Theoretical Distributions Chapter Notes | Quantitative Aptitude for CA Foundation PDF Download

Chapter Overview

Introduction

• In chapter thirteen, we talked about frequency distribution. Similarly, we can think about a probability distribution. Just like how we distribute total frequency across different class intervals, we distribute total probability (which is one) across various points for a discrete random variable or across class intervals for a continuous random variable.
• This type of distribution is called a Theoretical Probability Distribution because it exists in theory.
• Studying theoretical probability distribution is important for several reasons:
  • Observed frequency distribution: Often, an observed frequency distribution can be seen as a sample, meaning it represents part of a larger, unknown population. By fitting a theoretical probability distribution to, for example, the frequency distribution of lamps produced by a manufacturer, the manufacturer can estimate the average lifespan of the lamps with good accuracy.
  • In scientific studies, like assessing missile effectiveness, theoretical probability distribution can help estimate how many missiles are needed to achieve a specific military goal.
  • Understanding the distribution of smokers can help social activists inform local communities about the dangers of both active and passive smoking.
• Making projections: Theoretical probability distributions can be used to make short-term predictions about the future.
• Statistical analysis: Analysis in statistics relies on theoretical probability distribution. For example, setting confidence limits or testing statistical hypotheses about population parameters is based on the distribution of the population being analyzed.
• A probability distribution has all the same characteristics as an observed distribution. We define key measures like mean (μ), median, mode (m), and standard deviation (σ) in the same way as before.
• Probability distributions can be either discrete or continuous depending on the type of random variable being examined.
• Two important examples of discrete probability distributions are:
  • Binomial Distribution
  • Poisson Distribution

 Binomial Distribution 

 The Binomial Distribution is a crucial and widely used discrete probability distribution. It originates from a specific type of random experiment called the Bernoulli process, named after the renowned mathematician Jacob Bernoulli. In this context, a 'trial' refers to an attempt to achieve a particular outcome that is neither guaranteed nor impossible. The key characteristics of Bernoulli trials are outlined below: 

• Two Outcomes: Each trial has two mutually exclusive and exhaustive outcomes. One outcome is termed a 'success,' while the other is a 'failure.' For instance, in a coin toss, getting heads is considered a success, and getting tails is a failure. 
• Independence: The trials are independent of each other. 
• Constant Probability: The probability of success (denoted by p) and the probability of failure (denoted by q = 1 - p) remain constant throughout the trials. 
• Finite Trials: The number of trials is a finite positive integer. 

 A discrete random variable x is said to follow a binomial distribution with parameters n and p, denoted as x ~ B(n, p), if its probability mass function is given by: 

f(x) = P(X = x) = (n choose x) p^x q^(n-x) for x = 0, 1, 2, ..., n, and f(x) = 0 otherwise. 

• Non-negativity: Since n > 0 and p, q ≥ 0, it follows that f(x) ≥ 0 for every x. 
• Normalization: The sum of probabilities over all possible values of x equals 1, i.e., ∑ f(x) = 1. 
• Biparametric Distribution: Binomial distribution is characterized by two parameters, n and p. If the values of n and p are known, the distribution is fully defined. 
• Mean: The mean of the binomial distribution is given by μ = np. 
• Mode: The mode of the binomial distribution, depending on the values of n and p, can be unimodal or bimodal. It is denoted by μ₀ and is calculated as: 
• Variance: The variance of the binomial distribution is given by σ² = npq. Since p and q are less than or equal to 1, npq < np, meaning the variance is always less than the mean. The maximum variance occurs at p = q = 0.5, with a maximum value of n/4. 
• Additive Property: If X and Y are two independent binomial variables such that X ~ B(n₁, p) and Y ~ B(n₂, p), then (X + Y) ~ B(n₁ + n₂, p). 

