1. PROBABILITY DISTRIBUTIONS
A listing of all the values the random
variable can assume with their
corresponding probabilities.

2. PROBABILITY DISTRIBUTIONS
A random variable does not mean that the
values can be anything ( a random number)
Random variables have a well defined set of
outcomes and well defined probabilities for
the occurrence of each outcome.
The random refers to the fact that the
outcomes happen by chance -- that is,
you don’t know which outcome will occur
next.

3. PROBABILITY DISTRIBUTIONS
Example : Rolling of a die
X sum
p(x) 1/6 1/6 1/6 1/6 1/6 1/6 6/6=1

4. PROBABILITY DISTRIBUTIONS
There are two types of random variables:
that is , discrete and continuous.
DRV can assume a countable number
of values.
Example: The number of sales made by a
a salesperson in a given week.
The number of customers waiting
to be served in a restaurant at a
particular time.

5. PROBABILITY DISTRIBUTIONS
Two types of probability distributions
Discrete
Continuous

6. PROBABILITY DISTRIBUTIONS
Discrete variable can take only a limited
number of values, which can be listed.
Continuous variable can take on any
value within a given range, so we cannot
list all the possible values.

7. PROBABILITY DISTRIBUTIONS
The probability distribution of a discrete
random variable is a graph, table or formula
that specifies the probability associated with
each possible value the random variable can
assume.

8. PROBABILITY DISTRIBUTIONS
Example : Tossing two coins and let x be the
number of heads observed.
Solution: X
p(x) 1/4 1/2 1/4
p(x)
p(x)
1/2
1/2
1/4
1/4 x x

9. PROBABILITY DISTRIBUTIONS
The mean or expected value, of a discrete
random variable x is
= E(x) =
All x
The expected value is the mean of the
probability distribution, a measure of
its central tendency.

10. PROBABILITY DISTRIBUTIONS
The variance of a discrete random
variable x is
All x

11. PROBABILITY DISTRIBUTIONS
Expected values of discrete random variables
Example:
Examine the probability distribution for
x ( the number of heads observed in the
tossing of two fair coins)
In a large number of experiments, 1/4 should
result in x=0, 1/2 in x=1 and 1/4 in x=2
heads.
Therefore, the average number of heads is
=0(1/4) + 1(1/2) + 2(1/4)

12. Example 1: Consider a random variable X with the following probability distribution:
p(x) = 0.05x, x = 2, 3, 4, 5, or 6
Express the probability distribution in tabular form.
b) Find the following probabilities:
P(X≥4)
P(X > 4)
P(3≤X≤5)
P(2 < X < 4)
P(X = 4.5)

13. Solution - Example 1 a) X
p(x)
b) 0.75
0.55
0.60
0.15
0.00

14. Example 2
Let X and Y be two independent random variables with the following probability distributions: X
p(x) Y
p(Y)
Calculate E(X) and E(Y)
Calculate V(X) and V(Y)

15. PROBABILITY DISTRIBUTIONS
Binomial Probabilities
A binomial experiment is an experiment
which satisfies these four conditions:
1. A fixed number of trials
2. Each trial is independent of the
others.
3. There are only two outcomes
4. The probability of each outcome
remains constant from trial to
trial.

16. PROBABILITY DISTRIBUTIONS
Binomial Probabilities
A binomial experiment has a fixed number
of independent trials, each with only two
outcomes.
Examples: 1. Asking 300 people if they
watch TV news.
2. Rolling a die to see if a 5
appears

17. PROBABILITY DISTRIBUTIONS
Binomial Probabilities
The Binomial Formula( Probability Distribution)
Where:
p = probability of a success
q = 1-p = probability of failure
r = number of success desired
n = number of trials

18. PROBABILITY DISTRIBUTIONS
Binomial Probabilities
Mean:
Standard Deviation:

19. Example 3: An insurance representative has appointments with 4 prospective clients tomorrow. From the past experience she knows that the probability of making a sale on any appointment is 1 in 5 or What is the probability that she will sell a policy to 3 out of 4 prospective clients?

