1. Probability distributions
We extend the probability analysis by considering random variables (usually the outcome of a probability experiment)
These (usually) have an associated probability distribution
Once we work out the relevant distribution, solving the problem is usually straightforward

2. Random variables
Most statistics (e.g. the sample mean) are random variables
A random variable is a numeric event whose value is determined by a chance process or an experiment
The event should not be under the control of the observer; i.e., the value of the random variable is unknown before the experiment is carried out.

3. Random variables
Example 1: Suppose we roll two dice and take the sum of the numbers showing up.
This sum is clearly a random variable because its value is determined by chance
Such an experiment produces 11 possible values (what are they?)
Example 2: Consider a random experiment in which a coin is tossed three times. Let X be the number of heads. Let H represent the outcome of a head and T the outcome of a tail.

4. Random variables
The sample space for such an experiment will be: TTT, TTH, THT, THH, HTT, HTH, HHT, HHH.
Thus the possible values of X (number of heads) are x = 0,1,2,3.
Many random variables have well-known probability distributions associated with them.
To understand random variables, we need to know about probability distributions.

5. Discrete & continuous random variables
A discrete random variable is a variable that can assume only certain clearly separated values resulting from a count of some item of interest (countably finite or infinite).
Example: Let X be the number of heads when a coin is tossed 3 times. Here the values for X are x = 0,1,2,3.
A continuous random variable is a variable that can assume one of an infinitely large number of values.
Assumes any value within a designated range of values.
Example: Height of a student in this class.

6. Probability Distribution for Discrete RVs
Each value of a random variable has an associated probability
The probability distribution of a random variable lists all the possible values of the random variable and their corresponding probabilities
If P (X) is the probability that X is the value of the random variable, then

7. Discrete probability distributions
Suppose we toss a coin three times and want to observe the number of heads.
The possible values are 0, 1, 2, 3.
What is the probability distribution of the number of heads?

8. Discrete probability distributions
No. of heads X
Frequency
Probability
P (X) 1 2 3
1/8 = .125
3/8 = .375
8/8 = 1.00

9. Expected value of discrete random variable
The mean of a random value is called its expected value
For discrete random variables, it is the weighted mean of all possible values of the random variable, with weights being the probabilities

10. Expected value For the three tosses of a coin, we have
The expected value is not the value we “expect” on any single toss
Rather it is the long run average value (when the experiment is done a large number of times)

11. Variance for a random variable
The variance is given by
Or for computational ease, we use
Take square root to obtain standard deviation

12. Variance X P(X) X P(X) X-E(X) [X-E(X)]2 [X-E(X)]2P(X) X2 X2P(X) 1 2 31 2 3
.125
.375
.75
-1.5
-.5 .5
1.5
2.25
.25
.2813
.0938 4 9
1.125
.7502
3.0

13. Variance
= 3.0 – (1.5)2 = .75
Standard deviation = .87

14. 6-14
Practice
The Managing Director of Perfect Painters, a painting firm in Accra, has studied his records for the past 20 weeks and reports the following number of houses painted per week. Compute the mean and variance of the number of houses painted per week.

15. Probability Distribution:
6-15
Solution
Probability Distribution:

16. Mean number of houses painted per week:
6-16
Solution cont’d
Mean number of houses painted per week:
Variance of the number of houses painted per week:

17. Some standard probability distributions
Binomial distribution (discrete)
Normal distribution (continuous)
Poisson distribution (discrete)

18. When do they arise?
Binomial - when the underlying probability experiment has only two possible outcomes (e.g. tossing a coin)
Normal - when many small independent factors influence a variable (e.g. IQ, influenced by genes, diet, etc.)
Poisson - for rare events, when the probability of occurrence is low

19. The Binomial distribution
Applied when?
Only two mutually exclusive outcomes are possible in each trial (success or failure)
The outcomes in the series of trials are independent
The probability of success, denoted P, in each trial remains constant from trial to trial
The objective of using the BD is to determine the probability values for various possible number of successes (X), given the number of trails (n) and the known (and constant) probability of success (P)

20. The Binomial distribution
Consider five tosses of a coin. We can draw a tree diagram for this experiment, but we won’t.
Recall from last lecture, we can write the probability of 1 Head in 2 tosses as the probability of a head and a tail (in that order) times the number of possible orderings (# of times that event occurs).
P (1 Head) = ½  ½  2C1 = ¼  2 = ½
We can apply same technique to calculate the probability for the number of heads in five tosses of a coin.

21. The Binomial distribution
P (X Heads in five tosses of a coin)
P(X = 0) = (½)0  (½)5  5C0 = 1/32  1 = 1/32
P(X = 1) = (½)1  (½)4  5C1 = 1/32  5 = 5/32
P(X = 2) = (½)2  (½)3  5C2 = 1/32  10 = 10/32
P(X = 3) = (½)3  (½)2  5C3 = 1/32  10 = 10/32
P(X = 4) = (½)4  (½)1  5C4 = 1/32  5 = 5/32
P(X = 5) = (½)5  (½)0  5C5 = 1/32  1 = 1/32

22. Probability distribution of 5 tosses of a coin

23. Binomial distribution with different parameters
Eight tosses of an unfair coin (P = 1/6 )

24. The Binomial ‘family’
Like other distributions, the Binomial is a family of distributions, members being distinguished by their different parameters.
The parameters of the Binomial are:
P - the probability of ‘success’
n - the number of trials
Notation: X ~ B(n, P)
Read as “r follows a Binomial distribution with parameters n and P”.

