QBM 101 Business Statistics

Department of Business Studies

Faculty of Business, Economics & Accounting
HELP University
 SUBJECT OUTLINE:
 Module 1: Introduction; organizing
and graphing data; numerical
descriptive measures

 Module 2: Probability, discrete random

variables; continuous random variables
and the normal distribution

 Module 3: Sampling distributions;

estimation; hypothesis testing

 Module 4: Simple linear regression

 CHAPTER 5: DISCRETE
RANDOM VARIABLES
 5.1 Random variables
 5.2 Probability distribution of a
discrete random variable
 5.3 Mean and standard deviation
of a discrete random variable
 5.4 The binomial probability
distribution
 5.5 The Poisson probability
distribution
TABLE 5.1 FREQUENCY AND RELATIVE FREQUENCY
DISTRIBUTION OF THE NUMBER OF VEHICLES OWNED
BY FAMILIES
RANDOM VARIABLES

Definition
A random variable is a variable whose value is
determined by the outcome of a random
experiment.

A random variable that assumes countable values

is called a discrete random variable.

A random variable that can assume any value

contained in one or more intervals is called a
continuous random variable.
RANDOM VARIABLES

Examples of continuous random variables:

1. The length of a room

2. The time taken to commute from home to work
3. The amount of milk in a gallon (note that we do not
expect “a gallon” to contain exactly one gallon of
milk but either slightly more or slightly less than
one gallon)
4. The weight of a letter
Definition
The probability distribution of a discrete random
variable lists all the possible values that the random
variable can assume and their corresponding
probabilities.

The probability distribution of a discrete random

variable possesses the following two characteristics.

1. 0 ≤ P(x) ≤ 1 for each value of x

2. Σ P(x) = 1.
EXAMPLE 5-2

Each of the following tables lists certain values of x

and their probabilities. Determine whether or not
each table represents a valid probability
distribution.
EXAMPLE 5-2

(a) No, since the sum of all probabilities is not

equal to 1.0.
(b) Yes.
(c) No, since one of the probabilities is negative.
EXAMPLE 5-3
The following table lists the probability
distribution of the number of breakdowns per
week for a machine based on past data.

(a) Present this probability distribution graphically.

(b) Find the probability that the number of
breakdowns for this machine during a given week is
i. exactly 2
ii. 0 to 2
iii. more than 1
iv. at most 1
 EXAMPLE 5-3
Let X denote the number of breakdowns for this machine
during a given week.
 EXAMPLE 5-3
(b) Using Table 5.4,

i. P(exactly 2 breakdowns) = P(X = 2) = .35

ii. P(0 to 2 breakdowns) = P(0 ≤ X ≤ 2)

= P(X = 0) + P(X = 1) + P(X = 2)
= .15 + .20 + .35 = .70

iii. P(more then 1 breakdown) = P(X > 1)

= P(X = 2) + P(X = 3)
= .35 +.30 = .65

iv. P(at most one breakdown) = P(X ≤ 1)

= P(X = 0) + P(X = 1)
= .15 + .20 = .35
 EXAMPLE 5-4
According to a survey, 60% of all students at a large
university suffer from math anxiety. Two students are
randomly selected from this university. Let X denote the
number of students in this sample who suffer from math
anxiety. Develop the probability distribution of X.

Let us define the following two events:

N = the student does not suffer from math anxiety
M = the student suffers from math anxiety

P(X = 0) = P(NN) = .16

P(X = 1) = P(NM or MN) = P(NM) + P(MN)

= .24 + .24 = .48

P(X = 2) = P(MM) = .36

The mean of a discrete variable x is the value that is
expected to occur per repetition, on average, if an
experiment is repeated a large number of times. It is
denoted by µ, also known as the expected value, E(X), and it
is calculated as
µ = E(X) = Σ x P(x)
Discrete probability
distribution

Mean,   E ( X )   x  P( x)
E ( X )   x  P( x)
2 2

Variance,  E ( X )   E ( X ) 
2 2 2
EXAMPLE 5-5

Find the mean number of breakdowns per week for

this machine.
EXAMPLE 5-5

The mean is µ = Σx P(x) = 1.80

DISCRETE PROBABILITY FUNCTION
STANDARD DEVIATION

The standard deviation of a discrete random

variable X measures the spread of its probability
distribution and is computed as:

  x 2
P ( x )   2

  E ( X )   E ( X )
2 2
EXAMPLE 5-6

Compute the standard deviation of X.