20. Outline of Answer P(sale)=0.2 P(No sale)=1-0.2=0.8 n=4 P(3 sales)=
=0.0256

21. Example 4: A company is planning to sell a new product in four areas North, South, East and West. The probability that the product will be successful in an area is 0.3. Success in one area will be independent of success or failure in the other areas. What is the probability of success in no areas, one area, two areas, three areas, and four areas?

22. Outline of Answer P (success)=p=0.3; q=1-p=0.7; n=4
Use Binomial Tables.
P(0 successes)=P(0)=0.2401
P(1)=0.4116
P(2)=0.2646
P(3)=0.0756
P(4)=0.0081

23. Example 5: Of the total output produced in a factory, 8% is defective
Example 5: Of the total output produced in a factory, 8% is defective. In a batch of 500 units, what is the mean expected number of defective items, and what is the standard deviation?

24. Outline of Answer N=500; p=0.8 Mean = np = 500*0.8 = 40 units
Standard Deviation =

25. PROBABILITY DISTRIBUTIONS
Poisson Distribution
A probability distribution used when a
density of items is distributed over a
period of time.
The sample size needs to be large and
the probability of success to be small.

26. PROBABILITY DISTRIBUTIONS
Poisson Distribution
Characteristics:
1. The experiment consists of counting the
number of times a particular event occurs
during a given unit of time or in a given
area or volume.
2. The probability that an event occurs in a
given unit of time, area or volume is the
same for all the units.

27. PROBABILITY DISTRIBUTIONS
3. The number of events that occur in one
unit of time, area or volume is independent
of the number that occur in other units.
4. The mean (or expected) number of events
in each unit will be denoted by

28. PROBABILITY DISTRIBUTIONS
Poisson Formula(Probability Distribution)
Where:
= probability of exactly x
occurrences
= , constant.

29. Example 6: Alden and Associates write weekend trip insurance at a very nominal charge. Records show that the probability that a motorist will have an accident during weekend and file a claim is Suppose Alden wrote 400 policies for the forthcoming weekend. What is the probability that exactly two claims will be filed?

30. Outline of Answer p=0.0005 n=400 Mean=λ=np=0.2 Using Poisson Tables,

31. Example 7: It is thought that, on an average, 0
Example 7: It is thought that, on an average, 0.01% of the workforce in a particular industry will suffer from an acute form of industrial disease. Hot Flush Inc., employs a workforce of 2,000 men, and the senior medical adviser of the company has stated that if three people in the workforce suffer from the disease, the cases should be treated as a sign that there are unacceptable health hazards.
Calculate the probability that three or more men in the workforce will catch the disease.

32. Outline of Answer p=0.0001 n=2000 Mean=λ=np=0.2 Using Poisson Tables,
P(3 or more)=1-P(2 or less)
=1-[P(0)+P(1)+P(2)]
=1-[ ]=0.0012

33. PROBABILITY DISTRIBUTIONS
The Poisson distribution can be a reasonable
approximation of the binomial under the
following conditions:
1. n is large, and
2. p is small.
The rule most often used is when n is
greater than or equal to 20 and p is
less or equal to 0.05.

34. Normal Distribution
This is the most important continuous distribution.
Many distributions can be approximated by a normal distribution.
The normal distribution is the cornerstone distribution of statistical inference.

35. Normal Distribution
A random variable X with mean m and variance s2 is normally distributed if its probability density function is given by

36. Finding Normal Probabilities
Two facts help calculate normal probabilities:
The normal distribution is symmetrical.
Any normal distribution can be transformed into a specific normal distribution called…
“STANDARD NORMAL DISTRIBUTION”
Example
The amount of time it takes to assemble a computer is normally distributed, with a mean of 50 minutes and a standard deviation of 10 minutes. What is the probability that a computer is assembled in a time between 45 and 60 minutes?

