Random Variable Discrete Probability Distribution continuous Probability Distribution Probability Mass Function Probability Density Function Expected value variance Binomial Distribution poisson distribution normal distribution

1. PROBABILITY
DISTRIBUTIONS
BY DR. SARABJEET KAUR
1

2. LO 1 Distinguish between discrete and continuous random
variables.
LO 2 Describe the probability distribution for a discrete random
variable.
LO 3 Calculate and interpret summary measures for a discrete
random variable
LO 4 Describe the binomial distribution and compute relevant
probabilities.
LO 5 Describe the Poisson distribution and compute relevant
probabilities
LO 6 Describe the Continuous distribution ( Normal Distribution)
and compute relevant probabilities
LEARNING OBJECTIVES
(LOS)
2

3. INTRODUCTORY CASE:
AVAILABLE STAFF FOR
PROBABLE CUSTOMERS
3
□ Anne Jones is a manager of a local Starbucks. Due to a weak
economy and higher gas and food prices, Starbucks announced
plans in 2008 to close 500 U.S. locations.
□ While Anne’s store will remain open, she is concerned that
nearby closings might affect her business.
□ A typical Starbucks customer visits the chain between 15 and
18 times a month.
□ Based on all this, Anne believes that customers will average 18
visits to her store over a 30-day month.

4. INTRODUCTORY CASE:
AVAILABLE STAFF FOR
PROBABLE CUSTOMERS
4
□ Anne needs to decide staffing needs.
◦ Too many employees would be costly to the store.
◦ Not enough employees could result in losing customers who
choose not to wait.
□ With an understanding of the probability distribution of
customer arrivals, Anne will be able to:
◦ Calculate the expected number of visits from a typical Starbucks
customer in a given time period.
◦ Calculate the probability that a typical customer visits the store
a specific number of times in a given time period.

5. DISCR
ETE PROBABILITY
DISTRIBUTIONS
5
LO 1 Distinguish between discrete and continuous random variables.
▪ Random variable
✔ It assigns numerical values to the outcomes of a
random experiment.
✔ Denoted by uppercase letters (e.g., X ).
▪ Values of the random variable are denoted by
corresponding lowercase letters.
✔ Corresponding values of the random variable:
x1, x2, x3, . . .

6. RANDOM VARIABLES AND
DISCR
ETE PROBABILITY
DISTRIBUTIONS
6
▪ Random variables may be classified as:
✔ Discrete
▪ The random variable assumes a countable number
of distinct values.
✔ Continuous
▪ The random variable is characterized by (infinitely)
uncountable values within any interval.

7. DISCR
ETE PROBABILITY
DISTRIBUTIONS
7
• Consider an experiment in which two shirts are selected
from the production line and each is either defective (D)
or non-defective (N).
✔ Here is the sample space:
✔ The random variable X is
the number of defective shirts.
✔ The possible number of
defective shirts is the set {0, 1, 2}.
• Since these are the only a countable number of possible
outcomes, this is a discrete random variable.
(D,D)
(D,N)
(N,D)
(N,N)

8. Every random variable is associated with a
distribution that describes the variable completely
PROBABILITY
DISTRIBUTIONS
8
probability
□ Types of Probability Distribution
(i) Discrete
(ii) Continuous

9. If a random variable is a discrete variable, its probability
distribution is called a discrete probability distribution.
Ex , Suppose if we toss a coin are two times. This simple
experiment can have four possible outcomes: HH, HT, TH, and TT.
Now, let the random variable X represent the number of Heads
that result from this experiment. The random variable X can only
take on the values 0, 1, or 2, so it is a discrete random variable.
DISCRETE PROBABILITY
DISTRIBUTION
9
Possible values of x Probability of each value
0 1/4
1 2/4
2 1/4

10. • Two key properties of discrete probability distributions:
✔ The probability of each value x is a value between
0 and 1, or equivalently
✔ The sum of the probabilities equals 1. In other words,
PROBABILITY
DISTRIBUTIONS

11. Continuous Probability Distribution is that in which the
random variable can be any number within some given
range of values.
11
Continuous Probability Distribution

12. Weight of students (in kg) PROBABILITY
50-55 4/100
55-60 3/100
60-65 3/100
65-70 17/100
70-75 30/100
75-80 20/100
80-85 12/100
85-90 6/100
90-95 3/100
above95 2/100
12
Eg

