Probability Distributions - Viplav Kambli

Probability Distributions
The laws of probability allow us to obtain the chances of a given set of outcomes if we know the
probabilities of various possibilities in a certain situation. Sometimes the given situation may be generalised
so that standard rules can be used for calculating the required probabilities. If a generalised probability rule
may be obtained which covers all possible outcomes in a situation, we get a probability distribution
function, so called because it spels the probability of occurrence of each outcome—the proportion of time
an outcome is likely to occur or the proportion of cases likely to lie in a given class.
We now discuss some generalised distributions that are very widely obtained and are relevant to the
managers. They are based on theoretical considerations and in many cases frequency patterns observed in
real life conform to either of these. Using these distributions, predictions can be made on theoretical
grounds.
BINOMIAL DISTRIBUTION
A binomial distribution is applicable when
(i) an experiment involves n trials,
(ii) each of the trials can result in a set of dichotomous alternatives, one of which is arbitrarily termed
as success and the other as failure,
(iii) the trials are independent so that the probability of success is the same in each trial, and
(iv) the focus is on the occurrence of a certain number of successes.
In such a situation, the probabilities for the occurrence of 0, 1, 2, .. ., n successes are given by the
successive terms of the binomial expansion (q + p) n , where q is the probability of failure and p is the
probability of success in a trial, and there are n trials. In general,
P(x) = n C x q n – x p x
where x = the number of successes
Example 1 Output of a production process is known to be thirty per cent defective. What is the
probability that a sample of 5 items would contain 0, 1, 2, 3, 4, and 5 defectives?
If the appearance of a defective item is considered a success, then the probability of success in a trial,
p = 0.3. Thus, q = 1 – p = 1 – 0.3 = 0.7. With n = 5, and P(x) = n C x q n – x p x , we have
No. of successes (x) Probability p(x)
0 0.16807
1 0.36015
2 0.30870

Appendix A4: Probability Distributions 21
3 0.13230
4 0.02835
5 0.00243
Total = 1.00000
A binomial distribution has two parameters: n and p. It means that a binomial distribution can be
specified completely by its n and p.
The mean of a binomial distribution is np and its standard deviation equals npq . For the example
stated, it can be shown that mean = 5 ¥ 0.3 = 1.5 and standard deviation = 5¥ 03 . ¥ 07 . = 1.025.
If, for a binomial distribution, p = q = 0.5, it would be a symmetrical one. If p > q, the distribution would
be negatively skewed, whereas if p < q, the distribution would have a positive skewness. Further, greater
the divergence between p and q, for a given n, more pronounced would the skewness be.
POISSON DISTRIBUTION
The Poisson distribution is the distribution of rare events since it deals with situations where the chances of
occurrence of an event are very low. It is used where the events happen at random i.e one cannot predict
precisely when each would occur, nor how many will occur altogether. Examples of these include road
accidents, fire, arrival of customers at a shop and so on.
The distribution is used either as an approximation to binomial distribution when n is large and p is very
small, or in its own right.
According to this model,
P(x) = e –l l x
x!
where e is the exponential 2.7183, l is the mean value (equal to np when used as approximation to
binomial) and x is the number of occurrences, which may be any integer from 0 to any value.
Example 2 An insurance company receives, on an average, 2 telephone calls every 15 minutes.
Find the chance that (a) no calls, and (b) 3 calls be received in a 30-minute interval.
According to the given information, the average number of calls during 30 minute period, l = 4.
Probability of no calls, P(0) = e –l l
x! = 2.7183–4 ¥ 4 0! = 0.0183
Probability of 3 calls, P(3) = 2.7183 –4 ¥ 4 = 0.1954
3!
Important features of the Poisson distribution are:
(a) There is no theoretical maximum number of events that can occur. Whether the average value is
small or large, for example, we can theoretically conceive an infinite fires during a year. However,
the probabilities of the successes build up sharply around mean and then fall at a brisk rate so that
the probabilities of higher number of successes become extremely small and are negligible.
Further, whatever the mean value of a Poisson variable (x) be, the probabilities add upto 1.
Total Probability = P(0) + P(1) + P(2) + P(3) + . . .
3
0