EXAMPLE 5-6

  E ( X )   E ( X )
2 2

 7.7  2.5 2

 1.45  1.204
EXAMPLE 5-7
Loraine Corporation is planning to market a new
makeup product. According to the analysis made by
the financial department of the company, it will
earn an annual profit of $4.5 million if this product
has high sales and an annual profit of $1.2 million if
the sales are mediocre, and it will lose $2.3 million a
year if the sales are low. The probabilities of these
three scenarios are .32, .51 and .17 respectively.

(a) Let X be the profits (in millions of dollars) earned

per annum from this product by the company. Write
the probability distribution of X.
(b) Calculate the mean and the standard deviation
of X.
EXAMPLE 5-7

E ( X )     xP  x   $ 1.661 million
σ  Var ( X )   x P  x  
2 2

 E ( X 2 )   E ( X )  8.1137  (1.661) 2
2

 $ 2.314 million
EXERCISE 1

The following table records the probability

distribution of a discrete random variable X.

x 1 2 3 5 8 13
P(X = x) 0.10 0.25 r 2r 0.15 0.05

(a) Determine the value of r.

(b) Evaluate P(2  X  9).
(c) Determine  X and  X .
Conditions of a Binomial Experiment

A binomial experiment must satisfy the following four

conditions.

1. There are n identical trials.

2. Each trail has only two possible outcomes.
3. The probabilities of the two outcomes remain
constant.
4. The trials are independent.
TOSSING A COIN
Consider the experiment consisting of 10 tosses of a
coin. Determine whether or not it is a binomial
experiment.

1. There are a total of 10 trials (tosses), and they are

all identical. Here, n = 10.
2. Each trial (toss) has only two possible outcomes: a
head and a tail.
3. The probability of obtaining a head (a success) is ½
and that of a tail (a failure) is ½ for any toss. That is,
p = P(H) = ½ and q = P(T) = ½
4. The trials (tosses) are independent.

Consequently, the experiment consisting of 10 tosses

is a binomial experiment.
THE BINOMIAL PROBABILITY DISTRIBUTION
For a binomial experiment, the probability of
exactly x successes in n trials is given by the
binomial formula

x n x
P( X  x)  Cx p q n

where
n = total number of trials
p = probability of success
q = 1 – p = probability of failure
x = number of successes in n trials
n - x = number of failures in n trials
Binomial random
variables

X ~ B(n, p)
n x
P( X  x)  Cx p q
n x

  np
  npq
Fair vs biased coin
Fair coin: Equal probability of
getting a head or a tail.
P(Head) = P(Tail ) = 0.5
P(Head) + P(Tail) = 1.0

Biased coin: Different probability

of getting a head or a tail.
P(Head) = 0.7 ≠ P(Tail) = 0.3
P(Head) + P(Tail) = 1.0
Terminology

More than 4: P( X  4)
Less than 4: P( X  4)
At least 4: P( X  4)
At most 4: P( X  4)
Between 4 and 6: P(4  X  6)
Between 4 and 6 (inclusive): P(4  X  6)
EXAMPLE 5-10
Five percent of all DVD players manufactured by a
large electronics company are defective. A quality
control inspector randomly selects three DVD player
from the production line. What is the probability that
exactly one of these three DVD players is defective?
D = a selected DVD player is defective P(D) = .05
G = a selected DVD player is good P(G) = .95
P(DGG) = P(D) P(G) P(G)
= (.05)(.95)(.95) = .0451
P(GDG) = P(G) P(D) P(G)
= (.95)(.05)(.95) = .0451
P(GGD) = P(G) P(G) P(D)
= (.95)(.95)(.05) = .0451

P(1 DVD player in 3 is defective)

= P(DGG or GDG or GGD)
= P(DGG) + P(GDG) + P(GGD)
= .0451 + .0451 + .0451
= .1353
EXAMPLE 5-10
EXAMPLE 5-10
n = total number of trials = 3 DVD players
x = number of successes = number of defective DVD
players = 1
n – x = number of failures = number of good DVD
players = 3 - 1 = 2
p = P(success) = .05
q = P(failure) = 1 – p = .95
X ~ B(n  3, p  0.05)
EXAMPLE 5-11
At the Express House Delivery Service, providing high-
quality service to customers is the top priority of the
management. The company guarantees a refund of all
charges if a package it is delivering does not arrive at its
destination by the specified time. It is known from past
data that despite all efforts, 2% of the packages mailed
through this company do not arrive at their destinations
within the specified time. Suppose a corporation mails
10 packages through Express House Delivery Service on
a certain day.

