:
k
0
1
5
10
Pr {X = k } 0.1 0.5 0.1 0.3
But a continuous random variable can take any value in a
interval; we could never list all of these values in a table
Continuous Distributions
A continuous random variable has possible values over a
continuum.
The total probability of one is not in discrete chunks at
specific locations, but rather is ground up like a very fine
dust and sprinkled on the number line.
We cannot represent the distribution with a table of
possible values and the probability of each.
Continuous Distributions (continued)
Instead, we represent the distribution with a probability
density function which measures the thickness of the
probability dust.
Probability is measured over intervals as the area under
the curve.
A legal probability density f :
1
2
is never negative (f (x) ≥ 0 for −∞ < x <R ∞).
∞
has a total area under the curve of one ( −∞ f (x)dx = 1).
For You
Read 3.6 in your book
Outline
1
Introduction: Continuous Random Variables (3.6)
2
Introduction: Normal Distribution (4.1-4.2)
3
Areas Under a Normal Curve (4.3)
Standard Normal Table: Computing Probabilities
Standard Normal Table: Finding Quantiles
General Normal Distribution
Using the Computer
A Case Study
4
Assessing Normality (4.4)
The Normal Distribution
The Normal Distribution is the most important distribution
of continuous random variables.
The normal density curve is the famous symmetric,
bell-shaped curve.
The central limit theorem is the reason that the normal
curve is so important. Essentially, many statistics that we
calculate from large random samples will have
approximate normal distributions (or distributions derived
from normal distributions), even if the distributions of the
underlying variables are not normally distributed.
This fact is the basis of most of the methods of statistical
inference we will study in the last half of the course.
Chapter 4 introduces the normal distribution as a
probability distribution.
Chapter 5 culminates in the central limit theorem, the
primary theoretical justification for most of the methods of
statistical inference in the remainder of the textbook.
The famous bell curve
Used everywhere: very useful description for lots of biological
(and other) random variables.
Body weight,
Crop yield,
Protein content in soybean,
Density of blood components
σ
σ
Y ∼ N (µ, σ)
values can, in
principle, go to
infinity, both ways.
µ − 4σ µ − 2σ
µ
µ + 2σ µ + 4σ
The Normal Density
Normal curves have the following bell-shaped, symmetric
density.
f (x) = √
1
2πσ
− 21
e
x−µ
σ
2
,
(−∞ < x < ∞)
Parameters: The parameters of a normal curve are the
mean µ and the standard deviation σ.
Notation: a random variable is normal with mean µ and
standard deviation σ,
Y ∼ N (µ, σ)
Standard Normal. We often consider µ = 0 and σ = 1.
Z ∼ N (0, 1)
Recall: Empirical Rule
The 68–95–99.7 Rule.
For every normal curve:
1
2
3
≈ 68% of the area is within one SD of the mean
≈ 95% of the area is within two SDs of the mean
≈ 99.7% of the area is within three SDs of the mean
Standard Normal Density
Standard Normal Density
Area within 1 = 0.68
Area within 2 = 0.95
Density
Area within 3 = 0.997
−3
−2
−1
0
Possible Values
1
2
3
Outline
1
Introduction: Continuous Random Variables (3.6)
2
Introduction: Normal Distribution (4.1-4.2)
3
Areas Under a Normal Curve (4.3)
Standard Normal Table: Computing Probabilities
Standard Normal Table: Finding Quantiles
General Normal Distribution
Using the Computer
A Case Study
4
Assessing Normality (4.4)
Recall: the Normal Density
f (x) = √
1
2πσ
− 12
e
(can you integrate that?)
x−µ
σ
2
,
(−∞ < x < ∞)
Compute areas
There is no formula to calculate general areas under the
normal curve.
(The integral of the density has no closed form solution.)
We will learn to use normal tables for this.
We will also learn how to use R.
However, you will have to use normal tables on the exams.
The Standard Normal Table
We have tables for N (0, 1)
The standard normal table lists the area to the left of z
under the standard normal curve for each value from
−3.49 to 3.49 by 0.01 increments.
The normal table is located:
1
2
inside front cover of your textbook.
Table 3 (p. 675-676)
Numbers in the margins represent z.
Numbers in the middle of the table are areas to the left of z.
Warning!
Learn the normal table today.
Make it part of your being.
Standard Normal Table: Computing Probabilities
The standard normal Z ∼ N (0, 1)
Whole area = 1 (total probability rule)
Pr {Z ≤ 0} = by symmetry
Pr {Z = 0} = 0
Pr {Z < 1} = .8413 (table, front cover)
Pr {0 ≤ Z ≤ 1} =
Pr {−1 ≤ Z ≤ 1} =
Pr {Z > 1.5} =
−4
−2
0
1
2
3
4
Draw a picture and make use of symmetry.