22 Quantitative Techniques in Management
-l 2 -l 3 -l
-l
le l e l e
= e + + + + L
1! 2! 3!
= e –l 2 3
F l l I
1 + l + + + L
HG
2 3 K J
! !
The series in the bracket, being the exponential series, adds upto e l . Thus, total probability
= e –l ¥ e l = 1.
(b) A Poisson probability distribution is positively skewed. However, the skewness becomes less
pronounced as the mean value increases.
(c) The Poisson distribution has only one parameter—the mean l.
(d) For a Poisson distribution, mean and variance are each equal to l.
(e) We have, for this distribution, the following recursive relationship, P(x) = P(x – 1) ¥ l /x.
Negative exponential distribution Using Poisson’s rule, although we can predict chances of
occurrences but not the events because they occur at random. The time between events is variable and has
a distribution known as negative exponential distribution or simply exponential distribution. Thus, there
is a relation between Poisson and exponential distributions so that if the number of events occurring in a
specified time follows a Poisson distribution with mean l, then the waiting time until the first occurrence,
T, will follow an exponential distribution with a mean equal to 1/l and variance equal to 1/l 2 .
The problem of finding probabilities for events defined in terms of T can be solved by considering the
relationship between Poisson and exponential distributions. Since T is the waiting time until the first
occurrence of an event in which the total number of occurrences in a given time interval follows the
Poisson distribution, it should not be surprising that a relationship does exist. In particular, if we consider
a fixed amount of time t, then the event T > t is simply the event that the waiting time until the first
occurrence is longer than t units of time. If we know that the waiting time for the first occurrence is longer
time than t units, then there must be no occurrences during this interval. Otherwise the waiting time until
the first occurrence would be less than t. Consequently, the event T > t is equivalent to the event X = 0,
where X is the number of occurrences in t units of time. Since X has a Poisson distribution, it follows that
P(T > t) = P(X = 0) = e -ltal
t f 0
= e –lt
0!
Example 3 On an average, 2 calls are received in 15 minutes. Find the average time between
successive calls. What is the probability that the first call of the day would be received not before 10
minutes? Within 5 minutes? 7
2 1 minutes?
Average rate of calls, l = 2/15 calls/minute
\ Expected time between successive calls = 15/2 = 7.5 minutes. With t = 10 minutes,
P(T > t) = e –l t = 2.7183 –(2/15) (10) = 2.7183 –4/3 = 0.2636
When t = 5 minutes
P(T £ t) = 1 – e –l t = 1 – 2.7183 –(2/15) (5) = 0.4866
When t = 7.5 minutes,
P(T £ t) = 1 – 2.7183 –(2/15) (15/2) = 1 – 0.3679 = 0.6321.

Appendix A4: Probability Distributions 23
NORMAL DISTRIBUTION
Normal distribution is a very important and useful distribution for a manager because many phenomena
follow such a distribution or are close to it. In contrast to the binomial and Poisson distributions in which
our concern is with determining probabilities of some number of successes, which can assume only discrete
value of 0, 1, 2 and so forth, we deal in the normal distribution with characteristics which can assume any
value between two given limits. Height of an individual, to illustrate, shall not be only, say, 65 or 66
inches—it could be any value between these two.
A characteristics, or variable, is said to be distributed normally if its curve appears as shown in Fig. 1
and is represented by the following expression.
1 exp (– (x – m)
y(x) =
2 /2s 2 )
s 2p
Here y(x) depicts the height of the y-ordinate at a specific value of x, and p and e are, respectively, equal
to 3.1416 and 2.7183. Thus, if m and s, the mean and the standard deviation, are known, we can get the
height of the curve at any specific value of x.
It may be noted that:
(a) A normal curve is a unimodal, bell-shaped, and symmetrical about its mean. As seen in the figure,
the curve on either side of m is a mirror image of the other side. The mean, the median, and the
mode all coincide.
(b) The total area under the curve is divided evenly because of symmetry: 50% of area is to the right of
a perpendicular line drawn at the mean and 50% is to its left.
(c) It is assumed that the variable can take any value between – • and + •. As such, a normal curve
approaches closely, but never touches, the horizontal axis.
(d) If we construct vertical lines at a distance of one standard deviation from mean in both the
directions, the area under the curve enclosed by these lines is equal to 68.27% of the total area. If we
draw these lateral boundaries at two standard deviations from the mean in both the directions, they
would enclose 95.45% of the total area. Similarly, m ± 3s covers 99.73% area under the curve.
(e) A normal curve is defined completely by the mean, m, and standard deviation, s (> 0). That is, each
different value of m and s specifies a distinct normal distribution and curve. Thus, the normal
distribution is a family of distributions in which a member is distinguished from the others on the
basis of the twin values of m and s. A distinguished member of this family is the distribution which
has a zero mean and a standard deviation of 1. It is called the standard normal distribution.
y
yx ()= 1 =e
s ÷ 2p
-( x 2
- m) /2s 2
Fig. 1
m
Normal Curve
Variable x