13. ✔ A probability mass function is used to describe
discrete random variables.
✔ A probability density function is used to describe
continuous random variables.
PROBABILITY
DISTRIBUTIONS
13

14. • A discrete probability distribution may be viewed as a
table, algebraically, or graphically.
• For example, consider the experiment of rolling a
six-sided die. The probability distribution is:
• Each outcome has an associated probability of 1/6. Thus, the pairs
of values and their probabilities form the probability mass
function for X.
DISCRETE PROBABILITY
DISTRIBUTIONS

15. • Another way to look at a probability
• distribution is to examine its cumulative probability
distribution.
✔ For example, consider the experiment of rolling a
six-sided die. The cumulative probability distribution is:
The cumulative probability distribution gives the
probability that X is less than or equal to x.
For example,
DISCRETE PROBABILITY
DISTRIBUTIONS

16. • Example: Consider the probability distribution
that reflects the number of credit cards that
Bankrate.com’s readers carry:
✔ Is this a valid probability
distribution?
✔ What is the probability that a
reader carries no credit cards?
✔ What is the probability that a
reader carries fewer than two?
✔ What is the probability that a reader carries at least two credit
cards?
PROBABILITY
DISTRIBUTIONS

17. • Consider the probability distribution that reflects the
number of credit cards that Bankrate.com’s readers
carry:
✔ Yes, because 0 < P(X = x) < 1
and ΣP(X = x) = 1.
✔ P(X = 0) = 0.025
✔ P(X < 2) = P(X = 0) + P(X = 1)
= 0.025 + 0.098 = 0.123.
✔ P(X > 2) = P(X = 2) + P(X = 3)
+ P(P = 4*) = 0.166 + 0.165 + 0.546 = 0.877.
Alternatively, P(X > 2) = 1 − P(X < 2) = 1 − 0.123 = 0.877.
PROBABILITY
DISTRIBUTIONS

18. • Summary measures for a random variable include
the
✔ Mean (Expected Value)
✔ Variance
✔ Standard Deviation
5.2 EXPECTED VALUE,
VARIANCE, AND STANDARD
DEVIATION
LO 3 Calculate and interpret summary measures for a discrete random variable.

19. For a discrete random variable X with values x1, x2, x3, . . .
that occur with probabilities P(X = xi), the expected value
of X is the probability weighted average of the values:
E(X) = ∑ XP(X)
Variance and Standard Deviation
E(X2) = ∑ X2P(X)
Var(X) = σ2=E(X2) – [E(X) ] 2
EXPECTED VALUE AND
VARIAN
CE
19

20. DISTRIBUTION FOR THE
NUMBER OF TRAFFIC ACCIDENTS
DAILY IN TIER 2 CITY OF INDIA
20
Number of Accidents Daily (X) P (X = xi)
0 0.10
1 0.20
2 0.45
3 0.15
4 0.05
5 0.05
I. Compute the mean number of accidents per day
II. Compute the standard Deviation

21. Answer
I. 2
II. 1.1832 (1.4)
21

22. THE NUMBER OF ARRIVALS PER MINUTE AT AXIS BANK
LOCATED IN CENTRAL PART OF THE MUMBAI WAS
RECORDED OVER A PERIOD OF 200 MINUTES, WITH THE
FOLLOWING RESULTS
22
Arrivals Frequency
0 14
1 31
2 47
3 41
4 29
5 21
6 10
7 5
8 2
a. Compute the Expected number of arrivals per minute.
b. Compute the Standard Deviation.

23. • Example: Brad Williams, owner of a car dealership in
Chicago, decides to construct an incentive compensation
program based on
performance.
✔ Calculate the expected value of the annual bonus amount.
✔ Calculate the variance and standard deviation of the annual
bonus amount.
EXPECTED VALUE,
VARIANCE, AND STANDARD
DEVIATION

24. Solution: Let the random variable X denote the bonus amount (in
$1,000s) for an employee.
24
EXPECTED VALUE,
VARIANCE, AND
STANDARD DEVIATION

25. • Application of Expected Value to Risk
✔ Suppose you have a choice of receiving $1,000 in cash
or receiving a beautiful painting from your grandmother.
✔ The actual value of the painting is uncertain. Here is a
probability distribution
of the possible worth
of the painting. What
should you do?
EXPECTED VALUE,
VARIANCE, AND STANDARD
DEVIATION

