STAT 515 fa 2021 Lec 07 slides The Normal distribution in

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.

Karl B. Gregory (U. of South Carolina) STAT 515 fa 2021 Lec 07 slides 1 / 16

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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).


Carl Friedrich Gaussian

Gaußo Standard der

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


so Leer E µ

PEVoxEX M

a d µ

0 0

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.


X growth of randomly selected

loblolly pine from age 3 S

X Normal r S E Ya

is Ö is

iiiiiiiiiin

is Ö is

3

i i

s r s 20 Star Stir 4T

7

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.


Ö3 3 F

https://people.stat.sc.edu/gregorkb/z-table.pdf

zof standard deviations aboveor below

µ

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


o 2

5 2

µr p auf

O Mo

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


transform into

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.


pyxDiPELE Ipy.z E

O O

Ä Ein

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.


play x s pCE Es SE pG 8oz o

4641 0.6195

t.IS

X Normal m S 8 4

PG8 Zoo pnorm4 pnorm 1.8

gg

Tl

P 5.25 6 2 p sey6

p s.FI z 651

p z siEi p z 1019

3023

c 001

iiiX N µ S E k so

4

471.6

5.6

P X s.es p Ei SEE p ze E

E

ÄÄT

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.


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

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.


gotten

X N µ S E Ya

i

jog

norm ots

go.rs E 0.675 S

X wlm.se zu Nlo

A

Eüös

HÜTEInterval 5 1 0.675 5 4 0.6

Exercise: Redo some exercises with pnorm and qnorm functions in R.


The cumulative dist function of N o.o dist

gnormo g

Z NCO

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?


X N µ 120 8 42 o

ÄEAE

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

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

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.


i

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


Sam

ple

Qua

ntile

s

-2 -1 0 1 2

-3-2

-10

1

not Normal


Sam

ple

Qua

ntile

s


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.


254120

X N N µ 3000 8 25 16in i

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STAT 515 fa 2021 Lec 07 slides The Normal distribution in