Normal and Poisson
Distributions
GTECH 201
Lecture 14
Sampling

Population


Unit


Any individual member of the population
Sample


The entire group of objects about which
information is sought
A part or a subset of the population used to gain
information about the whole
Sampling Frame

The list of units from which the sample is chosen
Simple Random Sampling

A simple random sample of size n is a
sample of n units chosen in such a way
that every collection of n units from a
sampling frame has the same chance of
being chosen
Random Sampling in R


In R you can simulate random draws
For example, to pick five numbers at
random from the set 1:40, you can
> sample(1:40,5)
[1] 4 30 28 40 13
Sampling with Replacement

Default in R
is ‘without replacement’
sample(c("H", "T"), 10, replace=T)
[1] "T" "T" "T" "T" "T" "H" "H" "T" "H" "T“
prob=c(.9,.1)
sample(c("S", "F"), 10, replace=T, prob)
Random Number Tables

A table of random digits is:

A list of 10 digits 0 through 9 having the following
properties




The digit in any position in the list has the same chance
of being any of of 0 through 9;
The digits in different positions are independent, in that
the value of one has no influence on the value of any
other
Any pair of digits has the same chance of being any of
the 100 possible pairs, i.e., 00,01,02, ..98, 99
Any triple of digits has the same chance of being any of
the 1000 possible triples, i.e., 000, 001, 002, …998, 999
Using Random Number Tables

A health inspector must select a SRS of size 5
from 100 containers of ice cream to check for
E. coli contamination





The task is to draw a set of units from the
sampling frame
Assign a number to each individual
Label the containers 00, 01,02,…99
Enter table and read across any line
81486 69487 60513 09297
81, 48, 66, 94, 87, 60, 51, 30, 92, 97
Random Number
Generation in R




> rnorm(10)
> rnorm(10, mean=7, sd=5)
> rbinom(10, size=20, prob=.5)
We will revisit the meaning of the
parameters at the end of today’s session
Combinatorics 1




Back to draw five out of 40
sample(1:40,5)
The probability for any given number is
1/40 in the first sample,, 1/39 in the
second, and so on
 P(x ) = 1/(40*39*38*37*36*35)
> 1/prod(40:36)
[1] 1.266449e-08
But…
Combinatorics 2




We don’t care about the order of the
five numbers out of 40
There are 5*4*3*2*1 combinations for
the five drawn numbers
 > prod(1:5) / prod(40:36)
[1] 1.519738e-06
Shorthand for the above in
> 1/choose(40,5)
Binomial Distribution


Discrete probability distribution
Events have only 2 possible outcomes



binary, yes-no, presence-absence
Computing probability of multiple events or
trials
Examples


Probability that x number of people are alive at the
age of 65
Probability of a river reaching flood stage for three
consecutive years
When to Apply Binomial

If sample is less than 10% of a large
population in which a proportion p have a
characteristic of interest, then the
distribution X, the number in the sample
with that characteristic, is approximately
binomial (n, p), where n is the sample size
Geometric Distribution


Tossing a biased coin until the first head
appears pr(H) = p
pr(X = x) = pr(TT…T H) = pr(T1 ∩ T2 ∩ ..∩ Hx)
= (1 – p)x-1 p
The geometric distribution is the distribution
of the number of tosses of a biased coin up to
and including the first head
Poisson Distribution

Discrete probability distribution


Named in honor of Simeon Poisson
(1781-1840)
What is it used for?


To model the frequency with which a specified event
occurs over a period of time
The specified event occurs randomly



Independent of past or future occurrences
Geographers also use this distribution to model how
frequently an event occurs across a particular area
We can also examine a data set (of frequency counts
in order to determine whether a random distribution
exists
Poisson Distribution is used…

To analyze the number of patients arriving at a
hospital emergency room between 6 AM and 7
AM on a particular day




Obvious implications for resource allocation
To analyze the number of phone calls per day
arriving at a telephone switchboard
To analyze the number of cars using the drive
through window at a fast-food restaurant
To analyze hailstorm occurrence in one
Canadian province
The Poisson Probability Formula
P( X  x)  e


x
x!
Lambda () is a positive real number (mean frequency)
e = 2.718 (mathematical constant)
X = 0, 1, 2, 3, ….(frequency of an occurrence)
X!= X factorial
Example - 1

