1. Probability literally means ‘chance’. Simple examples of probability may refer to the following:
a. What is the probability that a product launched by a particular company will be successful or not?
b. What is the probability or chance that a company will exceed its sales turnover of the previous year?
Probability is basically a fraction or a proportion where the denominator is the total number of
possible cases and the numerator is the number of likely cases.
RANDOM VARIABLE:
Again, if we want to find out the number of employees(out of the total workforce of 1000) in a company
who will be chosen to work on a particular project, we will denote the number by ‘X’ as we do not know
with certainty what value ‘X’ will take prior to the experiment. Hence, ‘X’ is a random variable.
So a random variable
a. assumes different numerical values associated with an experiment but the values cannot be known
with certainty prior to the experiment.
b. Random variables are denoted by the letters, X, Y or Z and they can be either discrete or continuous.
A discrete random variable will assume certain specific or a countable number of values.
With such kind of variables, we are concerned with counting something.
A continuous random variable, on the other hand, is concerned with measuring something. A
continuous variable, can, in essence assume any value within a specified interval .
Next, we move on to the probability distributions. Probability distribution is a schedule which
enumerates the values (that the random variables can take) and their associated probabilities.
DISCRETE PROBABILITY DISTRIBUTION:
A discrete probability distribution can be generated when the values of the discrete
random variables and their associated probabilities are known. Such a probability distribution will
specify with what probability a random variable will assume a particular value.
The probability that a random variable X will assume a value ‘x’ is given by p(x).For e.g,the probability
that there will be 1 head when an unbiased coin is tossed, will be given by p(x=1) where X is the number
of heads and x=1 is a specific value that ‘X’ can assume.
There are two important rules for discrete probability distributions:
a.0≤p(x) ≤1 for all values of ‘x’ and
b.∑p(x) =1. i. e. all the probabilities will add up to 1.
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2. Let us now discuss two types of discrete probability distributions, viz., i) Binomial and ii)
Poisson.
BINOMIAL PROBABILITY DISTRIBUTION:
The binomial distribution deals with a binomial experiment which has only two outcomes, i.e.,
success or failure, present or absent, pass or fail etc. A binomial experiment is conducted in a repeated
number of trials. It deals with probability of getting defective items (success) out of a random sample of
100 items in a factory while inspection etc.
The conditions required for a binomial experiment are:
a. the number of experiments or trials to be conducted is ‘n’ and each experiment has only two
outcomes, i.e., success or failure.
b. the outcome of any experiment is independent from the outcome of any other experiment, the
probability of success is given by ‘p’ and the number of successes in any trial is denoted by the binomial
random variable ‘X’.
The formula of binomial distribution is:
. n!xn x
p( x) . p .q
x!(n x)!
where p is the probability of success, n is the number of trials, and x is the number of successes in the
trials. Thus, the parameters of the binomial distribution are ‘n’ and ‘p’. The binomial distribution varies
with the values of the parameters.
POISSON PROBABILITY DISTRIBUTION:
The Poisson Probability Distribution is also a Discrete Probability Distribution .This deals with
occurrence of an event within a specified interval and counting the number of events occurring within a
time interval, for e.g., the number of spelling mistakes one makes while typing a single page, the
number of calls that comes in a call centre within an hour, the number of persons who enter a queue in
a departmental store within an interval of one minute etc. The Poisson random variable ‘X’ may take any
values from 0 to ∞ and the mean and the variance of the distribution is equal.
The conditions required for a Poisson distribution are:
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3. 1. The Poisson distribution gives the occurrence of an event and not the non-occurrence of an event in a
given situation.
2. Events are random over a specified time interval i.e. the probability of occurrence of more than one
event within a specified time interval is very small.
3. Events are independent in a specified time interval i.e. the probability of past or future occurrence of
an event will not affect its probability of occurrence in the current or given period.