25. Formula for Calculating Binomial Probabilities
For the Binomial distribution, X ~ B(n, P), we can calculate the probability of X successes as
P(X) = nCx * Px * (1-P)(n-x)
E.g. X ~ B(5, ½ ) means P(X) = 5Cx(½)x (1- ½ )(5-x)
and from this we can work out P(X) for any value of X.

26. Example
Probability of obtaining two heads in three tosses of a coin: X=2 and n=3
X ~ B(3, ½ )
P(X=2) = 3C2 (.5)2(.5) = 3×.25×.5 = .375
Note that, nCX is the combination formula, which is equal to

27. Using complement rule
When the BD is used, it is typically because we wish to determine the probability or “X or more” successes [i.e., P(X ≥ xi)] or “ X or fewer successes [i.e., P (X ≤ xi)]
If the individual probabilities to be summed is large, it is easier to use the complement rule
Example: P (X ≥ xi) = 1 - P(X < xi)

28. Example
Suppose the probability is .05 that a randomly selected student of the University of Ghana owns a car. What is the probability of observing two or more student car-owners in a random sample of 20 students?
P(X≥2) = P(X=2) + P(X=3) + …..+ P(X=20)
P(X≥2) = 1 – P(X<2) = 1- P(X=0, 1)
= 1 – [P(X=0) +P(X=1)]
Now
P(X=0) = 20C0 (.05)0(.95)20 = .3585
P(X=1) = 20C1 (.05)(.95)19 = .3774
So P(X≥2) = 1 – ( ) = .2641

29. Mean and variance of the Binomial
Mean = E(X) = n  P
Variance = σ2 = n  P  (1-P)
On average, you would expect 10 Heads from (n = ) 20 tosses of a fair (P = ½) coin (10 = 20  ½)
It is not difficult to demonstrate the mean and variance formulae starting from E(X) = Sigma X P(X) and V(X) = E(X2) - E(X)2.

30. Practice at least three are unemployed.
6-21
Practice
The Ministry of Employment reports that 20% of the labour force in Ghana is unemployed. From a sample of 14 members of the labour force, calculate the following probabilities using the formula for the binomial probability distribution:
three are unemployed
at least three are unemployed.

31. At least three are unemployed:
Solution
Three are unemployed:
P(x=3)=.250
At least three are unemployed:
P(x ≥ 3) = 1 - P(x<3)
= 1 – [P(x=0) +P(x=1) + P(x=2)]
= 1- ( ) =.552

32. The Poisson distribution
It is a sampling process in which events occur over time or space.
The Poisson distribution is used to describe a number of processes or events such as
The distribution of telephone calls going through a switch board
The demand of patients for service at a health facility
The arrival of vehicles at a tollbooth
The number of accidents occurring at a road intersection, etc.
The Poisson is the limiting case of the Binomial when the probability of success is very small, i.e. BD becomes more and more skewed to the right as probability of success becomes smaller and smaller (rare events).

33. The Poisson distribution
Characteristics defining a Poisson random variable are
The experiment consists of counting the number of times a particular event occurs during a given time interval
The probability that the event occurs in one time interval is independent of the probability of the event occurring in another time interval
The mean number of events in each unit of time is proportional to the length of the time interval.

34. The Poisson distribution
The probability that the Poisson random variable will assume the value X is given by
For X = 0, 1, 2,3 …….
λ is the mean of the distribution (mean number of events occurring in a given unit of time)
e is approximately and is the base of the natural logarithms.

35. Example
Suppose an average of 2 calls per minute are received at a switchboard during a designated time interval, then the probability that exactly 3 calls are received in a randomly sampled minute is:

36. Poisson
As with the BD, the PD typically involves determining the probability of “X or more” or “X or fewer” number of events.
To calculate, we sum the appropriate probability values
The use of the complement rule may also come in handy
For example, we may want to calculate the probability of receiving 3 or more calls in a three-minute interval
That is P(X≥3) = P(X=3) + P(X=4) + ……………
Using the complement rule, we have
P(X≥3) = 1 - P(X<3) = 1 – [P(x=0) + P(x=1) + P(x=2)]

37. Poisson
From our proposition 3, we know the mean number of occurrences is proportional to the length of the time interval
So if we expect a mean of 2 calls per minute, in three minutes we must expect 6 calls
For an interval of 30 seconds, the mean number of calls is one
The probability of 5 calls in a three-minute interval implies λ = 6, so P(X=5 / λ=6) = .1606
The probability of no calls in an interval of 30 seconds implies that λ = 1, so P(X=0 / λ=1) = .3679

38. Expected value and variance
The expected value and variance for a Poisson random variable are both equal to the mean number of events for the time interval of interest
E (X) = λ
Var (X) = λ

39. Poisson approximation of Binomial
When the probability of occurrence (success) is very small (P<.05)
And the number of trials is large (n>20)
So that λ = nP and we apply the Poisson formula
Some say use the Poisson in place of the Binomial when nP < 5

40. Example
A manufacturer claims a failure rate of 0.2% for its hard disk drives. In an assignment of 500 drives, what is the probability that, none are faulty, one is faulty, etc?
On average, 1 drive (0.2% of 500) should be faulty, so λ = nP = 1.
Note nP = 1

41. Example (continued) The probability of no faulty drives is
The probability of one faulty drive is
and
Note: summing over all x values gives one, as expected. The Binomial distribution could be used instead, giving the exact answer. For x=0 this is whereas the Poisson gives