(a) Find the probability that exactly one of these 10

packages will not arrive at its destination within the
specified time.
(b) Find the probability that at most one of these 10
packages will not arrive at its destination within the
specified time.
EXAMPLE 5-11
n = total number of packages mailed = 10
p = P(success) = .02
q = P(failure) = 1 – .02 = .98

X ~ B(n  10, p  0.02)

(a) P( X  1)  10C1 (.02)1 (.98)9  10  (.02)1 (.98)9
 (10)(.02)(.83374776)  0.1667
(b) P( X  1)  P( X  0)  P( X  1)
 10C0 (.02)0 (.98)10  10C1 (.02)1 (.98)9
 (1)(1)(.81707281)  (10)(.02)(.83374776)
 0.8171  0.1667
= 0.9838
EXAMPLE 5-12
In a Pew Research Center nationwide telephone survey
conducted in March through April 2011, 74% of college
graduates said that college provided them intellectual
growth (Time, May 30, 2011). Assume that this result
holds true for the current population of college
graduates. Let X denote the number in a random sample
of three college graduates who hold this opinion. Write
the probability distribution of X and draw a bar graph
for this probability distribution.

n = total college gradates in the sample = 3

p = P(a college graduate holds the said opinion) = .74
q = P(a college graduate does not hold the said opinion)
= 1 - .74 = .26
EXAMPLE 5-12
P( X  0)  3C0 (.74)0 (.26)3  (1)(1)(.017576)  .0176
P( X  1)  3C1 (.74)1 (.26) 2  (3)(.74)(.0676)  .1501
P( X  2)  3C2 (.74) 2 (.26)1  (3)(.5476)(.26)  .4271
P( X  3)  3C3 (.74)3 (.26)0  (1)(.405224)(1)  .4052
EXAMPLE 5-13
In an NPD Group survey of adults, 30% of 50-year-old or
older (let us call them 50-plus) adult Americans said
that they would be willing to pay more for healthier
options at restaurants (USA TODAY, July 20, 2011).
Suppose this result holds true for the current population
of 50-plus adult Americans. A random sample of five 50-
plus adult Americans is selected.

(a) Find the probability that exactly three persons in

this sample hold the said opinion.
(b) Find the probability that at most two persons in this
sample hold the said opinion.
(c) Find the probability that at least three persons in
this sample hold the said opinion.
(d) Find the probability that one to three persons in this
sample hold the said opinion.
EXAMPLE 5-13
EXAMPLE 5-13
n  5, p  0.3, q  0.7
X ~ B(n  5, p  0.3)
(a) P( X  3)  0.1323
(b) P( X  2)  P( X  0)  P( X  1)  P( X  2)
 0.1681  0.3602  0.3087
= 0.8372
(c) P( X  3)  P( X  3)  P( X  4)  P( X  5)
= 0.1323 + 0.0283 + 0.0024
= 0.163
(d) P(1  X  3)  P( X  1)  P( X  2)  P( X  3)
 0.3602  0.3087  0.1323
 0.8012
PROBABILITY OF SUCCESS AND THE SHAPE
OF THE BINOMIAL DISTRIBUTION

1. The binomial probability distribution is symmetric

if p =.50.

2. The binomial probability distribution is skewed to

the right if p is less than .50.

3. The binomial probability distribution is skewed to

the left if p is greater than .50.
SYMMETRIC

Probability Distribution of x for n = 4 and p = .50

SKEWED TO THE RIGHT

Probability Distribution of x for n = 4 and p = .30

SKEWED TO THE LEFT

Probability Distribution of x for n = 4 and p = .80

MEAN AND STANDARD DEVIATION OF THE
BINOMIAL DISTRIBUTION
The mean and standard deviation of a binomial
distribution are, respectively,

  np and   npq

where n is the total number of trails, p is the

probability of success, and q is the probability of
failure.
EXAMPLE 5-14
In a 2011 Time magazine poll, American adults were
asked, “When children today in the U.S. grow up, do
you think they will be better off or worse off than
people are now?” Of these adults, 52% said worse.
Assume that this result is true for the current
population of U.S. adults. A sample of 50 adults is
selected. Let X be the number of adults in this sample
who hold the above-mentioned opinion. Find the mean
and standard deviation of the probability distribution
of X.
EXAMPLE 5-14
n = 50, p = .52, and q = .48
Using the formulas for the mean and standard
deviation of the binomial distribution,

X ~ B(n  50, p  0.52)

  np  50(.52)  26
  npq  (50)(.52)(.48)  3.5327
EXERCISE 2
Assume that 40% of all the students in a
mathematics department failed their statistic test.

