A variable that can assume any possible value between two points is called

Random Numbers:

1. In our every day life, we base many of our decisions on random outcomes, i.e., change occurrence.  For e.g., captains of two cricket teams toss a coin to decide as to which team will play first, or lotteries are drawn by spinning wheal, etc.
2. Random numbers are the numbers obtained by some random process (manually or mechanically).
3. These numbers are assumed to be randomly and uniformly (equally) distributed.  The basic random numbers are the 10 one-digit numbers, i.e., 0, 1, 2, ………. 9.  Each of these numbers has an equal change 1/10 of being selected.
4. Random numbers can be generated manually as well as mechanically.  Random numbers can be generated manually by drawing cards from playing cards or rotating spinning wheel, etc.  Mechanically generated random numbers are from calculators and computers.
5. The most common use of random numbers is for selection of samples.

Random Variables:

1. Experiments in which outcomes vary from trial to trial are called ‘Random Experiments’.
2. A variable whose values are determined by the outcomes of a random experiment is called a random variable.
3. In other words, random variable is a rule which assigns numbers to the outcomes of the possibility space and is denoted by X.
4. For example, throwing of a die is a random experiment and its outcomes, i.e., the occurrence of 1, 2, 3, 3, 4, 5 and 6 is a random variable.
5. A random variable is also called a ‘chance variable’, ‘stochastic variable’ or simply a ‘variable’.  Capital letters of X or Y are used to denote a variable and lower case letters x or y are used to denote its values.
6. Many random variables may be defined for one and the same possibility space.
7. When any characteristics of the individuals of a population (or a sample) are measured or counted, the characteristic itself is a random variable.
8. The random variables are further bifurcated into:

(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 1st 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:

1. An arrangement of all possible values of a random variable along with their respective probabilities is called a ‘probability distribution’ or a ‘probability function’.
2. Probability distribution can be further bifurcated into:

(a)    Discrete Probability Distribution, and

(b)   Continuous Probability Distribution.

(a)   Discrete Probability Distribution: Let a discrete random variable X assume values x1, x2, x3, ……….., xn with respective probabilities P(x1), P(x2), P(x3), …………, P(xn).  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(xi) ≤ 1

                            (ii)            ∑P(xi) = 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}

Example:

Determine whether the function

Solution:

(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 σ2 = V(X) = Σ X2 · 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: 

Example:

Calculate the mean and variance for the following probability distribution:

Solution:

μ = E(X) = Σ X · P(X) = 2.4

σ2 = V(X) = Σ X2 · P(X) – [E(X)]2 = 8.06 – (2.4)2 = 2.3

Binomial Probability Distribution:

1. Binomial probability is a mathematical formula to determine probabilities of the discrete values of a random variable called ‘Binomial Random Variable’.
2. The following are the conditions of Binomial Probability:

(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.

1. The number of success in n experiments is the Binomial Random Variable and is denoted by X.  The possible values of X are 0, 1, 2, 3, 4, ….., n.  The probabilities of the values of X are calculated by the following formula:

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:

Mean and Variance:

Hyper Geometric Probability Distribution:

1. It is a formula to determine the probabilities of the values for a random variable called ‘Hyper Geometric Random Variable’.
2. Following are the conditions of hyper geometric random variable:

(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

1. The probabilities of these values are calculated by the formula:

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:

The Hyper Geometric Probability Distribution of RV ‘No. of Women in the Committee’ is as follows:

Poisson Probability Distribution:

1. A random variable created by counting the number of items or events in a unit of either time or space is called a ‘Poisson Random Variable’.
2. Examples of Poisson random variable are the number of accidents per day on a highway, number of cars arriving at petrol pump in a five minute period of time, number of typing mistakes per page and number of defects in a painted surface, etc.
3. A Poisson probability distribution formula assigns probabilities to the values of the ‘Poisson Random Variable’:

Where x = 0, 1, 2, 3, ……..

1. Where λ (lambda) is the only parameter of the distribution and e is the mathematical constant 2.71828………..:

(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):

1. The concept of probability for continuous random variable is somewhat different with that of a discrete random variable.
2. The function or the formula of continuous probability distribution is generated and its curve is drawn on a graph paper such that:

(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’.

1. The probability of an interval from a to b is defined as the area under the probability curve between the two vertical lines erected on the x-axis at the points aand b.
2. The probability of an individual value under the continuous probability distribution is considered zero.
3. Probabilities of continuous random variable are represented by areas under the probability curve.

Normal Probability Distribution:

1. The most important and widely used probability density function is the ‘Normal Distribution’ where probability curve is a bell shaped symmetrical curve:

1. The most mathematical form of Normal Probability Density Function is:

Where – ∞ ≤ x ≤ ∞

1. A normal probability distribution or its probability curve characterised by two quantities μ and σ called the parameters of the distribution.
2. Two normal curves with different means μ and equal standard deviations σ are as below:

1. The normal curves with different standard deviations σ and equal means μ:

1. Two normal curves with different means μ and different standard deviations σ:

Area under Normal Curve:

1. The area between two limits of an interval under a normal probability curve cannot be determined analytically.
2. Tables of areas evaluated numerically could have been constructed but it would be impossible for an infinite number of normal curves for all values of μ and σ.
3. This problem is overcome by ‘Standard Normal Probability Distribution’ whose mean is zero (μ = 0) and standard deviation is one (σ = 1).  The standard normal variable is denoted by ‘x’:

1. The table of areas under the standard normal curve is used to find area under normal probability curve:
2. Following steps are involved in determining the area or probability of a particular interval of a normal distribution with μ and σ:

(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.

1. Precisely, a value of random variable ‘x’ can be converted to value ‘z’ by:

Where μ and σ are the mean and standard deviation of the random variable z.

1. Conversely, the z-value can be converted into random variable x by:

x = μ + σ · z

1. ‘z’ is the number of standard deviations from or to the mean.  All intervals containing the same number of standard deviations from mean will contain the same area under the curve for any normal distribution.
2. ‘Normal Area Table’ gives an idea under the curve to the left of a z-value.  For example, for z = 1.51, the Area under Normal Curve (as shown in the Table) is 0.9345; for z = – 2.69, the Area under Normal Curve (from the Table) is 0.0036.
3. Some of the rules should be remembered:

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:

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:

1. A population with unknown mean and standard deviation can be assumed a normal population of the frequency distribution of a sample is symmetrical.  The sample mean and sample standard deviation are used as estimates of population mean and population standard deviation respectively.
2. Observations or data are always discrete, recorded up to a certain degree of accuracy irrespective of whether the variable itself is discrete or continuous.
3. When the symmetrical distribution of any data is assumed to be normal, a continuity correction is applied to the observed values to make the data continuous.
4. If the data are recorded in whole numbers, data values are considered as mid-points of the intervals x ± 0.5, if the data are recorded up to one decimal place, data values are considered as mid points of the intervals x ± 0.05 and so on.  It should be cleared that the 0.5 and 0.05 should be subtracted from lower limit and added to upper limit or at most limit.

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:

(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