Chapter - 6
6. Random Variables and Probability Distributions
Variable: is any characteristic or attribute that can assume different values.
A random variable (r.v): is a variable whose values are determined by chance.
Definition: A random variable is a numerical description of the outcomes of an experiment or a
numerical valued function defined on sample space, usually denoted by capital letters.
Example: If X is a random variable, then it is a function from the elements of the
sample space to the set of real numbers. i.e.
X is a function X: S → R
A random variable takes a possible outcome and assigns a number to it.
Example: Flip a coin three times, let X be the number of heads in three tosses.
* +
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
* +
X assumes a specific number of values with some probabilities.
Random variables are of two types:
1. Discrete random variable: are variables which can assume only a specific number of values.
They have values that can be counted.
Examples:
Toss a coin n time and count the number of heads.
Number of children in a family.
Number of car accidents per week.
Number of defective items in a given company.
Number of bacteria per two cubic centimeter of water.
2. Continuous random variable: are variables that can assume all values between any two give
values.
Examples:
Height of students at certain college.
Mark of a student.
Life time of light bulbs.
Length of time required to complete a given training.
Definition: a probability distribution consists of a value a random variable can assume and the
corresponding probabilities of the values.
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Example: Consider the experiment of tossing a coin three times. Let X be the number of
heads. Construct the probability distribution of X.
Solution: First identify the possible value that X can assume.
Calculate the probability of each possible distinct value of X and express
X in the form of frequency distribution.
0 1 2 3
( ) 1/8 3/8 3/8 1/8
NB: Probability distribution is denoted by P for discrete and by f for continuous random
variable.
6.1 Requirements for probability distribution
( ) ∑ ( )
( ) ∫ ( )
Note:
1. If X is a continuous random variable then (
∫ ( ) )
2. Probability of a fixed value of a continuous random variable is zero.
( ) ( ) ( ) ( ) ∫ ( )
3. If X is discrete random variable then
( ) ∑ ( ) ( ) ∑ ( )
( ) ∑ ( ) ( ) ∑ ( )
4. Probability means area for continuous random variable.
Introduction to Expectation
1. Let a discrete random variable X assume the values with the probabilities
( ) ( ) ( ) respectively. Then the expected value of X ,denoted as ( ) is
defined as:
( ) ∑ ( ) ( ) ( ) ( )
2. Let X be a continuous random variable assuming the values in the interval (a, b) such that
, ∫ ( ) then ( ) ∫ ( )
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Mean and Variance of a Random Variable
Let X be random variable.
The expected value of X is its mean
( )
The variance of X is given by:
( ) ( ) , ( )- ( )
∑ ( )
( ) {
∫ ( )
Remark: For a random variable X and constant "c''.
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
Example: What is the mean and variance of a random variable X obtained by tossing a coin
two times where X is the number of tails.
Solution: First find the sample space and construct the probability distribution of X.
* + * +
0 1 2
( ) 1/4 2/4 ¼
( ) ∑ ( )
( ) ( ) , ( )- ( )
( ) ∑ ( )
( ) ( ) , ( )- ( ) ( )
6.3. Common discrete probability distributions
1. The binomial distributions
It is used to model situations where there are 2- possible outcomes, success and failure.
A binomial experiment is a probability experiment that satisfy the following four
requirements:
1. There must be a fixed number of trials (called n).
2. Each trial has only one of the two possible outcomes. i.e. success or failure.
3. The outcome of each trial must be independent of each other.
4. The probability of success must remain the same for each trial.
Examples of binomial experiments:
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Tossing a coin 20 times to see how many tails occur.
Registering a newly produced product as defective or non defective.
Asking 100 people if they favour the ruling party.
Definition: The outcome of a binomial experiment and the corresponding probabilities of these
outcomes are called a Binomial Distribution.
.
Then, the probability of getting "x successes" in "n trials" becomes:
( ) . /
And this is sometimes written as: ( )
Example 1: If a fair coin is tossed 3 times, then what is the probability of getting:
a. Exactly 2 heads? b. 2 heads or more? c. less than 3 heads?
Solution:
* +
( )
( ) ( )( ) ( )
( ) ( ) ( ) ( )( ) ( ) ( )( ) ( )
( ) ( ) ( ) ( ) ( )
( )( ) ( )
Example 2: In a certain developing country 30% of the children are under nourished. In a
random sample of 25 children from a country, what is the probability that the number of
under nourished will be:
a) Exactly 10? b) Less than 5? c) 5 or more?
b) Between 3 or 5 inclusive? d) Less than 7, but more than 4?
Solution: n = 25; Let X- be the number of children who are under nourished:
( )
) ( ) ( )( ) ( )
) ( ) ( ) ( ) ( ) ( ) ( )
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( )( ) ( ) ( )( ) ( ) ( )( ) ( )
( )( ) ( ) ( )( ) ( )
) ( ) ( )
) ( ) ( ) ( ) ( )
) ( ) ( ) ( )
Remark: If X is a binomial random variable with parameters n and p then
( ) ( )
2. The Poisson distributions
It is used to model situations where the random variable x is the number of occurrences of a
particular event over a given period of time (space). Together with this property, the
following conditions must also be fulfilled.
