Lecture 3 - Probability Distributions

7/29/2019 Lecture 3 - Probability Distributions

1/15

Statistical inference: probability

distributions and confidence intervals


2/15

We are now familiar with descriptivestatistics; but the main use of statisticalmethods is not description, but prediction

o i.e. we collect samples mostly to predict

characteristics of the whole population

The key instrument of extrapolation fromsample to population is the analysis ofprobability distributions:

o by assuming that our variables have a certaindistribution (normal, uniform, etc.), we can usesamples to infer population properties

In the following we examine the concept and

uses of statistical distributions 2


3/15

Most utilised statistical distribution is the

normal distribution (the Bell curve)

o also the most infamous due to certain misuses

o

http://crab.rutgers.edu/~goertzel/normalcurve.htm

However, there is nothing intrinsically wrong

with using probability distributions

o

well, anything in the wrong hands (from a breadknife to a fundamental law of nature proposed by

a pacifist) may become a weapon

3
http://crab.rutgers.edu/~goertzel/normalcurve.htmhttp://crab.rutgers.edu/~goertzel/normalcurve.htmhttp://crab.rutgers.edu/~goertzel/normalcurve.htmhttp://crab.rutgers.edu/~goertzel/normalcurve.htmhttp://crab.rutgers.edu/~goertzel/normalcurve.htmhttp://crab.rutgers.edu/~goertzel/normalcurve.htmhttp://crab.rutgers.edu/~goertzel/normalcurve.htmhttp://crab.rutgers.edu/~goertzel/normalcurve.htm


4/15

The first reason for popularity of the normal curve isdescriptive; i.e. we use it to model distribution ofcertain traits that look bell-shaped

What traits are bell-shaped? Typically, traits that are

optimised or established by biological or socialprocesses, and thus have a tendency to occur at anexpected valueo classic example: biological traits under natural selection

o A reason Darwin applied the principle of optimisation tonatural processes is that optimisation was a current

concept in Victorian society (especially in Economics)

4


5/15


6/15

The normal distribution is just a modified version ofour exponential

The curve

N(0,1) =

is thestandard normal distribution with

mean=0

sd=1

sum of frequencies=1

Distribution N(0, 1) is possibly the most used instatistical analyses

It says that for example:

the probability of being well above average (+3standard deviations above mean) is only 0.1%

probability of being one standard deviation

below average (-1 sd) is 0.1+2.1+13.6=15.8%(i.e. everything below -1) 6

-3 -2 -1 0 +1 +2 +3


7/15

However, real traits (body height,income, schooling years,number of social mediaaccounts) may have a normaldistribution (bell shape), butrarely with mean=0 andstandard deviation=1

That is not a problem: we canstandardise variables, i.e.

transform them so thateverything you measure hasmean=0 and sd=1

How is this done? With z-scores7


8/15

1) We take variable x and subtract themean from each caseo if mean height is 180 cm, someone 170 cm tall

now measures 170-180=-10

2) We take all residuals (case minus mean)and divide by standard deviationo if sd=10 and mean is 180cm, someone

measuring 190 cm deviates -10 cm/10 cm= -1standard deviation below the mean

In summary, standardisation or calculation

ofz-scores is simply convertinganymeasurements into standard deviationunitsz

=

-3 -2 -1 0 +1 +2 +3


9/15

So: if in a populationo mean height = 180 cm

o standard deviation=10

and you are 170cm, theno you measure 10 cm above the average

o you measure z = (170 180)/10 = -1

This means that the probability ofbeing shorter than 170 cm in thispopulation iso 0.1 + 2.1 + 13.6 = 15.8%

The reason for standardising is clear:it is the theoretical step that allows theapplication of the normal distributionto many quantifiable aspects of reality

9


10/15

We are interested in intervals of the normal

curve, not points

Why? What does it mean to ask what is the

probability of being a millionaire in the UK?(or their frequency)

o it does not mean the probability of having

exactly 1 million (thats a single point in

the curve)

o it means everyone havingover 1 million

(and thats an interval of the curve)

Cumulative probabilityis the probability of an

interval of values 10

a lower interval

an upper interval


11/15

It is easy to estimate cumulative probability of being

smaller than a value in RStudio

o you provide individual (test) value, mean, and sd,

and R calculates z-score and probability of the

interval defined by that value

Command pnorm(test value, mean, sd) calculates

cumulative probability from left to right, i.e. from to a value x (thats the blue area)

Example: if your height is 170 cm, average is 180

cm, and sd=10 cm, then probability of being shorter

than 170 cm is

o > pnorm(170,180,10)

o [1] 0.1586553 11

a lower interval


12/15

pnorm can estimate upper intervals too (i.e. the probability of

beingovera given value)

Example:

o what is the probability of being at least (i.e. taller than) 190

cm in the same population?

1) Probability of beingsmallerthan 190 cm (the WHITE area) is

> pnorm(190,180,10)

[1] 0.8413447

i.e. 0.841=84.1%

2) Thus probability of being over 190 cm is the rest of the curve

> 1-pnorm(190,180,10)

[1] 0.1586553

i.e.: probability of being taller than 190 cm is 1 (100%) minus the

probability of being smaller than 190 cm 12

an upper interval


13/15

Important: we can combine the two

things to calculate probability of extreme

values (i.e. too large or too small)

So what is the probability of being

shorter than 170cm OR taller than 190

cm, with N(180, 10)?

> 1pnorm(190, 180, 10)+pnorm(170, 180, 10)

(check why)

13


14/15

Now the most important case (well see why):

What about probability ofnot being extreme, i.e. of being

between 170 cm and 190 cm? (This means less than 10 cm

off average of 180 cm)

o > pnorm(190, 180, 10) pnorm(170, 180, 10)

14


15/15

Take the estimates of years at school by country (from the HDR2011

database); this is the variableschoolingyears:

How can we estimate the proportion of countries with children havinga) less than 3 years of schooling?

b) less than 5 years of schooling?

c) at least 7 years of schooling?

Hints:

-You need to use function pnorm

-To use pnorm you need the test value, the mean and the standard

deviation of variableschooling years 15

Documents

Lecture 3 - Probability Distributions