General Hospital, located in Phoenix, keeps records of
emergency room traffic. From these records, we find
that the number of patients arriving between 10 AM
and 12 Noon has a Poisson distribution of with
parameter
 =6.9
Determine the probability that, on any given day, the
number of patients arriving at that emergency room
between 10 AM and 12 Noon will be:
 Exactly four
 At the most two
Exactly four arrivals, x=4
4
(6.9)
P( X  4)  e
4!
 (6.9) 2266.7121
P( X  4)  2.718
24
 (6.9)
 0.095
P( X  2)
At the most, two arrivals…
P( X  2)  p( X  0)  p( X  1)  P( X  2)
0
1
2







(6.9)
(6.9)
(6.9)
 (6.9)
P( X  2)  e




 0!   1!   2!  
e
6.9
(1  6.9  23.805)
 0.032
Revisiting Mean and
Standard Deviation
Mean
Standard Dev.
 
 
What if…
We wanted to obtain a table of probabilities for the
random variable X, the number of patients arriving
between 10AM and 12 Noon?
Number
arriving, x
0
1
2
3
4
5
6
7
8
9
10
Probability
(X=x)
0.001
0.007
0.024
0.055
0.095
0.131
0.151
0.149
0.128
0.098
0.068
Discrete versus
Continuous Distributions


Moving from individual probabilities to total
number of successes or failures
Probability distribution f (x ) = P (X=x) for
discrete events:
n!
n x
x
f  x 
p 1  p 
x !(n  x)!

Probability distribution for continuous events:
2

x   

1
f  x 
exp  

2


2

2


Expected Values

Population standard deviation
square root of the average squared distance
of X from the mean 
sd ( X )  E [( X   )2 ]
Expected Values

Mean and Poisson distribution
E(X) = ixipr(xi)
= 0 x pr(0) + 1 x pr(1) + 2 x pr(2) + ...
e-0
e-1
e-2
= 0 x pr(
) + 1 x pr(
) + 2 x pr(
) + ...
0!
1!
2!


It can be shown that this adds to . Thus, for
Poisson-distributed populations E(X) = 
The standard deviation sd(X) for Poisson() is √
Probability Density Functions



Moving from the discrete to the continuous
Increasing the frequency of observations
results in an ever finer histogram
Total area under the curve = 1
Probability Density Functions

Population means and standard dev’s
•
x balances the distribution
f(x)
x
x
•
The standard deviation is calculated as for discrete
density functions
The Normal Distribution
Properties of a Normal
Distribution


Continuous Probability Distribution
Symmetrical about a central point




No skewness
Central point in this dataset corresponds to all three
measures of central tendency
Also called a Bell Curve
If we accept or assume that our data is normally
distributed, then,

We can compute the probability of different outcomes
Properties of a Normal
Distribution

Using the symmetrical property of the
distribution, we can conclude:





50 % of values must lie to the right, i.e. they are
greater than the mean
50% of values must lie to the left, i.e.
If the data is normally distributed, the probability
values are also normally distributed
The total area under the normal curve represents
all (100%) of probable outcomes
What can you say about data values in a normally
distributed data set?
Normal Distribution and
Standard Deviations
Approximating a Normal
Distribution
In reality,




If a variable’s distribution is shaped roughly
like a normal curve,
Then the variable approximates a normal
distribution
Normal Distribution is determined by



Mean
Standard Deviation
These measures are considered parameters of a
Normal Distribution / Normal Curve
Equation of a Normal Curve
 x  
1
f  x 
exp  
2

2

2

Mean =

; Standard Deviation =
e = 2.718 ;
 = 3.142

2




Areas Within
the Normal Curve

For a normally distributed variable,


the percentage of all possible observations
that lie within any specified range equals
the corresponding area under its
associated normal curve expressed as a
percentage
A college has an enrollment of 3264
female students. Mean height is 64.4
inches, standard deviation is 2.4 inches