Applications of Binomial Distribution:

• Coin Tossing Experiments: Used to analyze the outcomes of repeated coin tosses, where each toss has two possible outcomes: heads (success) or tails (failure). 
• Sampling Inspection Plans: Applied in quality control processes where items are inspected for defects. Each item can either be defective (success) or non-defective (failure). 
• Genetic Experiments: Used in studies involving inheritance patterns, where traits are passed down with specific probabilities. Each offspring can inherit a trait (success) or not (failure). 

Example 16.1: A coin is tossed 10 times. Assuming the coin to be unbiased, what is the probability of getting
(i) 4 heads?
(ii) at least 4 heads?
(iii) at most 3 heads?

Solution: We apply binomial distribution as the tossing are independent of each other. With every tossing, there are just two outcomes either a head, which we call a success or a tail, which we call a failure and the probability of a success (or failure) remains constant throughout.

Let X denotes the no. of heads. Then X follows binomial distribution with parameter n = 8 and p = 1/2 (since the coin is unbiased). Hence q = 1 – p = 1/2

The probability mass function of X is given by

(i) probability of getting 4 heads
= f (4)
= 10c₄ / 1024
= 210 / 1024
= 105 / 512

(ii) probability of getting at least 4 heads

= 210 + 252 + 210 + 120 + 45 + 10 + 1 / 1024

= 848 / 1024

(iii ) probability of getting at most 3 heads

= 176 / 1024 
= 11/64

Example 16.2: If 15 dates are selected at random, what is the probability of getting two Sundays?

Solution: If X denotes the number at Sundays, then it is obvious that X follows binomial distribution with parameter n = 15 and p = probability of a Sunday in a week = 1/7 and q = 1 – p = 6 / 7.

Then f(x) = ¹⁵c_x (1/7)^x . (6/7)^15–x .
for x = 0, 1, 2,……….. 15.
Hence the probability of getting two Sundays

Example 16.3: The incidence of occupational disease in an industry is such that the workmen have a 10% chance of suffering from it. What is the probability that out of 5 workmen, 3 or more will contract the disease?

Solution: Let X denote the number of workmen in the sample. X follows binomial with parameters n = 5 and p = probability that a workman suffers from the occupational disease = 0.1

Hence q = 1 – 0.1 = 0.9.

Thus f (x) = ⁵c_x . (0.1)^x . (0.9)^5-x

For x = 0, 1, 2,…….,5.

The probability that 3 or more workmen will contract the disease

Example 16.4: Find the probability of a success for the binomial distribution satisfying the following relation 4 P (x = 4) = P (x = 2) and having the parameter n as six.

Solution: We are given that n = 6. The probability mass function of x is given by

Example 16.5: Find the binomial distribution for which mean and standard deviation are 6 and 2 respectively.

Solution: Let x ~ B (n, p)
Given that mean of x = np = 6 … ( 1 )
and SD of x = 2
variance of x = npq = 4 ….. ( 2 )
2 Dividing ( 2 ) by ( 1 ), we get q = 2/3
Hence p = 1 – q = 1 / 3
Replacing p by 1 / 3 in equation ( 1 ), we get n × 1/3 = 6 
n = 18
Thus the probability mass function of x is given by

Example 16.6: Fit a binomial distribution to the following data:

Solution: In order to fit a theoretical probability distribution to an observed frequency distribution it is necessary to estimate the parameters of the probability distribution. There are several methods of estimating population parameters. One rather, convenient method is ‘Method of Moments’. This comprises equating p moments of a probability distribution to p moments of the observed frequency distribution, where p is the number of parameters to be estimated. Since n = 5 is given, we need estimate only one parameter p. We equate the first moment about origin i.e. AM of the probability distribution to the AM of the given distribution and estimate p.

For x = 0, 1, 2, 3, 4, 5.

A look at Table 16.1 suggests that the fitting of binomial distribution to the given frequency distribution is satisfactory.

Example 16.7: 6 coins are tossed 512 times. Find the expected frequencies of heads. Also, compute the mean and SD of the number of heads.