Pr {−0.5 ≤ Z ≤ 0.3} =
The standard normal Z ∼ N (0, 1)
Whole area = 1 (total probability rule)
Pr {Z ≤ 0} = by symmetry
Pr {Z = 0} = 0
Pr {Z < 1} = .8413 (table, front cover)
Pr {0 ≤ Z ≤ 1} =
Pr {−1 ≤ Z ≤ 1} =
Pr {Z > 1.5} =
−4
−2
0
1
2
3
4
Draw a picture and make use of symmetry.
Pr {−0.5 ≤ Z ≤ 0.3} =
The standard normal Z ∼ N (0, 1)
Whole area = 1 (total probability rule)
Pr {Z ≤ 0} = by symmetry
Pr {Z = 0} = 0
Pr {Z < 1} = .8413 (table, front cover)
Pr {0 ≤ Z ≤ 1} =
Pr {−1 ≤ Z ≤ 1} =
Pr {Z > 1.5} =
−4
−2
0
1
2
3
4
Draw a picture and make use of symmetry.
Pr {−0.5 ≤ Z ≤ 0.3} =
The standard normal Z ∼ N (0, 1)
Whole area = 1 (total probability rule)
Pr {Z ≤ 0} = .5 by symmetry
Pr {Z = 0} = 0
Pr {Z < 1} = .8413 (table, front cover)
Pr {0 ≤ Z ≤ 1} =
Pr {−1 ≤ Z ≤ 1} =
Pr {Z > 1.5} =
−4
−2
0
1
2
3
4
Draw a picture and make use of symmetry.
Pr {−0.5 ≤ Z ≤ 0.3} =
The standard normal Z ∼ N (0, 1)
Whole area = 1 (total probability rule)
Pr {Z ≤ 0} = .5 by symmetry
Pr {Z = 0} = 0
Pr {Z < 1} = .8413 (table, front cover)
Pr {0 ≤ Z ≤ 1} =
Pr {−1 ≤ Z ≤ 1} =
Pr {Z > 1.5} =
−5 −4
−2
0
1
2
4
Draw a picture and make use of symmetry.
Pr {−0.5 ≤ Z ≤ 0.3} =
The standard normal Z ∼ N (0, 1)
Whole area = 1 (total probability rule)
Pr {Z ≤ 0} = .5 by symmetry
Pr {Z = 0} = 0
Pr {Z < 1} = .8413 (table, front cover)
Pr {0 ≤ Z ≤ 1} =
Pr {−1 ≤ Z ≤ 1} =
Pr {Z > 1.5} =
−4
−2
0
1
2
4
Draw a picture and make use of symmetry.
Pr {−0.5 ≤ Z ≤ 0.3} =
The standard normal Z ∼ N (0, 1)
Whole area = 1 (total probability rule)
Pr {Z ≤ 0} = .5 by symmetry
Pr {Z = 0} = 0
Pr {Z < 1} = .8413 (table, front cover)
Pr {0 ≤ Z ≤ 1} = 0.8413 − 0.5 = .3413
Pr {−1 ≤ Z ≤ 1} =
Pr {Z > 1.5} =
−4
−2
0
1
2
4
Draw a picture and make use of symmetry.
Pr {−0.5 ≤ Z ≤ 0.3} =
The standard normal Z ∼ N (0, 1)
Whole area = 1 (total probability rule)
Pr {Z ≤ 0} = .5 by symmetry
Pr {Z = 0} = 0
Pr {Z < 1} = .8413 (table, front cover)
Pr {0 ≤ Z ≤ 1} = 0.8413 − 0.5 = .3413
Pr {−1 ≤ Z ≤ 1} =
Pr {Z > 1.5} =
−4
−2 −1 0
1
2
4
Draw a picture and make use of symmetry.
Pr {−0.5 ≤ Z ≤ 0.3} =
The standard normal Z ∼ N (0, 1)
Whole area = 1 (total probability rule)
Pr {Z ≤ 0} = .5 by symmetry
Pr {Z = 0} = 0
Pr {Z < 1} = .8413 (table, front cover)
Pr {0 ≤ Z ≤ 1} = 0.8413 − 0.5 = .3413
Pr {−1 ≤ Z ≤ 1} = 2 ∗ 0.3413 = .6826
Pr {Z > 1.5} =
−4
−2 −1 0
1
2
4
Draw a picture and make use of symmetry.
