Random
Numbers:
Random
Variables:
(a) Discrete Random Variable, and
(b)
Continuous Random Variable.
(a) Discrete Random Variable: A random variable which can assume only a finite number of values or a sequence of whole numbers is called a discrete random variable. For example, the number of spots on a die is a discrete random variable, number of persons enrolled for CSS examinations, number of students passed in 1^{st} division in a particular class, number of defective items in a lot, etc. are discrete random variables, which could assume any of the possible values, i.e., 1, 2, 3…….
(b)
Continuous Random Variable:
A random variable which can assume all possible values on a continuous scale in
a given interval is called a continuous random variable.
For example, height, weight, temperature, distance, life periods, speed,
etc. are continuous random variables.
Example:
A coin is tossed three times.
Find the possibility space and define two random variables for this
possibility space.
Solution:
S = {HHH, HHT, HTH, THH, HTT, TTH, THT,
TTT}
(i)
Let a random variable (X) the number of heads:
X = no. of heads.
Note: The same value may be assigned to different outcomes of the possibility space.
(ii) Let a random variable (X) head as +1 and tail as –1:
Probability
Distribution:
(a) Discrete Probability Distribution, and
(b)
Continuous Probability Distribution.
(a)
Discrete
Probability Distribution: Let a
discrete random variable X assume values x_{1}, x_{2},
x_{3}, ……….., x_{n} with respective
probabilities P(x_{1}), P(x_{2}), P(x_{3}),
…………, P(x_{n}). Since
the random variable takes a discrete set of values, it is also called a discrete
probability distribution. A
discrete probability distribution may take the form of a table, a graph or a
mathematical equation.
A
probability distribution is similar to a relative frequency distribution with
probabilities replacing relative frequencies.
A discrete
probability distribution must possess the following two properties:
(i) 0 ≤ P(x_{i}) ≤ 1
(ii)
∑P(x_{i}) = 1, which means that the sum of
probabilities is equal to one.
Example:
A coin is tossed three times.
Find the probability distribution of the random variable number of heads.
Solution:
S = {HHH, HHT, HTH, THH, HTT, THT, TTH,
TTT}
No.
of Heads X |
Probability
of X P(X) |
0 |
^{1}/_{8 } |
1 |
^{3}/_{8} |
2 |
^{3}/_{8} |
3 |
^{1}/_{8} |
Total |
1 |
Example:
Determine whether the function
for X = 1, 2, 3 and 4 can be a
probability distribution.
Solution:
X |
P(X) |
1 |
^{2}/_{14} |
2 |
^{3}/_{14} |
3 |
^{4}/_{14} |
4 |
^{5}/_{14} |
Total |
1 |
(b)
Continuous
Probability Distribution: As we known
that a random variable which can assume all possible values within a given
interval is called a continuous random variable.
Within a given interval, there are an infinite number of values.
For example, there may be an infinite number of weights between 69.5 kgs
and 70.5 kgs. In case of a
continuous random variable, therefore, we compute probabilities for various
intervals of continuous random variable, such as P(a ≤ X ≤ b)
or P(X ≥ c).
The probability
distribution of a continuous random variable cannot be presented in tabular
form. It can be represented by
means of a formula or through a graph. The
formula is necessarily in the form of a function of the numerical values of the
continuous random variable X. For
e.g., a continuous random variable can assume values between X = 2 and X = 4 and
the function is given by:
The continuous
probability distribution is further discussed in detail later.
Mean and
Variance of a Random Variable:
In a probability distribution of a
random variable X, the mean, also referred to as ‘Mathematical
Expectation’ or ‘Expected Value’, and the variance are defined
as:
μ = E(X) = Σ X · P(X)
and
σ^{2} = V(X) = Σ X^{2} · P(X) – [E(X)]^{2}
Distribution
Function:
A function showing probabilities that a
random variable X has a value less than or equal to x is called the ‘cumulative
distribution function’ or ‘distribution function of x’.
Symbolically, the cumulative
distribution function, denoted by f(x) is defined as:
The cumulative distribution function has
the following properties:
(i) f(– ∞) = 0 and f(∞) = 1, which means that f(x) is an increasing function ranging from 0 to 1.
(ii)
If a < b then f(a) < f(b)
for any real numbers a and b.
For a discrete random variable,
distribution function is obtained by cumulating probabilities just as we
obtained cumulative distribution.
