1. Probability & Statistical Inference Lecture 3
MSc in Computing (Data Analytics)

2. Lecture Outline A quick recap Solutions to last weeks question
Continuous distributions.

3. A Quick Recap

4. Probability & Statistics
Population
Representative Sample
Sample Statistic
We want to make decisions based on evidence from a sample i.e. extrapolate from sample evidence to a general population
To make such decisions we need to be able to quantify our (un)certainty about how good or bad our sample information is.
Make Inference
Describe

5. Some Definitions
An experiment that can result in different outcomes, even though it is repeated in the same manner every time, is called a random experiment.
The set of all possible outcomes of a random experiment is called the sample space of an experiment and is denote by S
A sample space is discrete if it consists of a finite or countable infinite set if outcomes.
A sample space is continuous if it contains an interval or real numbers.
An event is a subset of the sample space of a random experiment.

6. Some Definitions
A sample space is discrete if it consists of a finite or countable infinite set if outcomes.
A sample space is continuous if it contains an interval or real numbers.
An event is a subset of the sample space of a random experiment.

7. Probability
Whenever a sample space consists of n possible outcomes that are equally likely, the probability of the outcome 1/n.
For a discrete sample space, the probability of an event E, denoted by P(E), equals the sum of the probabilities of the outcome in E.
Some rules for probabilities:
For a given sample space containing n events E1, E2, E3, ,En
All simple event probabilities must lie between 0 and 1:
0 <= P(Ei) <= 1 for i=1,2, ,n
The sum of the probabilities of all the simple events within a sample space must be equal to 1:

8. Discrete Random Variable
A Random Variable (RV) is obtained by assigning a numerical value to each outcome of a particular experiment.
Probability Distribution: A table or formula that specifies the probability of each possible value for the Discrete Random Variable (DRV)
DRV: a RV that takes a whole number value only

9. Summary Continued…
For Discrete RV we often have a mathematical formula which is used to calculate probabilities,
i.e. P(x) = some formula
This formula is called the Probability Mass Function (PMF)
Given the PMF you can calculate the mean and variance by:
When the summation is over all possible values of x

10. Binomial Distribution – General Formula
This all leads to a very general rule for calculating binomial probabilities:
In General Binomial (n,p)
n = no. of trials
p = probability of a success
x = RV (no. of successes)
Where P(X=x) is read as the probability of seeing x successes.

11. Binomial Distribution
If X is a binomial random variable with the paramerters p and n then

12. Poisson Probability Distribution
Probability Distribution for Poisson
Where  is the known mean:
x is the value of the RV with possible values 0,1,2,3,….
e = irrational constant (like ) with value …
The standard deviation , , is given by the simple relationship;
 =

13. Question Time

14. Questions
A factory has two assembly lines, each of which is shut down (S), at partial capacity (P), or at full capacity (F). The following table gives the sample space
For where (S,P) denotes that the first assembly line is shut down and the second one is operating at partial capacity. What is the probability that:
Both assembly lines are shut down?
Neither assembly lines are shut down
At least one assembly line is on full capacity
Exactly one assembly line is at full capacity
Event A
P(A)
(S,S)
0.02
(S,P)
0.06
(S,F)
0.05
(P,S)
0.07
(P,P)
0.14
(P,F)
0.2
(F,S)
(F,P)
0.21
(F,F)
0.19

15. Answer A
A factory has two assembly lines, each of which is shut down (S), at partial capacity (P), or at full capacity (F). The following table gives the sample space
a) What is the probability both assembly lines are shut down = 0.02
Event A
P(A)
(S,S)
0.02
(S,P)
0.06
(S,F)
0.05
(P,S)
0.07
(P,P)
0.14
(P,F)
0.2
(F,S)
(F,P)
0.21
(F,F)
0.19

16. Answer B
A factory has two assembly lines, each of which is shut down (S), at partial capacity (P), or at full capacity (F). The following table gives the sample space
b) What is the probability that neither assembly lines are shut down
= = 0.74
Event A
P(A)
(S,S)
0.02
(S,P)
0.06
(S,F)
0.05
(P,S)
0.07
(P,P)
0.14
(P,F)
0.2
(F,S)
(F,P)
0.21
(F,F)
0.19

