Unformatted text preview: Lessons in Business Statistics
Prepared By
P.K. Viswanathan Chapter 5: Probability Distributions Introduction
Introduction:
Very often we want to describe and analyze
numbers in the form of a distribution. You
have seen this sharply in chapter 2 and
chapter 3. Distributions can be formed by
two methods; 1) from data collected and 2)
with the help of standard distributions that
have stood the test of time. This chapter
provides the conceptual framework and
applications of the three widely used
probability distributions namely The
Binomial Distribution, The Poisson
Distribution, and The Normal Distribution.
While the first two are discrete in nature,
the last one is a continuous distribution. 1) What is a
Probability Distribution?
You rotate a pair of dice and observe the sum by totaling the
numbers that turn up in both the dice. Let us designate this
sum by a random variable X. When the first die shows up
number 1, the second die could be 1, 2, 3, 4,5,6. Likewise
when the first die shows up number 2, the second die could
be 1, 2, 3, 4, 5, 6. If you continue this experiment this way
until all the possibilities are exhausted, the sample space will
contain 36 elements. The distribution that can be observed
by us will contain a sum =2, 3,4,5,6,7,8,9,10,11,12. Please
note that the starting sum will be =2 because when both the
dice show up number 1, the sum =1+1 =2. Likewise, the
maximum sum in the distribution will be 12 when both the
dice show up a number 6. If we write the values of the
random variable (sum) with associated probabilities, it will
become what is called a probability distribution. Probability Distribution for the Example
X=Sum
P(X)
2
1/36
3
2/36
4
3/36
5
4/36
6
5/36
7
6/36
8
5/36
9
4/36
10
3/36
11
2/36
12
1/36
Please note that the probability of the sample space
=1. (Adding all the probabilities above you get 1) Definition of Probability
Distribution
In precise terms, a probability distribution is a total listing of the
various values the random variable can take along with the
corresponding probability of each value.
A real life example would be the pattern of distribution of the
machine breakdowns in a manufacturing unit. The random variable in
this example would be the various values the machine breakdowns
could assume. The probability corresponding to each value of the
breakdown is the relative frequency of occurrence of the breakdo wn.
The probability distribution for this example is constructed by the
actual breakdown pattern observed over a period of time. Statisticians
use the term “observed distribution” of breakdowns. 2) Binomial Distribution
The Binomial Distribution is a widely used probability
distribution of a discrete random variable. It plays a major
role in quality control and quality assurance function.
Manufacturing units do use the binomial distribution for
defective analysis. An item is considered defective if it
has one or more types of defects. Reducing the number of
defectives using the proportion defective control chart (p
chart) is an accepted practice in manufacturing
organizations. Binomial distribution is also being used in
service organizations like banks, and insurance
corporations to get an idea of the proportion customers
who are satisfied with the service quality. Conditions for Applying Binomial
Distribution (Bernoulli Process) Trials are independent and random.
There are fixed number of trials (n trials).
There are only two outcomes of the trial
designated as success or failure.
The probability of success is uniform through
out the n trials Binomial Probability Function
Binomial Probability Function
Under the conditions of a Bernoulli process,
The probability of getting x successes out of n trials is indeed the definition of a Binomial
Distribution. The Binomial Probability Function is given by the following expression n x
P(x) (1 p) n x
p x Where P(x) is the probability of getting x successes in n trials
n is the number of ways in which x successes can take place out of n trials
x n!