The distribution function for the
probability distribution of the previous two examples is as below:
x |
f(x) |
x < 0 |
0 |
0 ≤ x < 1 |
^{1}/_{8 } |
1 ≤ x < 2 |
^{4}/_{8} |
2 ≤ x < 3 |
^{7}/_{8} |
x ≥ 3 |
1 |
x |
f(x) |
x < 1 |
0 |
1 ≤ x < 2 |
^{2}/_{14 } |
2 ≤ x < 3 |
^{5}/_{14} |
3 ≤ x < 4 |
^{9}/_{14} |
x ≥ 4 |
1 |
Example:
Calculate the mean and variance for the
following probability distribution:
X |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
P(X) |
0.11 |
0.23 |
0.34 |
0.16 |
0.10 |
0.06 |
0.04 |
0.01 |
Solution:
X |
P(X) |
XּP(X) |
X^{2}ּP(X) |
0 |
0.11 |
0 |
0 |
1 |
0.23 |
0.23 |
0.23 |
2 |
0.34 |
0.68 |
1.36 |
3 |
0.16 |
0.48 |
1.44 |
4 |
0.10 |
0.40 |
1.60 |
5 |
0.06 |
0.30 |
1.50 |
6 |
0.04 |
0.24 |
1.44 |
7 |
0.01 |
0.07 |
0.49 |
Total |
1 |
2.4 |
8.06 |
μ = E(X) = Σ X · P(X) = 2.4
σ^{2}
= V(X) = Σ X^{2} · P(X) – [E(X)]^{2} = 8.06 –
(2.4)^{2} = 2.3
Binomial
Probability Distribution:
(i) If an experiment contains only two possible outcomes, i.e., success or failure.
(ii) The probability of ‘success’ is denoted by ‘p’ and the probability of ‘failure’ is denoted by ‘q’ where q = 1 – p or p + q = 1.
(iii)
Such an experiment is repeated n times independently.
In independent repetitions, the probability p remains constant.
Where x
= 1, 2, 3, 4, ………, n
The above
formula is ‘Binomial Probability Distribution’.
The two constant quantities p and n are called the
parameters of a Binomial Distribution. The
quantity q is not a separate parameter because q = 1 – p.
Mean and
Variance of a Binomial Distribution:
The mean and variance of a binomial
distribution are directly evaluated in terms of its parameters p and n.
Example:
A coin is tossed 3 times.
‘Number of heads’ in 3 tosses is the random variable X.
Calculate probabilities of all possible values of X.
Also calculate mean and variance.
Solution:
Experiment: A coin is tossed for 3 times.
Success: Head
p = P(success) = P(head) = ½
n = number of times the coin is tossed = 3
x = 0,
1, 2, 3.
Now applying the Binomial Formula:
P(x=0) = =
P(x=1) = =
P(x=2) = =
P(x=3) =
=
Mean and Variance:
X |
P(X) |
X.P(X) |
X^{2}.P(X) |
0 |
0.125 |
0 |
0 |
1 |
0.375 |
0.375 |
0.375 |
2 |
0.375 |
0.75 |
1.5 |
3 |
0.125 |
0.375 |
1.125 |
Total |
1 |
1.5 |
3 |
Hyper
Geometric Probability Distribution:
(i) There are N items of which K are of first kind and the remaining (N – K) are of second kind,
(ii)
A sample of n items is randomly drawn without replacement from the
N items.
Number of items
of first kind in the sample is the random variable X:
Possible values of X are 0, 1, 2, ………, k when n ≥ K and
0,
1, 2, ……….., n when n < K
Where x = 0, 1, 2, 3, …….., k when n ≥ k
And x =
0, 1, 2, 3, ………, n when n< k
The above formula is called ‘Hyper Geometric Probability Distribution’. A schematic explanation of this formula may be given as:
Example:
A committee of 3 persons is to be formed
from among 3 men and 2 women. If
the selection of the committee members is random, construct the probability
distribution of the random variable ‘Number of women in the committee’.
Solution:
Where x = 0, 1, 2, 3, ……, k; when n ≥ k
And
x = 0, 1, 2, 3, …….., n; when n < k
The Hyper Geometric Probability
Distribution of RV ‘No. of Women in the Committee’ is as follows:
X |
P(X) |
0 |
0.1 |
1 |
0.6 |
2 |
0.3 |
Total |
1 |
Poisson
Probability Distribution:
Where x
= 0, 1, 2, 3, ……..
(i) The number of events per unit of time or space remains stable for a long period of time. This is the parameter of the distribution denoted by λ.
(ii)
The number of events in one time period is independent of the number of
events in another time period.
Example:
In an industry, the average number of
damaged output units per week is 10. What
is the probability that there will be (i) no damaged unit in the next week, (ii)
5 damaged units in the next week, and (iii) 15 damaged units in the next week.
Solution:
(i)
no damaged unit in the week:
X = number of damaged output units next week = 0
λ
= average number of damaged units per week = 10
(ii)
5 damaged units in the next week:
X = number of damaged output units next week = 5
λ
= average number of damaged units per week = 10
(iii)
15 damaged units in the next week:
X = number of damaged output units next week = 15
λ
= average number of damaged units per week = 10
Poisson
Approximation to Binomial Distribution:
The computations involved in the
binomial distributions become quite tedious when n is large.
