STAT 515 fa 2021 Lec 07 slides The Normal distribution in
Transcript of STAT 515 fa 2021 Lec 07 slides The Normal distribution in
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STAT 515 fa 2021 Lec 07 slides
The Normal distribution
Karl B. Gregory
University of South Carolina
These slides are an instructional aid; their sole purpose is to display, during the lecture,definitions, plots, results, etc. which take too much time to write by hand on the blackboard.
They are not intended to explain or expound on any material.
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Normal or Gaussian probability distributionA continuous rv X with pdf given by
f (x) =1p2⇡
1�exp
� (x � µ)2
2�2
�
has the Normal distribution with mean µ and variance �2
We write X ⇠ Normal(µ,�2).
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Carl Friedrich Gaussian
Gaußo Standard der
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The pdf of the Normal(µ,�2) distribution:
µµ� � µ+ �
area = .683
µ+ 2�µ� 2�
area = .954
µ+ 3�µ� 3�
area = .997
x
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so Leer E µ
PEVoxEX M
a d µ
0 0
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Mean and variance of Normal distributionIf X ⇠ Normal(µ,�2), then
EX = µ
VarX = �2
Exercise: Suppose growth in height (ft) of Loblolly pines from age three to five isNormal(µ = 5,�2 = 1/4). Give the probability that the growth of a randomlyselected Loblolly pine is
1 between 4.5 and 5.5 feet.2 more than 7 feet.3 less than 5.5 feet.4 between 3.5 feet and 5.5 feet.
Use the picture on the previous slide.
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X growth of randomly selected
loblolly pine from age 3 S
X Normal r S E Ya
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is Ö is
iiiiiiiiiin
is Ö is
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s r s 20 Star Stir 4T
7
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Get probabilities for X ⇠ Normal(µ,�2) like
P(a < X < b) =
Z b
a
1p2⇡
1�exp
� (x � µ)2
2�2
�dx
Conversion to the Standard Normal distributionIf X ⇠ Normal(µ,�2), then
Z =X � µ
�⇠ Normal(0, 1).
The Normal(0, 1) dist. is called the Standard Normal distribution and its pdf is
�(z) =1p2⇡
e�z2/2.
Can look up integrals over this pdf in a table.
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Ö3 3 F
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zof standard deviations aboveor below
µ
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The pdf of the Normal(0, 1) distribution:
0�1 1
area = 0.683
2�2
area = 0.954
3�3
area = 0.997
x
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o 2
5 2
µr p auf
O Mo
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The pdf of the Normal(0, 1) distribution:
0�1.282 1.282
area = 0.80
1.645�1.645
area = 0.90
1.96�1.96
area = 0.95
2.576�2.576
area = 0.99
x
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transform into
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If X ⇠ Normal(µ,�2), we can find P(a < X < b) in two steps:
1 Transform a and b to the Z -world (# of standard deviations world):
a 7! a� µ
�and b 7! b � µ
�,
2 FindP
✓a� µ
�< Z <
b � µ
�
◆
by using a “Z -table”�a table of Standard Normal probabilities.
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Ä Ein
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Exercise: Suppose growth in height (ft) of Loblolly pines from age three to five isNormal(µ = 5,�2 = 1/4). Give the probability that the growth of a randomlyselected Loblolly pine is
1 between 5.25 and 6.25 feet.2 more than 7.8 feet.3 less than 5.25 feet.4 between 4.1 feet and 5.2 feet.
Find the probabilities in the “Z -world” using a Z -table.
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t.IS
X Normal m S 8 4
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p s.FI z 651
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p z siEi p z 1019
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4
471.6
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5.6
P X s.es p Ei SEE p ze E
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Quantiles of a continuous random variableFor a continuous rv X with a strictly increasing cdf, the ✓th quantile of X is thevalue q✓ which satisfies
P(X q✓) = ✓,
where ✓ is a value in [0, 1].
A quantile is like a percentile, but not expressed as a percentage.
Example: If X is the weight of a fresh chicken egg:With probability 0.90, a randomly selected egg has weight q0.90.With probability 0.25, a randomly selected egg has weight exceeding q0.75.The median weight is q0.50.
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X f x
Ä
80.90 90 quatik of X
90 tile of X
Find 0 quantile of X Z Nla
X N µ
ink
Find O quantik of X Stütz Z Nla
iii
fypU
z
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If X ⇠ Normal(µ,�2), we can find q✓ such that P(X q✓) = ✓ in two steps:
1 Find qZ✓ such that P(Z < qZ✓ ) = ✓ using a “Z -table”.2 Get the corresponding quantile in the X -world as
q✓ = µ+ �qZ✓
Exercise: Suppose growth in height (ft) of Loblolly pines from age three to five isNormal(µ = 5,�2 = 1/4). Let X denote the height of a randomly selectedLoblolly pine and find
1 the 75%-tile of growth.2 the median of the growths, i.e. the 50%-tile of X .3 an interval, centered at the mean, within which X lies with probability 0.50.
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X N µ S E Ya
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jog
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go.rs E 0.675 S
X wlm.se zu Nlo
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Eüös
HÜTEInterval 5 1 0.675 5 4 0.6
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Exercise: Redo some exercises with pnorm and qnorm functions in R.
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The cumulative dist function of N o.o dist
gnormo g
Z NCO
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Exercise: You sell jars of baby food labelled as weighing 4oz ⇡ 113g. Supposeyour process results in jar weights with the Normal(µ = 120,�2 = 42) distributionand that you will be fined if more than 2% of your jars weigh less than 113g.
1 What proportion of your jars weigh less than 113g?2 To what must you increase µ to avoid being fined?3 Keeping µ = 120g, to what must you reduce � to avoid being fined?
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X N µ 120 8 42 o
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p s p F E P za E
We need P X 113 40.02
Write PL XF F 0.02
P z 113J E 0.02
113 µ 4 GoÄH
Gnorm0.02
Euböa 2.054
Need to kamµ 113 4 2.0541 121.22
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X N µ 120 E
P X 113 0.02
p JE E 0.02
p z j 0.02
tüte 113 120 E T Go
113 170 We need re Ii 3.4
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Do my data come from a Normal distribution?
Example: These are the commute times (sec) to class of a sample of students.
1832 1440 1620 1362 577 934 928 998 1062 9001380 913 654 878 172 773 1171 1574 900 900
Check with a Q-Q plot whether the data quantiles match those of a Normal distribution.
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i
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Some more Q-Q plots:
-2 -1 0 1 2
01
23
not Normal
Theoretical Quantiles
Sam
ple
Qua
ntile
s
-2 -1 0 1 2
-2-1
01
2
Normal
Theoretical Quantiles
Sam
ple
Qua
ntile
s
-2 -1 0 1 2
-3-2
-10
1
not Normal
Theoretical Quantiles
Sam
ple
Qua
ntile
s
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Sum of independent Normal random variablesIf X1 ⇠ Normal(µ1,�2
1), . . . ,Xn ⇠ Normal(µ1,�2n) are independent random
variables, thennX
i=1
Xi ⇠ Normal
nX
i=1
µi ,nX
i=1
�2i
!.
In the above, independent means that the values of the rvs don’t affect one other.
Exercise: Consider boxes containing 25 jars of baby food (from previous).1 What is the expected weight of the boxes?2 What is the standard deviation of the box weights?3 Give the probability that the box weighs less than 2,975 grams.
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254120
X N N µ 3000 8 25 16in i