=
x!(n x)! p is the probability of success, which is the same through out the n trials.
p is the parameter of the Binomial distribution
x can take values 0, 1, 2, ………., n Example for Binomial Distribution Example:
A bank issues credit cards to customers under the
scheme of Master Card. Based on the past data, the
bank has found out that 60% of all accounts pay on
time following the bill. If a sample of 7 accounts is
selected at random from the current database,
construct the Binomial Probability Distribution of
accounts paying on time. Solution
This problem can be structured as a Bernoulli Process where an account
paying on time is taken as success and an account not paying on time is
taken as failure. The random variable x represents here an account
paying on time, which can take values 0,1,2,3,4,5,6,7. You need to
prepare a table containing x and P(x) for all the values of x. Performing
calculations using Binomial Probability Function
n x
P(x) (1 n for x = 0,1,2,3,4,5,6,7 is very tedious. p
p) x x The best option is to use the Electronic Spread sheet of Microsoft Excel
to calculate the Binomial Probabilities both for individual values and
for the cumulative position. This facility is available under the option
“Paste Function”. The form of the function is: =BINOMDIST(x, n, p, 0
or 1) where x is the number of successes, n is the number of trials, and
p is the probability of success in each trial. The last term 0 or 1
performs a logical operation. If you enter 0, the computer returns the
individual probability value; if 1 is entered, the computer gives the
cumulative probability. Spreadsheet Showing the Solution Mean and Standard Deviation of
the Binomial Distribution
The mean of the Binomial Distribution is
given by = E(x)= np
The Standard Deviation is given by
=
np(1 p)
For the example problem in the previous two slides,
Mean =7 =4.2.
0.6
4.2(1 0.60) = 1.30
Standard Deviation = 3) Poisson Distribution
Poisson Distribution is another discrete distribution
which also plays a major role in quality control in the
context of reducing the number of defects per standard
unit such as number of defects per item, number of
defects per transformer produced, number of defects
per 100 m2 of cloth, etc. Other real life examples would
include 1) The number of cars arriving at a highway
check post per hour; 2) The number of customers
visiting a bank per hour during peak business period. Poisson Process The probability of getting exactly one success
in a continuous interval such as length, area,
time and the like is constant The probability of a success in any one interval
is independent of the probability of success
occurring in any other interval. The probability of getting more than one
success in an interval is 0. Poisson Probability Function
Poisson Distribution Formula
x
e λλ
P(x) x! where
P(x) = Probability of x successes given an idea of λ
λ Average number of successes
=
e = 2.71828(based on natural logarithm)
x = successes per unit which can take values 0, 1, 2, 3,………….. λis the Parameter of the Poisson Distribution.
Mean of the Poisson Distribution is = λ Standard Deviation of the Poisson Distribution is = λ Example
If on an average, 6 customers arrive every two
minutes at a bank during the busy hours of working,
a) what is the probability that exactly four customers
arrive in a given minute? b) What is the probability
that more than three customers will arrive in a given
minute? Solution
λ 3 ( 6 customers every two minutes implies 3 customers per
=
minute)
a) You are required to find out P(x =4). Substituting in the
formula the value of x =4, you can get the answer after
e λ x
simplifying the expression P(x) λ
x!
However, you are strongly encouraged to use the Microsoft Excel.
Standard Poisson table or manual calculation cannot match
“Excel”. Use the Paste Function option and then click Poisson.
Follow the prompts, you get the answer. Just like the Binomial
Distribution, here also the logical operator 0 gives the individual
probabilities, and 1 gives the cumulative probabilities. The output
from Excel is given in the next slide for the values of x ranging
from 0 to 10 Spreadsheet showing Solution Look at the table and you will find that corresponding to x =4 in column B, the probability
P(x=4) is 0.168031 found in column C.