37. Finding Normal Probabilities
Solution
If X denotes the assembly time of a computer, we seek the probability P(45<X<60).
This probability can be calculated by creating a new normal variable the standard normal variable.
Every normal variable
with some m and s, can
be transformed into this Z.
Therefore, once probabilities for Z
are calculated, probabilities of any
normal variable can be found.
E(Z) = m = 0
V(Z) = s2 = 1

38. Finding Normal Probabilities
Example - continued 45
- 50 X
- m 60
- 50
P(45<X<60) = P( < < ) 10 s 10
= P(-0.5 < Z < 1)
To complete the calculation we need to compute
the probability under the standard normal distribution

39. Finding Normal Probabilities
Example - continued
P(45<X<60) = P( < < ) 45 X 60
- m
- 50 s 10
= P(-.5 < Z < 1)
We need to find the shaded area
z0 = 1
z0 = -.5

40. Finding Normal Probabilities
Example - continued
P(-.5<Z<1) = P(-.5<Z<0)+ P(0<Z<1) = = .5328
.1915
.1915
.1915
.1915
.3413
-.5 .5
1.0

41. Example 8
The amount of soda pop in each bottle is normally distributed with a mean of 32.2 ounces and a standard deviation of .3 ounces.
Find the probability that a bottle bought by a customer will contain more than 32 ounces.
Solution
The random variable X is the amount of soda in a bottle.
0.7486
x = 32
m = 32.2

42. Finding Values of Z
Sometimes we need to find the value of Z for a given probability
We use the notation zA to express a Z value for which P(Z > zA) = A A zA

43. Finding Values of Z Example 9 Solution
Determine z exceeded by 5% of the population
Determine z such that 5% of the population is below
Solution
z.05 is defined as the z value for which the area on its right under the standard normal curve is .05.
0.45
0.05
0.05
-Z0.05
Z0.05
1.645

44. Characteristics of Normal Probability Distribution
The curve is bell shaped, that is, it has the same shape on either side of the vertical line from mean.
It has a single peak. As such it is unimodal.
The mean is located at the centre of the distribution.
The distribution is symmetrical.
The two tails of the distribution extend indefinitely but never touch the horizontal axis (asymptotic).

45. The mean, median and mode have the same
value, that is, mean = median = mode.
The percentage distribution of area under
standard normal curve is broadly as follows:
± 1s =68.27%; ± 2s =95.44% and ± 3s =99.73%.
The units for the standard normal distribution curve are denoted by Z and are called the Z values or Z scores. They are also called standard units or standard scores. The Z score is known as a ‘standardised’ variable because it has a zero mean and a standard deviation of one.

46. Bell-shaped curve

47. Area under the normal curve, between  ± k 
k Area in %

48. Normal approximation to the Binomial
Normal approximation to the binomial works best when
the number of experiments (sample size) is large, and
the probability of success, p, is close to 0.5.
For the approximation to provide good results two conditions should be met:
np 5; n(1 - p) 5

49. Example 10
If X is a normal random variable with a mean of 100 and a standard deviation of 10, find the following probabilities:
P(X ≥ 128)
P(X ≤ 113)
P(87 ≤ X ≤ 98)

50. If Z is a standard normal random variable, find the value z for which:
Example 11
If Z is a standard normal random variable, find the value z for which:
the area between 0 and z is
the area to the right of z is
the area to the left of z is
the area to the left of z is
the area to the right of z is
the area between –z and z is

51. Examples
12. Suppose the owner of a bakery knows that the daily demand for his whole meal bread is a random variable having the mean of 400 loaves and the standard deviation is 20. What is the probability that the demand for its bread will exceed 450 loaves?
13. The average monthly sales of 5000 firms are normally distributed with mean and standard deviation of Rs.36,000 and Rs.10,000 respectively. Find
a) the number of firms having sales over Rs.40,000
b) the percentage of firms having sales between Rs.38,500 and Rs.41,000
c) the number of firms having sales between Rs.30,000 and Rs.40,000