17. Answer C
A factory has two assembly lines, each of which is shut down (S), at partial capacity (P), or at full capacity (F). The following table gives the sample space
c) What is the probability at least one assembly line is on full capacity
= = 0.71
Event A
P(A)
(S,S)
0.02
(S,P)
0.06
(S,F)
0.05
(P,S)
0.07
(P,P)
0.14
(P,F)
0.2
(F,S)
(F,P)
0.21
(F,F)
0.19

18. Answer D
A factory has two assembly lines, each of which is shut down (S), at partial capacity (P), or at full capacity (F). The following table gives the sample space
d) What is the probability exactly one assembly line is at full capacity
= = 0.52
Event A
P(A)
(S,S)
0.02
(S,P)
0.06
(S,F)
0.05
(P,S)
0.07
(P,P)
0.14
(P,F)
0.2
(F,S)
(F,P)
0.21
(F,F)
0.19

19. Exercise: There is more that one way to skin a cat!
If two fair die are thrown what is the probability that at least one score is a prime number (2, 3, 5)?
What is the compliment of the event?
What is its probability?
There are three ways (at least) that we can approach this problem

20. Solution 1: Brute Force Approach
Enumerate the sample space and select those outcomes that satisfy the desired conditions
36 possible combinations of 2 die

21. Solution 1: Brute Force Approach
Enumerate the sample space and select those outcomes that satisfy the desired conditions
36 possible combinations of 2 die
27 combinations include a prime number

22. Solution 1: Brute Force Approach
Enumerate the sample space and select those outcomes that satisfy the desired conditions
36 possible combinations of 2 die
27 combinations include a prime number
Probability of at least one prime is 27/36 = 0.75

23. Solution 1: The Compliment
What is the compliment of the event?
That neither score is a prime number (2, 3, 5) when two fair dice are thrown
What is its probability?
Let the event be E and its probability be P(E)
Then the compliment of E is E’ and the probability of E`, P(E`), is equal to 1 – P(E)
In our case P(E) = => P(E`) = 1 – = 0.25

24. Solution 2: Find the Probability of the Compliment
To start let’s work out, if we throw a single dice what is the probability of not getting a prime number?

25. Solution 2: Find the Probability of the Compliment
The brute force approach is fine for two dice, but cumbersome as the number of dice increases – i.e. 3 dice, 4 dice…..12 dice….1,247 dice!
Our question can be slightly rearranged to reveal a possible solution
If two fair die are thrown what is the probability that at least one score is a prime number (2, 3, 5)?

26. Solution 2: Find the Probability of the Compliment
The brute force approach is fine for two dice, but cumbersome as the number of dice increases – i.e. 3 dice, 4 dice…..12 dice….1,247 dice!
Our question can be slightly rearranged to reveal a possible solution
If two fair die are thrown what is the probability that at least one score is a prime number (2, 3, 5)?

27. Solution 2: Find the Probability of the Compliment
The brute force approach is fine for two dice, but cumbersome as the number of dice increases – i.e. 3 dice, 4 dice…..12 dice….1,247 dice!
Our question can be slightly rearranged to reveal a possible solution
If two fair die are thrown what is the probability that at least one score is a prime number (2, 3, 5)?
What is the probability of one or more primes from two dice throws?

28. Solution 2: Find the Probability of the Compliment
The brute force approach is fine for two dice, but cumbersome as the number of dice increases – i.e. 3 dice, 4 dice…..12 dice….1,247 dice!
Our question can be slightly rearranged to reveal a possible solution
If two fair die are thrown what is the probability that at least one score is a prime number (2, 3, 5)?
What is the probability of one or more primes from two dice throws?
What is the probability of one or more of outcome O from X trials?