Pr {−0.5 ≤ Z ≤ 0.3} =
The standard normal Z ∼ N (0, 1)
Whole area = 1 (total probability rule)
Pr {Z ≤ 0} = .5 by symmetry
Pr {Z = 0} = 0
Pr {Z < 1} = .8413 (table, front cover)
Pr {0 ≤ Z ≤ 1} = 0.8413 − 0.5 = .3413
Pr {−1 ≤ Z ≤ 1} = 2 ∗ 0.3413 = .6826
Pr {Z > 1.5} =
−4
0
1.5
4
5
Draw a picture and make use of symmetry.
Pr {−0.5 ≤ Z ≤ 0.3} =
The standard normal Z ∼ N (0, 1)
Whole area = 1 (total probability rule)
Pr {Z ≤ 0} = .5 by symmetry
Pr {Z = 0} = 0
Pr {Z < 1} = .8413 (table, front cover)
Pr {0 ≤ Z ≤ 1} = 0.8413 − 0.5 = .3413
Pr {−1 ≤ Z ≤ 1} = 2 ∗ 0.3413 = .6826
Pr {Z > 1.5} = 1 − Pr {Z ≤ −1.5}
= 1 − 0.9332 = .0668
−4
0
1.5
4
5
Draw a picture and make use of symmetry.
Pr {−0.5 ≤ Z ≤ 0.3} =
The standard normal Z ∼ N (0, 1)
Whole area = 1 (total probability rule)
Pr {Z ≤ 0} = .5 by symmetry
Pr {Z = 0} = 0
Pr {Z < 1} = .8413 (table, front cover)
Pr {0 ≤ Z ≤ 1} = 0.8413 − 0.5 = .3413
Pr {−1 ≤ Z ≤ 1} = 2 ∗ 0.3413 = .6826
Pr {Z > 1.5} = 1 − Pr {Z ≤ −1.5}
= 1 − 0.9332 = .0668
−4
−1.5
0
1.5
4
Draw a picture and make use of symmetry.
Pr {−0.5 ≤ Z ≤ 0.3} =
The standard normal Z ∼ N (0, 1)
Whole area = 1 (total probability rule)
Pr {Z ≤ 0} = .5 by symmetry
Pr {Z = 0} = 0
Pr {Z < 1} = .8413 (table, front cover)
Pr {0 ≤ Z ≤ 1} = 0.8413 − 0.5 = .3413
Pr {−1 ≤ Z ≤ 1} = 2 ∗ 0.3413 = .6826
Pr {Z > 1.5} = 1 − Pr {Z ≤ −1.5}
= 1 − 0.9332 = .0668
−4
−0.50.3
4
Draw a picture and make use of symmetry.
Pr {−0.5 ≤ Z ≤ 0.3} = Pr {Z ≤ 0.3} − Pr {Z ≤ −0.5}
The standard normal Z ∼ N (0, 1)
Whole area = 1 (total probability rule)
Pr {Z ≤ 0} = .5 by symmetry
Pr {Z = 0} = 0
Pr {Z < 1} = .8413 (table, front cover)
Pr {0 ≤ Z ≤ 1} = 0.8413 − 0.5 = .3413
Pr {−1 ≤ Z ≤ 1} = 2 ∗ 0.3413 = .6826
Pr {Z > 1.5} = 1 − Pr {Z ≤ −1.5}
= 1 − 0.9332 = .0668
−4
−0.50.3
4
Draw a picture and make use of symmetry.
Pr {−0.5 ≤ Z ≤ 0.3} = Pr {Z ≤ 0.3} − Pr {Z ≤ −0.5}
= .6179 − .3085 = .3094
Standard Normal Table: Finding Quantiles
Basic Idea
This is the inverse problem.
Before. Question: here is the interval; Answer: Probability
of the interval
Now. Question: Here is the probability. Answer: What is
the interval that matches with the probability.
Problem: many intervals have the same area.
We ask for a certain type of interval.
The Reverse Question: Finding quantiles
Pr {Z ≤?} = .975
the value ? is a quantile.
Use Table 3 (or front cover) backward
−3
0
?
3
Pr {Z ≤?} = 0.20
Pr {Z ≥?} = 0.80
−3
?
0
3
The Reverse Question: Finding quantiles
Pr {Z ≤?} = .975
? = 1.96
the value ? is a quantile.
Use Table 3 (or front cover) backward
−3
0
?
3
Pr {Z ≤?} = 0.20
Pr {Z ≥?} = 0.80
−3
?
0
3
The Reverse Question: Finding quantiles
Pr {Z ≤?} = .975
? = 1.96
the value ? is a quantile.
Use Table 3 (or front cover) backward
Pr {Z ≤?} = 0.20
Pr {Z ≥?} = 0.80
−3
0
?