In such cases the binomial distribution can be approximated to a Poisson
distribution with λ = n ּ p under the following
conditions:
(i) n is very large,
(ii) p is very small, and
(iii)
n ּ p
is finite.
A frequently used rule of thumb is that
the approximation is appropriate when p ≤ 0.05 and n ≥
20. However, the Poisson
distribution sometimes provides close approximations even in cases where n
is not large nor p is very small.
Example:
In a village, the local government
approximated that 2% of the population are infected with seasonal flu due to
absence of proper medication. What
is the probability that the number of infected persons in a random sample of 50
will be 4?
Solution:
Using binomial distribution with:
n = 50, p = 0.02 and x = 4
Using Poisson approximation to the
binomial with:
λ
= n ּ p = 50 × 0.02 = 1
The Poisson probability is close to the binomial probability.
Mean and
Variance of Poisson Distribution:
The mean of a Poisson Random Variable is
the parameter of the Poisson distribution λ, that is:
E(X)
= λ
The variance is also the parameter λ:
V(X)
= λ
Thus mean and variance of Poisson
distribution are equal to λ.
Continuous
Probability Distribution (In Detail):
(i) the function is non-negative for all possible values of the random variable, and
(ii)
the total area under the curve of the function is one.
This function
is called ‘probability density function’ and its curve a ‘probability
curve’.
Normal
Probability Distribution:
Where – ∞
≤ x ≤ ∞
Area
under Normal Curve:
(i) Determine the z-values for each limit of interval,
(ii) From the normal area table, determine the area for each z-value,
(iii)
Subtract the smaller area from the larger one.
Where μ
and σ are the mean and standard deviation of the random variable
z.
x
= μ + σ · z
(i) Area to the left of z = 0 is 0.5000
(ii) Area to the left of z = is 0
(iii)
Area to the left of z =
is 1.000
Example:
A normal random variable x has mean µ
= 24 and standard deviation σ = 1.8.
Determine z values for x = 14, 15.9, 29.2 and 33.
Also show these values on normal curve.
Solution:
For x = 14;
For x = 15.9;
For x = 29.2;
For x = 33;
Example:
A normal random variable x has
mean μ = 36 and standard deviation 2.05, determine the values of x
for z = – 3.36, – 1.8, 0.95 and 2.75.
Solution:
x = μ + σ · z
For z = – 3.36; x = 36 +
2.05 × (– 3.36) = 29.112 ≈ 29.11
For z = – 1.8; x = 36 +
2.05 × (– 1.8) = 32.31
For z = 0.95; x = 36 +
2.05 × 0.95 = 37.9475 ≈ 37.95
For z = 2.75; x = 36 +
2.05 × 2.75 = 41.6375 ≈ 41.64
Example:
The mean and SD of a normal random
variable are 34.5 and 5.8 respectively. Find
the following areas:
(i) to the left of 19.5
(ii) to the right of 40
(iii)
between 19.5 and 40
Solution:
(i)
to the left of 19.5, i.e., P(x
≤ 19.5):
P(– ∞ ≤ x ≤
19.5) = P(– ∞ ≤ z ≤ –2.59) = 0.0048
Where
(ii)
to the right of 40, i.e., P (x
≥ 40):
P(40 ≤ x ≤ ∞) =
P(0.95 ≤ z ≤ ∞) = 0.8289
Where
(ii)
between 19.5 and 40, i.e., P(19.5 ≤ x
≤ 40):
P(19.5 ≤ x ≤ 40) = P(– 2.59 ≤ z ≤ 0.95) = 0.8289 – 0.0048 = 0.8241
Continuity
Correction:
Normal
Approximation to Binomial Distribution:
A Binomial Distribution with large n
and moderate p can be approximated to a Normal Distribution with mean μ
= nּp and
:
μ
= nּp
Example:
A pair of dice is rolled for 800 times.
What is the probability that a total of 6 occur:
(i) at least 100 times, and
(ii)
between 150 to 300 times.
Solution:
n = 800
p = E = {15, 24, 33, 42, 51}
q =
(i)
Probability of at least 100 times, i.e.,
P(100 ≤ x ≤ 800) or P(99.5 ≤ x ≤ 800.5):
P(–1.19 ≤ z ≤
70.49)
From ‘Normal Area Table’ the Normal Area
corresponding to – 1.19 is 0.1170
= 1 – 0.1170 = 0.8830
(ii)
Probability of between 150 and 300 times,
i.e., P(130 ≤ x ≤ 300) or P(149.5 ≤ x ≤
300.5):
P(1.88 ≤ z ≤ 19.37)
From ‘Normal Area Table’ the Normal Area
corresponding to 1.88 is 0.9699
= 1 – 0.9699 = 0.0301