26. • Application of Expected Value to Risk
✔ First calculate the
expected value:
✔ Since the expected value is more than $1,000 it may seem
logical to choose the painting over cash.
✔ However, a risk adverse person might not agree.
EXPECTED VALUE,
VARIANCE, AND STANDARD
DEVIATION

27. 1.Binomial
2. Poisson
3. Normal
TYPES OF THEORETICAL /PROBABILITY
DISTRIBUTION
27

28. □ It is the theoretical distribution of discrete random
variables. It was discovered by mathematician James
bernouli.
BINOMIAL
DISTRIBUTION
28
□ It is the discrete Probability Distribution.

29. 1.A Bernoulli process consists of a series of n independent and identical
trials of an experiment such that on each trial:
Note: Each trial is independent i.e. probability of an outcome for any particular
trial is not influenced by the outcomes of the other trials.
2.There are only two possible outcomes:
Note: outcomes are mutually exclusive, such as ‘success’ or ‘failure’, ‘good’ or
‘defective’, ‘hit’ or ‘miss’, survive or die.
p= probability of a success
1−p = q = probability of a failure
3.Each time the trial is repeated, the probabilities of success and failure
remain the same.
Note: p and q remains fixed from trial to trial .
ASSUMPTIONS/FEATURES OF BINOMIAL
DISTRIBUTION
29

30. A fair coin is tossed 3 times and we are interested in finding the
probability of exactly two heads. Therefore we will consider head as
success and tail as failure with corresponding probabilities p and q
respectively.
□ Total outcomes 8: HHH, TTT, HTT , TTH, THT, HHT,HTH THH
□ Favorable Outcomes
□ HHT,HTH THH
P = 3/8.
EXAMPLE OF BINOMIAL
DISTRIBUTION
30

31. As the number of tosses increases(say 20 0r 50 times), it becomes
more and more difficult to calculate the probability. Here an easy
method is required and hence we use binomial formula.
31
Binomial Distribution

32. P(x)= n C p x q n-x
x
n = the number of trials
p= the probability of a success on a trial
q = the probability of a failure on a trial
X = the number of successes in n trials
X= 0, 1, 2, . . ., n
BINOMIAL
DISTRIBUTION
32

33. Q1.A coin is tossed six times , what is the probability of (a)
obtaining four heads?
(b) Four or more heads?
Q2. The incidence of a certain disease is such that on the
average 20% of workers suffer from it. (Assuming the
distribution fits binomial) If 10 workers are selected at
random, find the probability that
i) exactly 2 workers suffer from the disease
ii) not more than 2 workers suffer from the disease.
EXAMPL
ES
33

34. Q3. It is believed that 20% of the employees in an office are usually
late. If 10 employees report for duty on a given day, what is the
probability that:
(a) Exactly 3 employees are late.
(b) At most 3 employees are late.
(c) At least 3 employees are late.
SOL
VE
34
0.2, 0.88, 0.32.

35. Q4.A company manufactures motor parts. The market practice is
such that goods are sold on one month’s credit in the domestic
market. The limit of credit sales of different buyers is decided by
the manufacturer based upon the perception of the goodwill of
the individual buyer. The manufacturer observed that 30% of the
buyers take more than a month in making the payment. In a
particular city if goods are sold to 5 buyers on credit, what is the
probability that
(i) Exactly 3 buyers will delay the payment beyond one month
(ii) At most 2 buyers will delay ?
35
0.1323,0.837

36. Q5. After the privatisation of the power sector in Delhi,
consumers often complain that new meters installed by the
private power companies are defective and run faster. On
testing of meters it was found that 10% of the meters were
defective and run faster. In a housing society, a test check was
conducted on 6 meters, what is the probability that (i) one
meter is defective; (ii) at least one meter is defective?
BINOMIAL
DISTRIBUTION
36
0.354,0.47

37. • For a binomial distribution:
✔ The expected value
E(X) is:
✔ The variance Var(X) is:
✔ The standard deviation
SD(X) is:
BINOMIAL
DISTRIBUTION

38. Q1.In eight throws of a die 1 or 6 is considered as success. Find
the mean number of success and the SD.
Q2.The mean of a binomial distribution is 40 and standard
deviation is 6. Calculate n, p and q.
Q3.Bring out the fallacy , if any, in the following statement. The
mean is 10 and its s.d. is 6.
EXAMPL
ES
38

39. Q4. The probability of a bomb hitting a target is 2/5. Two
bombs are enough to destroy a bridge. If 7 bombs are
dropped at the bridge , find the probability that the bridge
is destroyed.
BINOMIAL
DISTRIBUTION
39
Q5. A student appearing in a multiple choice test answers 10
questions, purely by guessing. If there are 5 choices for
each question, What is the probability that 6 or more
answers will be correct.
0.007.