3
? = −0.84
−3
?
0
3
The Reverse Question: Finding quantiles
Pr {Z ≤?} = .975
? = 1.96
the value ? is a quantile.
Use Table 3 (or front cover) backward
Pr {Z ≤?} = 0.20
Pr {Z ≥?} = 0.80
−3
0
?
3
? = −0.84
? = −0.84
−3
?
0
3
zα Notation
Instead of ? we introduce a fancy notation.
Let zα be the number such that
Pr {Z < zα } = 1 − α
Draw Figure 4.19 (α to the right; 1 − α to the left)
Example. z.025 = 1.96
Draw a picture!
(Note: your book uses a capital, Zα )
General Normal Distribution
General Normal?
We are good if Y ∼ N (0, 1)
What if Y ∼ N (10, 5) or N (11, 3), or ...
(Directions from “Home”)
Standardization
All normal curves have the same shape, and are simply
rescaled versions of the standard normal density.
Consequently, every area under a general normal curve
corresponds to an area under the standard normal curve.
The key standardization formula is
z=
x −µ
σ
Solving for x yields
x = µ + zσ
which says algebraically that x is z standard deviations
above the mean.
Probability Example
If X ∼ N(100, 2), find Pr {X > 97.5}.
Solution:
X − 100
97.5 − 100
Pr {X > 97.5} = Pr
>
2
2
= Pr {Z > −1.25}
= 1 − Pr {Z < −1.25}
= ?
Quantile Example
If X ∼ N(100, 2), find the cutoff values for the middle 70%
of the distribution.
Solution: The cutoff points will be the 0.15 and 0.85
quantiles.
From the table, 1.03 < z < 1.04 and z = 1.04 is closest.
Thus, the cutoff points are the mean plus or minus 1.04
standard deviations.
100 − 1.04(2) = 97.92,
100 + 1.04(2) = 102.08
General form N (µ, σ)
All normal distributions have the same shape.
Transformation
Y −µ
∼ N (0, 1)
Z =
σ
Systolic blood pressure in healthy adults has a normal
distribution with mean 112 mmHg and standard deviation 10
mmHg, i.e. Y ∼ N (112, 10).
One day, I have 92 mmHg.
92 − 112
Y − 112
Pr {Y ≤ 92} = Pr
≤
10
10
= Pr {Z ≤ −2} = 0.0227
General form N (µ, σ)
µ = 112 mmHg, σ = 10 mmHg.
Y − 112
122 − 112
102 − 112
≤
≤
Pr {102 ≤ Y ≤ 122} = Pr
10
10
10
= Pr {−1 ≤ Z ≤ 1} = .6826
68.3% of healthy adults have systolic blood pressure between
102 and 122 mmHg.
A patient’s systolic blood pressure is 137 mmHg.
Y − 112
137 − 112
Pr {Y ≥ 137} = Pr
≥
10
10
= Pr {Z ≥ 2.5} = 1 − .9938 = 0.0062
This patient’s blood pressure is very high...
General form N (µ, σ)
µ = 112 mmHg, σ = 10 mmHg.
What is “High blood pressure”? For instance, it could be the
value BP∗ such that
Pr {Y ≤ BP∗ } = .95
We need z∗ such that Pr {Z ≤ z∗ } = .95. Table in back cover
(last line): z∗ = 1.645. Thus BP∗ lies 1.645 standard deviations
above the mean:
BP∗ = 112 + 1.645 ∗ 10 = 112 + 16.45 = 128.45 mmHg
Formal approach:
.95 = Pr {Y ≤ BP∗ } = Pr
with
BP∗ − 112
Y − 112
≤
10
10
BP∗ − 112
z∗ =
i.e. BP∗ = 112 + 10 z∗
10
= Pr {Z ≤ z∗ }
Using the Computer
Other Options
Remember you need the tables for the exam.
Statistical software (e.g. R)
Online Calculators. See course website, handouts.
Calculators for Various Statistical Problems
Doing calculations with R
norm
binom
p
q
d
normal distribution
binomial
probability: Pr {Y ≤ . . .}
quantile
density, or probability mass
function: Pr {Y = . . .}
> pnorm(1)
[1] 0.8413447
> pnorm(2)-pnorm(-2)
[1] 0.9544997
> pnorm(3)-pnorm(-3)
[1] 0.9973002
> 1- pnorm(137, mean=112, sd=10)
[1] 0.006209665
> qnorm(.95, mean=112, sd=10)
[1] 128.4485
> pbinom(1, size=6, prob=1/6)
[1] 0.7367755
> dbinom(0:6, size=6, prob=1/6)
[1] 0.335 0.402 0.201 0.054 0.008 0.001 0.000
A Case Study
Exam Hint
This how exam questions will often look.