(a) What is the probability that exactly half of a

group of 10 students failed their test?
(b) What is the probability that between 6 and 10
(inclusive) students out of a group of 20 students
passed their statistics test?
(c) What is the probability that none of a group of 8
students failed their statistics test?
Conditions to Apply the Poisson Probability Distribution

The following three conditions must be satisfied to apply

the Poisson probability distribution.
1. X is a discrete random variable.
2. The occurrences are random.
3. The occurrences are independent.

Examples:
1. The number of accidents that occur on a given highway
during a 1-week period.
2. The number of customers entering a grocery store
during a 1–hour interval.
3. The number of television sets sold at a department store
during a given week.
THE POISSON PROBABILITY DISTRIBUTION

Poisson Probability Distribution Formula

According to the Poisson probability

distribution, the probability of x occurrences in an
interval is

e  x
P( X  x) 
x!

where λ (pronounced lambda) is the mean number of

occurrences in that interval and the value of e is
approximately 2.71828.
Mathematical constants

  3.142...
Euler's number, e  2.71828...
Euler's constant,   0.5772...
THE POISSON PROBABILITY DISTRIBUTION

X ~ Po( )
e   x
P( X  x) 
x!
 
 
EXAMPLE 5-17
On average, a household receives 9.5 telemarketing
phone calls per week. Using the Poisson
distribution formula, find the probability that a
randomly selected household receives exactly 6
telemarketing phone calls during a given week.

e    x e 9.5 (9.5)6
P( X  6)  
x! 6!
(.00007485)(735, 091.8906)

720
 0.0764
Example: On average, two new accounts are opened per
day at an Imperial Savings Bank branch. Assume that the
events of account opening follow a Poisson distribution,
find the probability that:
(a) exactly 6 accounts will be opened during a one-day
period.
(b) at most 3 accounts will be opened during a one-day
period.
(c) less than 3 accounts will be opened during a two-day
period.
(d) between 4 and 6 (inclusive) accounts will be opened
during a three-day period.
(e) Find the mean and standard deviation of the number of
accounts opened during a five-day working week.
Let X be the number of new accounts.
(a) X ~ Po( =2)
e 2  26
P( X  6)   0.01203
6!

(b) P( X  3)
 P( X  0)  P( X  1)  P( X  2)  P( X  3)
e 2  20 e 2  21 e 2  22 e 2  23
   
0! 1! 2! 3!
2  2 0
21
2 2
2 3

e     
 0! 1! 2! 3! 
 0.8571
Let X be the number of new accounts.
(c) X ~ Po( =2  2=4)
P( X  3)
 P( X  0)  P( X  1)  P( X  2)
4 4 4
e 4 e 4 e 4
0 1 2
  
0! 1! 2!
4  4 0
41
4 2

e    
 0! 1! 2! 
 0.2381
Let X be the number of new accounts.
(d) X ~ Po( =3  2=6)
P(4  X  6)
 P( X  4)  P( X  5)  P( X  6)
6 6 6
e 6 e 6 e 6
4 5 6
  
4! 5! 6!

6 6
4
6 6 
5 6
e    
 4! 5! 6! 
 0.4551
(e) Let X be the number of new accounts.
X ~ Po( =5  2=10)
Mean,     10
Standard deviation,     10  3.162
EXAMPLE 5-20
On average, two new accounts are opened per day
at an Imperial Saving Bank branch. Find the
probability that on a given day the number of new
accounts opened at this bank will be
(a) exactly 6 ; (b) at most 3; (c) at least 7; (d) at
least 2
EXAMPLE 5-20

(a) P(X = 6) = .0120

(b) P(at most 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

=.1353 +.2707 + .2707 + .1804 = .8571

(c) P(at least 7) = P(X = 7) + P(X = 8) + P(X = 9)

= .0034 + .0009 + .0002 = .0045

(d) P(at least 2) = 1-P(X = 0) – P(X = 1) = 1 – .1353 - .2707 = .594

EXAMPLE 5-21
An auto salesperson sells an average of .9 car per
day. Let X be the number of cars sold by this
salesperson on any given day. Using the Poisson
probability distribution table, write the probability
distribution of X. Draw a graph of the probability
distribution.
EXERCISE 3
The number of students arriving at an information
counter from 8am to 5pm follows a Poisson
distribution with a mean of 18.

(i) What is the probability that no student arrives

in the next hour?
(ii) What is the probability that more than 5
students arrive at the counter from 9am to 11am?
(iii) What is the probability that there are more
than 7 but less than 12 students who arrive at the
counter from 12noon to 4:45pm?
 SUMMARY
 Discrete probability
 Binomial
 Poisson
 Calculate the probability of given
condition (more than 3, less than 4,
at most 5, less than 6 but at least 2,
between 7 and 9)
 Calculate the mean, variance, and
standard deviation