Frequency and relative frequency are
presented
Frequency and
Relative Frequency
Table
0.0735, i.e. 7.35 % of the
students are between 67 and
68 inches tall
Height (in) Freq
Rel. Freq
56 < 57
3
0.0009
57 < 58
6
0.0018
58 < 59
26
0.0080
59 < 60
74
0.0227
60 < 61
147
0.0450
61 < 62
247
0.0757
62 < 63
382
0.1170
63 < 64
483
0.1480
64 < 65
559
0.1713
65 < 66
514
0.1575
66 < 67
359
0.1100
67 < 68
240 0.0735
68 < 69
122
0.0374
69 < 70
65
0.0199
70 < 71
24
0.0074
71 < 72
7
0.0021
72 < 73
5
0.0015
73 < 74
1
0.0003
3264
1
Relative Frequency Histogram with Normal Curve
0.0735 = the area that has
been cross-hatched
Shaded area under the normal
curve approximates the
percentage of students who
are between 67-68 inches tall
Standardizing a Normal Variable





Once we have mean and standard deviation of a
curve, we know its distribution and the associated
normal curve
Percentages for a normally distributed variable are
equal to the areas under the associated normal
curve
There could be hundreds of different normal curves
(one for each choice of mean or std. dev. value
How can we find the areas under a standard
normal curve?
A normally distributed variable with a mean of 0
and a standard deviation of 1 is said to have a
standard normal distribution
Z Score
z
x

xx
z
s
The variable z is called the standardized version of x, or
the standardized variable corresponding to x, with the
mean = 0 and standard deviation = 1
Almost all observations in a dataset will lie within three
standard deviations to either side of the mean, i.e.,
almost all possible observations will have z scores
between – 3 and + 3
Normal Curve Properties




The total area under the standard
normal curve is equal to 1
The standard normal curve extends
infinitely in both directions, approaching
but never touching the horizontal axis
Standard normal curve is symmetric
about 0
Most of the area under a standard
normal curve lies between –3 and + 3
Using the Standard Normal Table
The times taken for runners to complete a local 10 km race is normally
distributed with a mean of 61 minutes and a standard deviation of 9
minutes. Let x be the finish time of a randomly selected runner. Find the
probability that x > 75 minutes
Step 1
Calculate the standard score
z = 75-61/9; z = 1.56
Step 2
Determine the probability from the normal table
For z of 1.56, p = 0.4406
Step 3
Interpret the result
p (x>75) = 0.5 – 0.446
= 0.054 or 5.4% chance
Using the Standard Normal Table

In the previous example, what is the probability that someone
finishes in less than 45 minutes?
Step 1
Calculate the standard score
z = 45-61/9; z = -1.78
Step 2
Determine the probability from the normal table
For z of -1.78, area= 0.4625
Step 3
Interpret the result
p (x<45) = 1- (0.5+0.4625)
= 0.038 or 3.8 % of the runners finish in less than 45 minutes
Three Distributions
Distribution
Parameters
Binomial
number of events or trials probability of success
Poisson
mean number of events
Normal
mean
standard deviation
Normal Approximations
for Discrete Distributions

Approximation of the Binomial




Binomial is used for large n and small p
If p is moderate (not close to 0 or 1), then
the Binomial can be approximated by the
normal
Rule of thumb: np (1-p) ≥ 10
Other normal approximations

If X ~ Poisson(), normal works well for  ≥
10
Built-in Distributions in

Four fundamental items can be
calculated for a statistical distribution:





Density or point probability
Cumulated probability, distribution function
Quantiles
Pseudo-random numbers
In
there are functions for each of
these
Density of a
Normal Distribution
0.4
0.3
0.2
0.1
0.0

> x = seq(-4, 4, 0.1)
> plot (x, dnorm(x), type="l")
dnorm(x)

-4
-2
0
x
2
4
For Discrete Distributions..
> x = 0:50

> plot (x, dbinom(x, size=50, prob=.33, type="h")
0.08
0.06
0.04
0.02
0.00
dbinom(x, size = 50, prob = 0.33)
0.10
0.12

0
10
20
30
x
40
50
Cumulative
Distribution Functions


Could be graphed but is not very
informative
Example


Blood sugar concentration in the US
population has a mean of 132 and a
standard deviation of 13.
How special is a patient with a value 160?
1 – pnorm(160, mean=132, sd=13)
[1] 0.01562612 or 1.5%
Random Number
Generation in R

> rnorm(10)
> rnorm(10, mean=7, sd=5)
> rbinom(10, size=20, prob=.5)

Now you understand the parameters…

