SRI RAMAKRISHNA INSTITUTE OF TECHNOLOGY
(AN AUTONOMOUS INSTITUTION) COIMBATORE- 641010
UICM006 & PROBABILITY AND STATISTICS
Random Variables and Distributions History"A gambler's dispute in 1654 led to the creation of a mathematical theory of probability by two famous French mathematicians, Blaise Pascal and Pierre de Fermat. Antoine Gombaud, Chevalier de Méré, a French nobleman with an interest in gaming and gambling questions, called Pascal's attention to an apparent contradiction concerning a popular dice game. The game consisted in throwing a pair of dice 24 times; the problem was to decide whether or not to bet even money on the occurrence of at least one "double six" during the 24 throws. A seemingly well-established gambling rule led de Méré to believe that betting on a double six in 24 throws would be profitable, but his own calculations indicated just the opposite.
This problem and others posed by de Méré led to an exchange of letters between Pascal and Fermat in which the fundamental principles of probability theory were formulated for the first time. Although a few special problems on games of chance had been solved by some Italian mathematicians in the 15th and 16th centuries, no general theory
was developed before this famous correspondence.
The Dutch scientist Christian Huygens, a teacher of Leibniz, learned of this correspondence and shortly thereafter (in 1657) published the first book on probability; entitled De Ratiociniis in Ludo Aleae, it was a treatise on problems associated with gambling. Because of the inherent appeal of games of chance, probability theory soon became popular, and the subject developed rapidly during the 18th century. The major
contributors during this period were Jakob Bernoulli (1654-1705) and Abraham de Moivre (1667-1754).
In 1812 Pierre de Laplace (1749-1827) introduced a host of new ideas and mathematical techniques in his book, Théorie Analytique des Probabilités. Before Laplace, probability theory was solely concerned with developing a mathematical analysis of games
of chance. Laplace applied probabilistic ideas to many scientific and practical problems. The theory of errors, actuarial mathematics, and statistical mechanics are examples of some of the important applications of probability theory developed in the 19th century.
Like so many other branches of mathematics, the development of probability theory has been stimulated by the variety of its applications. Conversely, each advance in the theory has enlarged the scope of its influence. Mathematical statistics is one important branch of applied probability; other applications occur in such widely different fields as genetics, psychology, economics, and engineering. Many workers have contributed to the theory since Laplace's time; among the most important are Chebyshev, Markov, von Mises, and Kolmogorov.
One of the difficulties in developing a mathematical theory of probability has been to arrive at a definition of probability that is precise enough for use in mathematics, yet comprehensive enough to be applicable to a wide range of phenomena. The search for a widely acceptable definition took nearly three centuries and was marked by much controversy. The matter was finally resolved in the 20th century by treating probability
theory on an axiomatic basis. In 1933 a monograph by a Russian mathematician A. Kolmogorov outlined an axiomatic approach that forms the basis for the modern theory. Since then the ideas have been refined somewhat and probability theory is now part of a more general discipline known as measure theory."
Statistics
Statistics is about data (observations). In statistics we organize and analyze data. We come up with statements that somehow summarize what the data looks like all together, like "the average score on the test was 81." That statement doesn't reveal anything about anyone's individual test scores, but it still tells something about how the class did as a whole. The kid who got a 47 can run home to mom and dad and brag that his class averaged an 81, and they might be so proud they take him out for ice cream. His parents are apparently not statistics experts.
Probability
Probability is about planning experiments and comparing the number of possible outcomes to the number of outcomes you want. If we roll a die and want to roll either a 1 or a 6, there are 6 possible outcomes (any of the 6 faces of the die could land facing up), but only 2 that we actually want to happen (either the 1 or the 6 facing up). Of course, if you have money on it your odds decrease substantially, as you'll then have some anti-gambling karma working against you.
Basic Definitions Sample Space
The set of all possible outcomes of a random phenomenon
Event
Any set of outcomes of interest
Probability of an event
The relative frequency of this set of outcomes over an infinite number of trials ( ) is the probability of event A
Mutually exclusive events
If and are mutually exclusive if both cannot occur at the same time.
Independent events
If and independent events if and only if
( ) ( ) ( )
Mean, Median, Mode, and Range The Mean
To calculate the mean, simply add all of your numbers together. Next, divide the sum by however many numbers you added. The result is your mean or average score.
For example, let's say you have four test scores: 15, 18, 22, and 20. To find the average, you would first add all four scores together, and then divide the sum by four. The resulting mean is 18.75. Written out, it looks something like this:
(15 + 18 + 22 + 20) / 4 = 75 / 4 = 18.75
The Median
The median is the middle value in a data set. To calculate it, place all of your numbers in increasing order. If you have an odd number of integers, the next step is to find the middle number on your list. In this example, the middle or median number is 15:
3, 9, 15, 17, 44
If you have an even number of data points, calculating the median requires another step or two. First, find the two middle integers in your list. Add them together, and then divide by two. The result is the median number. In this example, the two middle numbers are 8 and 12:
3, 6, 8, 12, 17, 44
Written out, the calculation would look like this:
(8 + 12) / 2 = 20 / 2 = 10
In this instance, the median is 10.
The Mode
In statistics, the mode in a list of numbers refers to the integers that occur most frequently. Unlike the median and mean, the mode is about the frequency of occurrence. There can be more than one mode or no mode at all; it all depends on the data set itself. For example, let's say you have the following list of numbers:
3, 3, 8, 9, 15, 15, 15, 17, 17, 27, 40, 44, 44
In this case, the mode is 15 because it is the integer that appears most often. However, if there were one fewer 15 in your list, then you would have four modes: 3, 15, 17, and 44.