24 Quantitative Techniques in Management
Calculation of probabilities For a normal distribution, the probability that the given variable would
take a value in a certain range, say between X 1 and X 2 , is calculated as the proportion of the area under the
normal curve between X 1 and X 2 , to the total area under it. For the purpose of calculating the probabilities,
the given distribution is expressed in terms of the standard normal distribution. This is done by stating the
variable X as the variable z, where
z = X -m
s
At X = m, z would equal zero, z would be positive for values of X > m, and negative for values X < m.
The proportion of areas under the normal curve between the mean and particular values of z are
tabulated and shown in Table B1 at the end of the book or in the tables given on web. To illustrate, some of
the areas are given here.
Area
(i) Between m and z = 1.00 0.3413
(ii) Between m and z = 1.45 0.4265
(iii) Between z = 1.2 and z = 2.8 0.4974 – 0.3849 = 0.1125
(iv) Between z = – 1.2 and z = 2.8 0.3849 + 0.4974 = 0.8823
(v) Beyond z = 1.2 0.5000 – 0.3849 = 0.1125
It may be noted that since the curve is symmetrical, the area between the mean and a particular value of
z is the same whether z is positive or negative.
Example 4 A machine is set to fill in coffee powder in tins, with an average of 200 gms, and a
standard deviation of 4 gms. Find the probability that a coffee tin selected at random shall contain
(a) at least 200 gms, (b) between 200 and 206 gms, (c) between 195 and 205 gms, and (d) less than
196 gms.
The given distribution has m = 200 gms and s = 4 gms.
(a) Area to the right of m being 0.50, this is the probability that a tin would contain at least 200 gms of
coffee.
(b) For X = 206, z = (206 – 200)/4 = 1.5.
From the normal area table, area between m and z = 1.5 is 0.4332. Thus P(200 £ X £ 206)
= 0.4332.
(c) Area between X = 195 and X = 205 equals area between m and X = 195, plus area between m and
X = 205. For X = 205, z = (205 – 200)/4 = 1.25 while for X = 195, z = (195 – 200)/4 = – 1.25. Area
between m and z = 1.25 is equal to 0.3944.
\ P(195 £ X £ 205) = 2 ¥ 0.3944 = 0.7888.
(d) For X = 196, z = (196 – 200)/4 = – 1. Area between m and z = – 1 equals 0.3413.
\ P(X < 196) = 0.5000 – 0.3413 = 0.1687.
Example 5 A manufacturer of batteries wishes to give a guarantee for free replacement of the
batteries whose life is less than a certain time period. If he desires to replace no more than 5% of the
batteries, what should be the guarantee period, if the lives of batteries are known to be normally
distributed with mean of 1200 hours and a standard deviation of 100 hours?
The given information is depicted in Fig. 2. Here the value of X is to be determined. Since the area
between m and X is 0.45, we observe from the normal area table that z corresponding to this area is
– 1.645.

Appendix A4: Probability Distributions 25
0.05
Fig. 2
m = 1200 hrs.
Determination of X
Hours
Thus, – 1.645 = X - 1200 or X = 1200 – 100 ¥ 1.645 = 1035.5
100
A guarantee of 1036 hours may, therefore, be given by the manufacturer.