I give you some question in the context of a scientific area;
you figure out the probabilty calculations to use.
(word problems)
Case Study
Example
Body temperature varies within individuals over time (it can be
higher when one is ill with a fever, or during or after physical
exertion). However, if we measure the body temperature of a
single healthy person when at rest, these measurements vary
little from day to day, and we can associate with each person an
individual resting body temperture. There is, however, variation
among individuals of resting body temperture.
The Question
Example
In the population, suppose that:
the mean resting body temperature is 98.25 degrees
Fahrenheit;
the standard deviation is 0.73 degrees Fahrenheit;
resting body temperatures are normally distributed.
Let X be the resting body temperature of a randomly chosen
individual. Find:
1
Pr {X < 98}, the proportion of individuals with temperature
less than 98.
2
Pr {98 < X < 100}, the proportion of individuals with
temperature between 98 and 100.
3
The 0.90 quantile of the distribution.
4
The cutoff values for the middle 50% of the distribution.
Outline
1
Introduction: Continuous Random Variables (3.6)
2
Introduction: Normal Distribution (4.1-4.2)
3
Areas Under a Normal Curve (4.3)
Standard Normal Table: Computing Probabilities
Standard Normal Table: Finding Quantiles
General Normal Distribution
Using the Computer
A Case Study
4
Assessing Normality (4.4)
Normal Probability Plots
A standard question begins, “Assuming that variable Y has
a normal distribution,. . . ”. But how do we know the
distribution is approximately normal?
Sometimes we can rely on the central limit theorem.
If given data, there are tests for normality, but there are
reasons not to do these.
It is generally better to make a plot that sheds light on the
question, “Is the data so far from normality as to bias a
method that assumes normality?”
There is no easy answer to the question, but a normal
probability plot is much more informative than the result of
a test.
What is a Normal Probability Plot?
A normal probability plot is a plot of the sorted sample data
versus something close to the expected z-score for the
corresponding rank of a random normal sample of the
same size.
For example, the expected z-score of the minumum from a
random sample of size 10 from a normal population is
about −1.54.
If the plotted points are close to a straight line, there is
evidence that the distribution is close to normal.
If the plotted points are far from a straight line, there is
evidence of non-normality.
The Basic Idea
A normal probability plot plots the ordered observations on
the y-axis (y(1) , . . . , y(n) ) against some standard “normal
scores” x1 , . . . , xn on the x-axis.
If the data were truly from a normal distribution it would lie
on a “straight” line.
Why a straight line? Y = µ + σZ
Normal Example: All three plots are normal (n = 50)
−1
0
1
●
−2
−1
0
1
2
●
2
−2
−1
0
1
2
Theoretical Quantiles
Theoretical Quantiles
Histogram of x
Histogram of x
Histogram of x
−3 −2 −1
0
x
1
2
3
8
0
4
Frequency
15
0 5
5
10
Frequency
12
Theoretical Quantiles
0
Frequency
1
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Sample Quantiles
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Normal Q−Q Plot
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Sample Quantiles
Normal Q−Q Plot
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Sample Quantiles
Normal Q−Q Plot
−3 −2 −1
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x
1
2
3
−2
−1
0
1
x
2
Normal Example: All three plots are normal (n = 500)
0
1
2
−3 −2 −1
0
1
2
2
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Histogram of x
Histogram of x
Histogram of x
2
3
0
0
x
1
40
80
80
−1 0
3
80
Theoretical Quantiles
Frequency
Theoretical Quantiles
Frequency
Theoretical Quantiles
40
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Sample Quantiles
3
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3
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Frequency
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Normal Q−Q Plot
40
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Normal Q−Q Plot
Sample Quantiles
3
1
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Sample Quantiles
Normal Q−Q Plot
−3 −2 −1
0
x
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3
Non-normal Example: One plot is not normal
(n = 100)
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Normal Q−Q Plot
Sample Quantiles
Normal Q−Q Plot
Sample Quantiles
0 2 4 6 8
Sample Quantiles
Normal Q−Q Plot
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Theoretical Quantiles
Theoretical Quantiles
Histogram of x
Histogram of x
Histogram of x
0
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4
6
x
8
−2
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x
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3
15
0 5
Frequency
15
0 5
Frequency
20
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Frequency
40
Theoretical Quantiles
−2
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For You
Look at Figures 4.26 - 4.32 (p.136 - 139)
See how different shaped histograms lead to different
normal probability plots
Warning!
Learn the normal table today.
Make it part of your being.