40. SYSKA, A LED MANUFACTURING COMPANY REGULARLY
CONDUCTS
QUALITY
40
checks at a specified periods on the products it manufactures.
Historically, the failure rate for LED light bulbs that the company
manufactures is 5%. Suppose a random sample of 10 LED light bulbs
is selected. What is the probability that
III.
I. None of the LED light bulbs are defective?
II. Exactly one of the LED light bulbs is defective?
Two or fewer LED light bulbs are defective?
IV. Three or more of the LED light bulbs are defective?

41. Here p = 0.05, q = 1- p = 1- 0.05 = 0.95 and n = 10,
P(X = 0) = 0.5987
P(X = 1) = 0.3151
P(X ≤ 2) = 0.9885
P(X ≥ 3) = 0.0115
41

42. Poisson Distribution
A second important discrete probability Distribution is
the Poisson Distribution, named after the French
mathematician S. Poisson.
42
This distribution is used to describe the behaviour of
events, where the total no of observation (n)is large and
their chance of success (p) is low.

43. THE POISSON
DISTRIBUT
ION
• A binomial random variable counts the number of successes
in a fixed number of trials.
• In contrast, a Poisson random variable counts the number
of successes over a given interval of time or space.
• Examples of a Poisson random variable include
✔ With respect to time—the number of car accidents took place
while crosing the Chirag Delhi flyover between 9:00 am and
10:00 am on a Monday morning.
✔ With respect to space—the number of defects in a
50-yard roll of fabric.

44. • A random experiment is a Poisson process if:
✔ The number of successes within a specified time or
space interval equals any integer between zero and
infinity.
✔ The numbers of successes counted in non-overlapping
intervals are independent.
✔ The probability that success occurs in any interval is
the same for all intervals of equal size and is
proportional to the size of the interval.
THE POISSON
DISTRIBUTION

45. ▪ The number of accidents that occur on a given highway during
a given time period.
▪ The number of printing mistakes in a page of a book
▪ The number of earthquakes in Delhi in a decade.
▪ The number of deaths of the policyholders recorded by the
LIC in the last year 2019.
▪ It is used by the quality control departments of manufacturing
industries to count the number of defects found in a lot.
EXAMPLES OF POISSON
DISTRIBUTION
45

46. For a Poisson random variable X, the probability
of x successes over a given interval of time or
space is
where μ (np) is the mean number of successes and
e = 2.718 is the base of the natural logarithm.
POISSON
DISTRIBUTION
46

47. • For a Poisson distribution:
✔ The expected value E(X) is:
✔ The variance Var (X) is:
✔ The standard deviation
SD(X) is:
THE POISSON
DISTRIBUTION

48. The chances of defective is one in 400 items . If 100 items
are packed in each box, what is the probability that
any given box will contain :
i) no defective
ii) less than two defectives
iii) one or more defectives
iv) more than 3 defectives
ILLUSTRATION 1
48
(Note: e -0.25
=0.7787)

49. Here, p = 1/400, probability of defective item which is very low.
n= 100 ,no of items packed in the box which is quite large.
μ = np = 0.25 ; average number of defectives in a box of 100 items.
i) P (x=0) =
ii) ii) P (x<2) = P (x=0) + P (x=1)
SOLUTI
ON
49

50. Customers arrive randomly at a retail counter at an
average rate of 10 per hour. Assuming a Poisson
Distribution, calculate
i) No customer arrives
ii) Exactly one customer arrives
ILLUSTRATION 2
50

51. Assuming that the probability of a fatal accident in a
factory during a year is 1/1200, calculate the probability
that in a factory employing 300 workers, there will be at
least two fatal accidents in a year.
ILLUSTRATION 3
51
Hint: e -0.25
=0.7787