Solution: If x denotes the number of heads, then x follows binomial distribution with parameters n = 6 and p = prob. of a head = ½, assuming the coins to be unbiased. The probability mass function of x is given by

for x = 0, 1, …..6.

The expected frequencies are given by Nf ( x ).

Thus mean

Applying formula for mean and SD, we get

μ = np = 6 x 1/2 = 3

Example 16.8: An experiment succeeds thrice as after it fails. If the experiment is repeated 5 times, what is the probability of having no success at all ?

Solution: Denoting the probability of a success and failure by p and q respectively, we have,

p = 3q
p = 3 (1 – p)
p = 3/4
q = 1 – p = 1/4 when n = 5 and p = 3/4, we have

for n = 0, 1, .......... , 5.
So probability of having no success
= f ( 0 )
= ⁵c₀ (3/4)⁰ (1/4 )^5–0
= 1/1024

Example 16.9: What is the mode of the distribution for which mean and SD are 10 and 5 respectively.

Solution: As given np = 10 .......... (1)

npq = 5 ...................... (2)

on solving (1) and (2), we get n = 20 and p = 1/2
Hence mode = Largest integer contained in (n+1)p
= Largest integer contained in (20+1) × 1/2
= Largest integer contained in 10.50
= 10.

Example 16.10: If x and y are 2 independent binomial variables with parameters 6 and 1/2 and 4 and 1/2 respectively, what is P ( x + y ≥ 1 )?

Solution: Let z = x + y.
It follows that z also follows binomial distribution with parameters
( 6 + 4 ) and 1/2
i.e. 10 and 1/2
Hence P ( z ≥ 1 )
= 1 – P ( z < 1 )
= 1 – P ( z = 0 )
= 1 – ¹⁰c₀ (1/2 )⁰ . (1/2 )^10–0
= 1 – 1 / 2¹⁰
= 1023 / 1024

Poisson distribution

Poisson distribution is a theoretical discrete probability distribution which can describe many processes. Simon Denis Poisson of France introduced this distribution way back in the year 1837.
Poisson Model
Let us think of a random experiment under the following conditions:
I. The probability of finding success in a very small time interval ( t, t + dt ) is kt, where k (>0) is a constant.
II. The probability of having more than one success in this time interval is very low.
III. The probability of having success in this time interval is independent of t as well as earlier successes.
The above model is known as Poisson Model. The probability of getting x successes in a relatively long time interval T containing m small time intervals t i.e. T = mt. is given by

for x = 0, 1, 2, ......… ∞ …… (16.7)
Taking kT = m, the above form is reduced to

for x = 0, 1, 2, ......∞ …..... (16.8)

Definition of Poisson Distribution

A random variable X is defined to follow Poisson distribution with parameter , to be denoted by X ~ P (m) if the probability mass function of x is given by

Here e is a transcendental quantity with an approximate value as 2.71828.

Application of Poisson distribution

Poisson distribution is applied when the total number of events is pretty large but the probability of occurrence is very small. Thus we can apply Poisson distribution, rather profitably, for the following cases:
a) The distribution of the no. of printing mistakes per page of a large book.
b)The distribution of the no. of road accidents on a busy road per minute.
c) The distribution of the no. of radio-active elements per minute in a fusion process.
d) The distribution of the no. of demands per minute for health centre and so on.

Example 16.11: Find the mean and standard deviation of x where x is a Poisson variate satisfying the condition P (x = 2) = P ( x = 3).

Solution: Let x be a Poisson variate with parameter m. The probability max function of x is then given by

now, P (x = 2) = P (x = 3)

Thus the mean of this distribution is m = 3 and standard deviation = √3 ≅ 1.73

Example 16.12: The probability that a random variable x following Poisson distribution would assume a positive value is (1 – e^–2.7). What is the mode of the distribution?