29. Solution 2: Find the Probability of the Compliment
The brute force approach is fine for two dice, but cumbersome as the number of dice increases – i.e. 3 dice, 4 dice…..12 dice….1,247 dice!
Our question can be slightly rearranged to reveal a possible solution
If two fair die are thrown what is the probability that at least one score is a prime number (2, 3, 5)?
What is the probability of one or more primes from two dice throws?
What is the probability of one or more of outcome O from X trials?
If questions are of this form we can work out the answer by working out the compliment first

30. Solution 2: Find the Probability of the Compliment
What is the probability that neither score is a prime number (2, 3, 5) when two fair dice are thrown?
This is an easier probability to calculate as we can consider throwing each dice as an independent event and combine the probabilities that neither results in a prime
It is the “one or more” in the previous problem that makes things tricky as we cannot consider each dice throw as an independent event

31. Solution 2: Find the Probability of the Compliment
To start let’s work out, if we throw a single dice what is the probability of not getting a prime number?
Sample space: {1, 2, 3, 4, 5, 6}
Primes: {2, 3, 5}
Non-primes: {1, 4, 6}
So, probability is 3/6

32. Solution 2: Find the Probability of the Compliment
If the probability of getting no prime if we throw one dice is 3/6, what is the probability of getting no primes if we throw two dice in a row?

33. Solution 2: Find the Probability of the Compliment
If the probability of getting no prime if we throw one dice is 3/6, what is the probability of getting no primes if we throw two dice in a row?
Dice rolls are independent events
Remember our intersection rule for independent events
So, the probability of getting no primes if we throw two dice in a row is:

34. Solution 2: Find the Probability of the Compliment
Our event, E, was that neither score is a prime number (2, 3, 5) when two fair dice are thrown
So the complement of this event, E`, is that at least one score is a prime number (2, 3, 5) when two fair dice are thrown
We know that given the probability of event E, P(E), we can work our the probability of the complement of this event, P(E`), as 1 – P(E)
So for our dice example

35. Solution 2: Find the Probability of the Compliment
The great thing is that this works for any number of dice
The probability, P(E), of getting no primes if we throw n dice in a row is:
So, for three dice the probability of getting no primes is
This means that the probability of getting at least one prime from 3 dice rolls is 1 – 1/8 = 7/8

36. Solution 3: Use the Binomial Distribution
Problems that can be stated as:
what is the probability of seeing x successes in n independent binary trials
can be solved using the Binomial distribution.
For example:
what is the probability of seeing 1 prime in 2 dice throws

37. Solution 3: Use the Binomial Distribution
The Binomial probability, P(X=x), (read as the probability of seeing x successes) is given by:
where n is the number of trials, p is the probability of a success and , known as a combination, is the number of ways of getting x successes from n trials

38. Solution 3: Use the Binomial Distribution
So, what is the probability of seeing 1 prime in 2 dice throws
n = p = 1/ x = 1

39. Solution 3: Use the Binomial Distribution
Exercise: What is the probability of seeing 2 primes in 2 dice throws

40. Solution 3: Use the Binomial Distribution
Exercise: What is the probability of seeing 2 primes in 2 dice throws

41. Solution 3: Use the Binomial Distribution
So, what is the probability of seeing one or more primes in 2 dice throws?
P(1 ≤ X ≤ 2) = P(X = 1) + P(X = 2) = ½ + ¼ = ¾
More generally then we can say that the probability of seeing one or more primes in n dice throws is:

42. Continuous Probability Distributions

43. Continuous Probability Distributions
Experiments can lead to continuous responses i.e. values that do not have to be whole numbers. For example: height could be 1.54 meters etc.
In such cases the sample space is best viewed as a histogram of responses.
The Shape of the histogram of such responses tells us what continuous distribution is appropriate – there are many.

44. Normal Distribution (AKA Gaussian)
The Histogram below is symmetric & 'bell shaped'
This is characteristic of the Normal Distribution
We can model the shape of such a distribution (i.e. the histogram) by a Curve

45. Normal Distribution µ = mean,  = standard deviation,
The Curve may not fit the histogram 'perfectly' - but should be very close
Normal Distribution - two parameters,
µ = mean,  = standard deviation,
The mathematical formula that gives a bell shaped symmetric curve
f(x) = Height of curve at x =

46. Normal Distribution Why Not P(x) as before?
=> because response is continuous
What is the probability that a person sampled at random is 6 foot?
Equivalent question: what proportion of people are 6 foot?
=> really mean what proportion are
'around 6 foot' ( as good as the measurement device allows) - so not really one value, but many values close together.