52. A manufacturer , who produces medicine bottles,
finds that 0.1% of the bottles are defective. The
bottles are packed in boxes containing 500 bottles.
A drug manufacturer buys 100 boxes from the
producer of bottles. Using Poisson distribution find
out how many boxes will contain
(i) No defective
(ii) At least 2 defectives (e - 0.5 = 0.6065)
EX
4
52
Ans61 &9

53. If the probability of a defective bolt is 0.2, find
(i) The mean
(ii) The standard deviation in a total of 900 bolts.
Q5
53

54. • Example: Returning to the Starbucks example, Ann believes
that the typical Starbucks customer averages 18 visits over a
30-day month.
✔ How many visits should Anne expect in a 5-day period from a typical
Starbucks customer?
✔ What is the probability that a customer visits the chain five
times in a 5-day period?
THE POISSON DISTRIBUTION
(CASE STUDY )

55. THE POISSON DISTRIBUTION
(CASE STUDY )
55

56. When no of trials become infinite, X takes continuous values, the
curve becomes smooth, called normal distribution.
Hence, normal distribution is approximation to binomial
distribution.
μ = Mean of the normal distribution.
σ= standard deviation of the normal distribution.
A random variable follows normal distribution with mean μ and standard
deviation σ.
X ~ N (μ , σ )
NORMAL
DISTRIBUTION
56

57. - ∞ <x<∞
σ = standard deviation of the normal distribution
Π = constant, 22/7=3.1416. √2 Π =2.5066
e=2.7183
μ = Mean of the normal distribution
PROBABILITY DENSITY FUNCTION OF
NORMAL DISTRIBUTION
57

58. X∽N(μ,σ)
A random variable X can be transformed to a standardized
normal variable Z by subtracting the mean and divided by
standard deviation.
Z=x- μ .
σ
PROBABILITY DENSITY
FUNCTION OF NORMAL
DISTRIBUTION
58
0 1
-1 2
-2 3
-3

59. □ Bell-shaped
□ Symmetric about mean
□ Continuous
□ Never touches the x-axis
□ Total area under curve is 1.00
□ Approximately 68% lies within 1 standard deviation of the mean,
95% within 2 standard deviations, and 99.7% within 3 standard
deviations of the mean. This is the Empirical Rule mentioned
earlier.
□ Data values represented by x which has mean µ and standard
deviation σ.
□ Probability Function given by
FEATURES OF NORMAL
DISTRIBUTION
59

60. THE STANDARD DEVIATION AND THE NORMAL CURVE
60

61. Suppose the salary of workers in a company follows normal
distribution. If the average salary is Rs.500 with a standard
deviation of Rs.100; Find the probability that workers earns a
salary between Rs.400 and Rs.650.
APPLICATI
ON
61
11618

62. EX 3
THE AVERAGE DAILY SALES OF 500 BRANCH
OFFICES IS 150 THOUSAND AND THE
STANDARD DEVIATION RS.
15THOUSAND. ASSUMING THE DISTRIBUTION
TO BE NORMAL , INDICATE HOW MANY
BRANCHES HAVE SALES BETWEEN:
62
(i) Rs.120 thousand and Rs.145 thousand.
(ii) Rs.140 thousand and Rs.165 thousand
174&295

63. A workshop produces 2000 units per day. The average weight
of units is 130 kg with a standard deviation of 10 kg.
Assuming the distribution to be normal , find out how many
units are expected to weigh less than 142 kg?
EX
4
63
1770

64. As a result of tests on 20,000 electric fans manufactured by a
company, it was found that lifetime of the fans was normally
Distributed with an average life of 2,040 hours and standard
deviation of 60 hours. On the basis of the information estimate the
number of fans that is expected to run for (a) more than 2,150
Hours (b) less than 1,960 hours.
EX
5
64
672 &1836

65. .0918
EX
6
65
Delhi’s Traffic police claims that whenever any rally is
organized in the city, traffic in the city is seriously
disrupted. On the day of rally, city’s traffic is disrupted
for about 3 hours( 180 minutes) on an average with a
standard deviation of 45 minutes. It is believed that the
disruption of traffic is normally distributed. If on a
certain day, a rally is organized in the city what is the
probability that:
(a) Traffic was disrupted up to 2 hours.
(b) Traffic was disrupted up to 5 hours. 0.9962