Solution: If x ~ P (m), then its probability mass function is given by

The probability that x assumes a positive value
= P (x > 0)
= 1– P (x ≤ 0)
= 1 – P (x = 0)
= 1 – f(0)
= 1 – e^–m
As given,

1 – e^–m = 1 – e^–2.7
e^–m = e^–2.7
m = 2.7
Thus μ₀ = largest integer contained in 2.7

= 2

Example 16.13: The standard deviation of a Poisson variate is 1.732. What is the probability that the variate lies between –2.3 to 3.68?

Solution: Let x be a Poisson variate with parameter m.

Then SD of x is √m .

As given √m = 1.732

⇒ m = (1.732)² ≅ 3.

The probability that x lies between –2.3 and 3.68

≅ 0.65

Example 16.14: X is a Poisson variate satisfying the following relation:
P (X = 2) = 9P (X = 4) + 90P (X = 6).
What is the standard deviation of X?

Solution: Let X be a Poisson variate with parameter m. Then the probability mass function of X is

Thus the standard deviation of X is 1 = 1

Example 16.15: Between 9 and 10 AM, the average number of phone calls per minute coming into the switchboard of a company is 4. Find the probability that during one particular minute, there will be,
1. no phone calls
2. at most 3 phone calls (given e–4 = 0.018316)

Solution: Let X be the number of phone calls per minute coming into the switchboard of the company. We assume that X follows Poisson distribution with parameters m = average number of phone calls per minute = 4.

1. The probability that there will be no phone call during a particular minute

= P (X = 0)

= e^{– 4}

= 0.018316

2. The probability that there will be at most 3 phone calls

Example 16.16: If 2 per cent of electric bulbs manufactured by a company are known to be defectives, what is the probability that a sample of 150 electric bulbs taken from the production process of that company would contain
1. exactly one defective bulb?
2. more than 2 defective bulbs?

Solution: Let x be the number of bulbs produced by the company. Since the bulbs could be either defective or non-defective and the probability of bulb being defective remains the same, it follows that x is a binomial variate with parameters n = 150 and p = probability of a bulb being defective = 0.02. However since n is large and p is very small, we can approximate this binomial distribution with Poisson distribution with parameter m = np = 150 x 0.02 = 3.

1. The probability that exactly one bulb would be defective

2. The probability that there would be more than 2 defective bulbs

=

Example 16.17: The manufacturer of a certain electronic component is certain that two per cent of his product is defective. He sells the components in boxes of 120 and guarantees that not more than two per cent in any box will be defective. Find the probability that a box, selected at random, would fail to meet the guarantee? Given that e^–2.40 = 0.0907.

Solution: Let x denote the number of electric components. Then x follows binomial distribution with n = 120 and p = probability of a component being defective = 0.02. As before since n is quite large and p is rather small, we approximate the binomial distribution with parameters n and p by a Poisson distribution with parameter m = n.p = 120 × 0.02 = 2.40. Probability that a box, selected at random, would fail to meet the specification = probability that a sample of 120 items would contain more than 2.40 defective items.

Example 16.18: A discrete random variable x follows Poisson distribution. Find the values of
(i) P (X = at least 1)
(ii) P (X ≤ 2/ X ≥ 1)
You are given E (x) = 2.20 and e–2.20 = 0.1108.

Solution: Since X follows Poisson distribution, its probability mass function is given by

Fitting a Poisson distribution

As explained earlier, we can apply the method of moments to fit a Poisson distribution to an observed frequency distribution. Since Poisson distribution is uniparametric, we equate m, the parameter of Poisson distribution, to the arithmetic mean of the observed distribution and get the estimate of m.

i.e.