47. Example: What proportion of graduates earn €35,000?
Would we exclude €35, or €34,999.99?
Round to the nearest €, €10, €100, €1000?
Continuous measure => more useful to get proportion from €35,000 - €40,000
Some Mathematical Jargon:
The formula for the normal distribution is formally called the normal probability density function (pdf)

48. The Shaded portion of the Histogram is the Proportion of interest
Can visualise this using the histogram of salaries.
The Shaded portion of the Histogram is the Proportion of interest

49. = proportional area under the curve = a probability of interest Need;
Since the histogram of salaries is symmetric and bell shaped, we model this in statistics with a Normal distribution curve.
Proportion = the proportion of the area of the curve that is shaded
So proportions
= proportional area under the curve
= a probability of interest
Need;
To know , 
To be able to find area under curve

50. Area under a curve is found using integration in mathematics.
In this case would need a technique called numerical integration.
Total area under curve is 1.
However, the values we need are in Normal Probability Tables.

51. Corresponding Value from Normal
Standard Normal
The Tables are for a Normal Distribution with
 = 0 and  = 1
this is called the Standard Normal
Can 'convert' a value from any normal to the standard normal using standard scores (Z scores)
Corresponding Value from Normal
 = 0
 = 1
Value from any Normal
Distribution
Standardize

52. Z-Score Example
Z scores are a unit-less quantity, measuring how far above/below  a certain score (x) is, in standard deviation units.
Example: A score of 35, from a normal distribution with
 = 25 and  = 5.
Z = ( 35 − 25) / 5 => 10/5 = 2
So 35 is 2 standard deviation units above the mean
What about a score of 20 ?
Z = ( ) / 5 => − 5 / 5 = − 1
So 20 is 1 standard unit below the mean

53. Z-Score Example Positive Z score => score is above the mean
Negative Z score => score is below the mean
By subtracting  and dividing by the  we convert any normal to  = 0, =1, so only need one set of tables!

54. Example:
From looking at the histogram of peoples weekly receipts, a supermarket knows that the amount people spend on shopping per week is normally distributed with:
 = €58
 = €15.

55. What is the probability that a customer sampled at random will spend less than €83.50 ?
Z = ( x −  ) / 
= ( € €58 ) / €15 => 1.7
Area from Z=1.7 to the left can be read in tables
From tables area less than Z = 1.7 =>
So probability is
Or 95.54%

57. What is the probability that a customer sampled at random will spend more than €83.50 ?
Z = ( x −  ) / 
= ( € €58 ) / €15 => 1.7
Area from Z=1.7 to the right can be read in tables
From tables area greater than Z = 1.7 => =
So probability is
Or 4.46%

58. Exercise Find the proportion of people who spend more than €76.75
Find the proportion of people who spend less than €63.50
Note: The tables can also be used to find other areas
(less than a particular value, or the area between two points)

59. Characteristics of Normal Distributions
Standard Deviation has particular relevance to Normal distribution
Normal Distribution => Empirical Rule
Between Z
(lower, upper)
%Area
-1,1
68 %
-2,2
95 %
-3,3
99.7 %
-∞, +∞
100%

60. The normal distribution is just one of the known continuous probability distributions.
Each have their own probability density function, giving different shaped curves.
In each case, we find probabilities by calculating areas under these curves using integration.
However, the Normal is the most important – as it plays a major role in Sampling Theory.

61. Other important continuous probability distributions include
Exponential distribution – especially positively skewed lifetime data.
Uniform distribution.
Weibull – especially for ‘time to event’ analysis.
Gamma distribution – waiting times between Poisson events in time etc.
Many others…..

62. Summary – Random Variables
There are two types – discrete RVs and continuous RVs
For both cases we can calculate a mean (μ) and standard deviation (σ)
μ can be interpreted as average value of the RV
σ can be interpreted as the standard deviation of the RV

63. Summary Continued…
For Discrete RV we often have a mathematical formula which is used to calculate probabilities,
i.e. P(x) = some formula
This formula is called the Probability Mass Function (PMF)
Given the PMF you can calculate the mean and variance by:
When the summation is over all possible values of x

64. Summary Continued…
For continuous RVs, we use a Probability Density Function (PDF) to define a curve over the histogram of the values of the random variables.
We integrate this PDF to find areas which are equal to probabilities of interest.
Given the PDF you can calculate the mean and variance by:
Where f(x) is usual mathematical notation for the PDF

65. Question Time

66. Next Week
Next week we will start with the practical part of the course. We will move to Lab