Events are independent of each other.
Events occur singly.
Events occur at a constant rate (i.e. the mean number of occurrences is
proportional to the length of the interval).
The poison distribution is used as a distribution of rare events such as:
Number of telephone calls per minute.
Number of misprints per page.
Number of bacteria per slide.
Number of road accidents per day.
The processes that given rise to such events are called poison processes.
The probability that the number of occurrences, x, over a given period of time is equal to:
( )
Where: λ = is the average number of events in a given period of time.
The Poisson distribution depends only on the average number of occurrences per unit time
(space).
The mean and variance of a poison distribution is given by
( ) ( )
Example: Suppose that the bank customers arrive randomly and independently on an average of 3.2
customers every 4 minutes. What is the probability that
a. Exactly two customers arrive in every 4 minutes?
b. Exactly two customers will arrive in every 8 minutes interval?
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c. One or more customers will arrive in every 12 minutes?
Solution: Let X- be number of customers arriving at the bank within 4 minutes.
λ = 3.2 within 4 minutes.
( )
( )
( )
( )
( ) ( ) ( ) ( )
( )
* ( )+ * ( )+ { }
6.4. Common continuous Probability distributions
1. Normal distribution:
It is the most popular and the most commonly used distribution.
It is a continuous, symmetric, bell shaped distribution of a variable.
A random variable X is said to have a normal distribution if its probability density function is given
by
. /
( )
√
( ) ( ) ( )
Note: If then the distribution obtained is standard normal distribution and
( )
its pdf is given by: ( )
√
Properties of normal distribution
1. It is a bell shaped and is symmetrical about its mean and it is mesokurtic. The maximum
ordinate is at
2. The curve never touches the x axis. Theoretically, no matter how far in either direction or
the curve extends, it never meets the x axis- but it gets increasingly closer.
3. The total area under the curve sums to 1, i.e., the area of the distribution on each side of the
mean is 0.5. ⇒∫ ( )
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4. The mean, median, and mode are equal and are located at the center of the distribution.
5. The probability that a random variable will have a value between any two points is equal to
the area under the curve between those points.
Note: As the normal distribution curve is a continuous distribution, the probability that "x" lies
between "a" and "b" can be calculated by finding the area under the curve b/n the values "a" and
"b". i.e.
( )
. /
( ) ∫ ( ) ∫
√
Evaluating this integral may not be simple, due to this fact, we use the normal table for
calculating probabilities for normally distributed variables.
But, this integral is a definite integral which is tedious to compute, to overcome this problem,
we standardize the value and we use the table of standard normal distribution to compute the
probabilities.
This is done by calculating the value of "Z" using: ( )
Areas under the standard normal distribution curve have been tabulated in various ways.
The most common ones are the areas between:
Given a normally distributed random variable "X" with
( ) ( ) Then
( ) ( ) ( ) ( )
Procedure to find the area under the standard normal distribution curve
1. Draw the picture
2. Shade the area desired.
3. Find the correct figure.
4. Follow the direction.
Note: The table gives the areas between 0 and any z value to the right of 0, and all areas are
positive.
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Example: Find the area under the standard
normal distribution which lies. e.
a. .
( ) ( )
( )
b. ( )
( ) ( ) ( )
( )
f.
( ) ( ) ( )
c.
( )
( )
( ) ( )
( ) ( )
( ) ( )
( )
d. ( )
( )
g.
( )
( ) (
) ( )
( )
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( ) ( )
Example 2: Find Z if
a) The normal curve area between 0 and z (positive) is 0.4726.
Solution: ( )
( )
b) The normal curve area to the left of Z is equal to 0.9868.
Solution: ( ) ( ) ( )
( )
( )
( )
Example 3: A random variable x has a normal distribution with mean 80 and standard deviation
4.8. What is the probability that it will take a value
a. Less than 87.2? b. Greater than 76.4? c. Between 81.2 and 86.0?
d. less than 70? e. between 70 and 75? f. Between 75 and 85?
Solution: ( ) ( )
a. ( ) . / ( ) ( ) ( )
.
b. ( )
. / ( ) ( ) ( )
( ) ( )
c. ( )
. / . /
( ) ( )– ( )
d. ( )
. / ( ) ( ) ( ) ( )
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e. ( )
. / . /
( ) ( )
( ) ( )
f. ( )
. / . /
( ) ( ) ( )
( ) ( ) ( )
Example 4: A normal distribution has mean 62.4, find its standard deviation if 20.05% of the area
under the normal curve lies to the right of 72.9.
Solution : ( ) ( )
. /
( )
( )
( ) ( )
( )
( )
( )
( )
Exercise: A random variable has a normal distribution with standard deviation 5. Find it’s mean if
the probability that the random variable will assume a value less than 52.5 is 0.6915.
Ans :
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