The fitted Poisson distribution is then given by

Example 16.19: Fit a Poisson distribution to the following data :

Solution: The mean of the observed frequency distribution is

Normal OR Gaussian Distribution

The two distributions discussed so far, namely binomial and Poisson, are applicable when the random variable is discrete. In case of a continuous random variable like height or weight, it is impossible to distribute the total probability among different mass points because between any two unequal values, there remains an infinite number of values. Thus a continuous random variable is defined in term of its probability density function f (x), provided, of course, such a function really exists, f (x) satisfies the following condition:

The most important and universally accepted continuous probability distribution is known as normal distribution. Though many mathematicians like De-Moivre, Laplace etc. contributed towards the development of normal distribution, Karl Gauss was instrumental for deriving normal distribution and as such normal distribution is also referred to as Gaussian Distribution.
A continuous random variable x is defined to follow normal distribution with parameters μ and σ² , to be denoted by

If the probability density function of the random variable x is given by

where μ and σ are constants, and σ > 0
Some important points relating to normal distribution are listed below:
(a) The name Normal Distribution has its origin some two hundred years back as the then mathematician were in search for a normal model that can describe the probability distribution of most of the continuous random variables.
(b) If we plot the probability function y = f (x), then the curve, known as probability curve, takes the following shape:

A quick look at figure 16.1 reveals that the normal curve is bell shaped and has one peak, which implies that the normal distribution has one unique mode. The line drawn through x =  has divided the normal curve into two parts which are equal in all respect. Such a curve is known as symmetrical curve and the corresponding distribution is known as symmetrical distribution. Thus, we find that the normal distribution is symmetrical about x = μ. It may also be noted that the binomial distribution is also symmetrical about p = 0.5. We next note that the two tails of the normal curve extend indefinitely on both sides of the curve and both the left and right tails never touch the horizontal axis. The total area of the normal curve or for that any probability curve is taken to be unity i.e. one. Since the vertical line drawn through x = μ divides the curve into two equal halves, it automatically follows that,
The area between – ∞ to μ = the area between μ to ∞ = 0.5

When the mean is zero, we have
the area between – ∞ to 0 = the area between 0 to ∞ = 0.5

(c) If we take μ = 0 and σ = 1 in (18.17), we have

The random variable z is known as standard normal variate (or variable) or standard normal deviate. The probability that a standard normal variate X would take a value less than or equal to a particular value say X = x is given by

Φ (x) is known as the cumulative distribution function.
We also have Φ (0) = P ( X ≤ 0 ) = Area of the standard normal curve between – ∞ and 0 = 0.5 …….. (16.20)

(d) The normal distribution is known as biparametric distribution as it is characterised by two parameters μ and σ² . Once the two parameters are known, the normal distribution is completely specified.

Properties of Normal Distribution

1. Since π = 22/7 , e^–θ = 1 / e^θ > 0, whatever θ may be,

it follows that f (x) ≥ 0 for every x.

It can be shown that

2. The mean of the normal distribution is given by μ. Further, since the distribution is symmetrical about x = μ, it follows that the mean, median and mode of a normal distribution coincide, all being equal to μ.

3. The standard deviation of the normal distribution is given by σ

Mean deviation of normal distribution is

Applications of Normal Distribution

• The use of normal distribution is not limited to just statistics; it is relevant in many fields such as science, social sciences, management, and commerce.
• Many continuous variables, like height, weight, wages, and profits, tend to follow a normal distribution pattern.
• If a variable does not follow normal distribution, often a simple transformation of that variable can result in a normal distribution.
• In the case of a binomial distribution, when the number of trials (n) is large and the probability of success (p) is moderate (not too high or too low), it also resembles a normal distribution.
• The Poisson distribution approaches a normal distribution when the value of m is large.
• Transformations to achieve normality are important because it is easier to calculate probabilities when assuming a normal distribution.
• Not only discrete random variables, but also the probability distributions of t, chi-square, and F can approximate a normal distribution under certain conditions.
• To make inferences about an unknown population, we use sampling, which relies on the assumption of normality to draw conclusions about the population.
• Additionally, the distributions of many sample statistics tend to approach a normal distribution when the sample size is large.