You want P(x > 3) = 1 P(x < = 3) = 1 0.647232 = 0.352768 ( Please note that P(x< = 3) is
found in Column D corresponding to the value of x =3 in column B. You could see
0.647232 in row 8 and column D. 4) Normal Distribution The Normal Distribution is the most widely used continuous distribution. It occupies
a unique place in the field of statistics. In fact, the entire quality control function that
employs the statistical techniques heavily will come to a grinding halt without the
use of the normal distribution. The control charts for reducing and stabilizing
variation relies on the normal distribution. Process capability studies to meet the
customer specifications needs the normal distribution. The whole subject matterinferential statistics is based on the normal distribution. In all management functions
including the human side, the observed frequency distributions encountered are all
fairly close to the normal distribution when the sample size is reasonably large. Properties of Normal Distribution The normal distribution is a continuous distribution looking like a bell. Statisticians use
the expression “Bell Shaped Distribution”. It is a beautiful distribution in which the mean, the median, and the mode are all equal to
one another. It is symmetrical about its mean. If the tails of the normal distribution are extended, they will run parallel to the horizontal
axis without actually touching it. (asymptotic to the xaxis) The normal distribution has two parameters namely the mean and the standard deviation Normal Probability Density Function
In the usual notation, the probability density function of the normal distribution is given below: f ( x) 1
e
σ 2π (x ) 2
μ 2
2σ x is a continuous normal random variable with the property x meaning x can take all real numbers in the interval x . You need not worry about this complicated
function because all calculations can be done without the help of this expression. What is important is that you should understand the normal distribution and you should be
able to apply it to practical problems. For this purpose, the knowledge of the actual probability
density function is not required.
If x follows a normal distribution with mean and standard deviation it is a
,
general practice among st atisticians to symbolize this expression as x~ N ().
, Property of Symmetry Property of SymmetryContinues Property of SymmetryContinues Standard Normal Distribution
For practical problems, which require the help of the normal distribution
for solutions, it is not possible to compute the probability in a
straightforward manner because each of those normal variables may
have different units of measurements. One problem may involve the unit
meter, another may involve kilograms, and so on. Therefore you need a
probability distribution that can tackle any unit of measurement.
Fortunately, you have the Standard Normal Distribution defined as
follows:
x
μ
z
σ z is called the standard normal variable. Please note that z is a pure
number independent of the unit of measurement. The random variable z
follows a normal distribution with mean=0 and standard deviation =1.
You first convert the original variable in a given problem into z. It is
only a transformation of scale. The probability table for z is available for
computing the necessary probabilities for a given situation. The elegant
way of getting the probability value based on z is to take advantage of
the Microsoft Excel. Example Problem
The mean weight of a morning breakfast cereal pack is
0.295 kg with a standard deviation of 0.025 kg. The
random variable weight of the pack follows a normal
distribution.
a)
b)
c) What is the probability that the pack weighs less than
0.280 kg?
What is the probability that the pack weighs more than
0.350 kg?
What is the probability that the pack weighs between
0.260 kg to 0.340 kg? Solutiona) x
μ
z
= (0.2800.295)/0.025 = 0.6. Click “Paste Function” of Microsoft Excel,
σ
then click the “statistical” option, then click the standard normal distribution option
and enter the z value. You get the answer. Excel accepts directly both the negative
and positive values of z. Excel always returns the cumulative probability value. When
z is negative, the answer is direct. When z is positive, the answer is =1 the
probability value returned by Excel. The answer for part a) of the question =
0.2743(Direct from Excel since z is negative). This says that 27.43 % of the packs
weigh less than 0.280 kg. Solutionb) x
μ
z
= (0.350 0.295)/0.025 =2.2. Excel returns a value of 0.9861. Since z is
σ
positive, the required probability is = 10.9861 =0.0139. This means that 1.39% of the
packs weigh more than 0.350 kg. Solutionc) For this part, you have to first get the cumulative probability up to 0.340 kg and then
x
μ
subtract the cumulative probability up to 0.260. z = (0.340 0.295)/0.025
σ
x
μ
=1.8(up to 0.340). z = (0.260 0.295)/0.025 =  1.4(up to 0.260). These two
σ
probabilities from Excel are 0.9641 and 0.0808 respectively. The answer is = 0.96410.0808 = 0.8833. This means that 88.33% of the packs weigh between 0.260 kg and
0.340 kg. ...
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This note was uploaded on 02/24/2012 for the course BUSINESS 281 taught by Professor Gray during the Spring '12 term at Florida State College.
 Spring '12
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 Business

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