Thus the probability of occurrence of a ‘discrete’ or a given number of occurrences in a
specified time interval is given by the following formula:
Where x is the number of occurrences=0….n; lamda is the mean or expected number of occurrences
within a given time period and >0; e is the base of natural logarithm where e=2.71828.
x
e
p( x)
x!
NORMAL DISTRIBUTION:
Next, we come to the Continuous Probability Distribution. There are gaps between
observations in a discrete probability distribution, for e.g., the number of defective items in a
shipment can be 0,1,2 etc. and not 1.4 or 2.6.,whereas in a continuous distribution, there is no
such gap between values, for e.g., the temperature of different cities in a country may be
10.40C,25.90C etc. The distribution is given by a function of the form f(x).
Now, the probability density function, or f(x) must satisfy two conditions, viz.
a. f(x) > 0, i.e. non-negative and
b. The total area under the curve is 1.
Now, the most important continuous probability distribution is the normal distribution. The
reason is that, the assumption of normality is very important in drawing conclusions from a
sample about the population from which the sample has been drawn.
In the case of a variable following a normal distribution, the normal probability plot is a straight line
and the data points in the QQ Plot do not seriously deviate from the fitted line.
For a non-normal distribution, the observations will be just the reverse.
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4. Some of the properties of the normal distribution are:
a. It is bell shaped and symmetrical about mean (which is also equal to median and mode)
b. Most of the observations are closer to the mean with only few observations being far away.
c. The distribution can be entirely determined by the values of mean (µ) and standard deviation (σ).
d. The spread of the distribution is given by the standard deviation (either large or small); but in each
case, approximately,
• 68.3 % of all observations lie within µ+σ, (i.e. 1 standard deviation of the mean)
approximately two thirds observations
• 95.4 % of all observations lie within µ+2σ
• 99.7 % of all observations lie within µ+3σ
A typical normal distribution would look like the following, i.e.
0.6
0.5
0.4
f(x)0.3
0.2
0.1
0
-4 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 4
X
Here, X, is the normal variable which follows a normal distribution with mean µ and variance
2
σ2., i.e, X follows N(µ,σ ).A suitable example may be as follows:
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5. X is plotted on the horizontal axis and the function (known as the probability density function)
f(x) is plotted on the vertical axis. Here, X has a mean of 40 and a standard deviation of 10.
A special case of the normal distribution is the standard normal distribution which has a mean of
zero and a standard deviation of one. This concept is specifically used to compare between
variables which do not have equal variance or which do not come from the same base and
scale. The standard normal variable is denoted by ‘z’ and can be represented as follows
(here Z is plotted on the horizontal axis and the function (known as the probability density
function) f (z) is plotted on the vertical axis.
0.6
0.5
0.4
f(z) 0.3
0.2
0.1
0
-4 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 4
Z
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6.  Any normal distribution can be converted to the Standard Normal Distribution,
simply by converting it’s mean to 0 and its standard deviation to 1.
i.e. Subtracting µ from each observation and dividing by σ, i.e.
X
Z
As the data are symmetrical, then we know that 50% of observations lie above and
below the mean.
Let us now consider a few examples on calculating the respective probabilities from a standard
normal curve:
Suppose the time taken to produce and dispatch an assignment of 10,000 commodities follows a
normal distribution with a mean value of 62.5 days with a standard deviation of 2.5 days.
Find the number of commodities whose production and dispatchment time is a).less than 65
days but greater than 57.5 days.
Ans: The area under a standard normal curve can be divided as follows:
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7. The two arrows in the grey colored regions above pertain to 2% of area under the standard
normal curve.
A) Area under standard normal curve between the values:
z= 57.5 -62.5 = -2 and z= 65-62.5 = 1.
2.5 2.5
Therefore, the required area under the standard normal curve is = (14%+34%+34%) = 82%.This
means 82% of 10000 commodities (i.e. = 8200 commodities) are supposed to have
processing time within this range.
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