The Range
The "range" of a list a numbers is just the difference between the largest and smallest values.
Problem: 1
Find the mean, median, mode and range of the following numbers 10, 39, 71, 42, 39, 76, 38, 25.
Answer:
Mean
( )
Median
First we arrange the given data in ascending order 10, 25, 38, 39, 39, 42, 71, 76
Mode
Mode (most repeated values)
Range
Range
Estimating Mean, Variance and Standard deviation for frequency distribution
̅ ∑ ∑ ∑( ̅) ∑ √∑( ̅) ∑ ̅ Problem: 2
A computer repair service received the following number of calls per day over a period of 30 days.
6 5 6 9 7 4 2 4 7 8 3 4 9 8 2 3 5 9 7 8 9 7 5 6 7 7 4 6 2 4
Using a frequency table, find the average number of calls per day. Also find the standard deviation, variance and co efficient of variation.
Solution: Data value Tally Frequency ( ̅) 2 *** 3 6 42.64 3 ** 2 6 15.35 4 ***** 5 20 15.66 5 *** 3 15 1.78 6 **** 4 24 0.21 7 ****** 6 42 9.08 8 *** 3 24 14.92 9 **** 4 36 41.73 Total ∑ ∑ ∑ ̅ ∑ ∑ ∑( ̅) ∑ √∑( ̅) ∑ √ ̅ Problem: 3
Find the mean, variance, standard deviation and coefficient of variation from the following table.
Class 0-30 30-60 60-90 90-120 120-150 150-180 180-210 Frequency 2 3 5 10 3 5 2
Data value Mid value Frequency ( ̅) 0-30 15 2 30 16928 30-60 45 3 135 11532 60-90 75 5 375 5120 90-120 105 10 1050 40 120-150 135 3 405 2352 150-180 165 5 825 16820 180-210 195 2 390 15488 Total ∑ ∑ ∑ ̅ ∑ ∑ ∑( ̅) ∑ √ √ ̅ Problem: 4
Life of bulbs produced by two factories A and B are given below:
Length of life (in hours) Factory A (No. of bulbs) Factory B (No. of bulbs)
550-650 10 8 650-750 22 60 750-850 52 24 850-950 20 16 950-1050 16 12 Total 120 120
The bulbs of which factory are more consistent from the point of view of length of life? Answer:
For factory A
Data value Mid value Frequency ( ̅) 550-650 600 10 6000 78021.889 650-750 700 22 15400 2996.1558 750-850 800 52 41600 648449.8228 850-950 900 20 18000 896083.778 950-1050 1000 16 1600 5538115.022 Total ∑ ∑ ∑ ̅ ∑ ∑ ∑( ̅) ∑ √ √ ̅ For factory B
Data value Mid value Frequency ( ̅) 550-650 600 8 4800 51200 650-750 700 60 42000 24000 750-850 800 24 19200 345600 850-950 900 16 14400 774400 950-1050 1000 12 1200 4036800 Total ∑ ∑ ∑ ̅ ∑ ∑
∑( ̅) ∑ √ √ ̅ Since C.V. of factory B > C.V. of factory A
This shows Factory B has more variability which means bulbs of factory A are more consistent.
Axioms:
Probability of any event , ( ) is assigned in such a way that it satisfies certain conditions. These conditions for assigning probability are known as Axioms of probability. There are three such axioms. All conclusions drawn on probability theory are either directly or indirectly related to these three axioms.
Axiom: 1
For any event belongs to the sample space , the value of probability of the event lies between zero and one. Mathematically expressed as: ( ) . Thus, Axiom 1 states that probabilities of events for a particular sample space are real numbers on the interval , -.
Axiom: 2
Probability of all the events in a sample space or the sample space in total is equal to one. Mathematically denoted as ( ) .
Axiom: 3
For any two mutually exclusive events and , the probability of union of them is equal to simple sum of the probabilities of individual events. It is mathematically denoted as: ( ) ( ) ( ).
Conditional Probability
If and are two events in sample space and ( ) , then the probability of given than has already occurred is denoted as ( ⁄ ) and mathematically expressed as:
( * ( ) ( )
Definition: Random Variables
A real valued function defined on the outcome of a probability experiment is called random variables.
Experiment:
Consider an experiment of throwing a coin at only once. The outcomes are either Head or Tail. * + where and are random variable.
Types of random variable:
Discrete random variable
Continuous random variable
Discrete random variable (DRV)
A random variable whose set of possible values is either finite or countably infinite is called discrete random variable.
Probability mass function (PMF)
If is a discrete random variable then the function ( ) ( ), then the probability is called Probability mass function of .
The number ( ) should satisfy the following conditions
Probability of any outcome from any experiment must be positive or zero. ( )
Sum of Probability of all outcomes from any experiment must be equal to one. ∑ ( )
Cumulative distribution function (or) Distribution function
The cumulative distribution function ( ) of a random variable with probability distribution ( ) is defined as ( ) ( ) Important results: ( ) ( ) , - ( ) ( ) ( ) ( ) ( ) ( )
Mean or Average or Expectation of
( ) ∑ Variance of ( ) ( ) , ( )- ( ) ∑ Note: i. , - ii. , - iii. , - ( ) iv. ( ) v. ( ) ( ) ( ) ( ) , ( ) - Problem: 5
Construct the distribution function of the discrete random variable whose probability distribution is given below:
1 2 3 4 5 6 7 Total
( ) 0.10 0.15 0.25 0.20 0.15 0.10 0.05 1
1 2 3 4 5 6 7 Total ( ) 0.10 0.15 0.25 0.20 0.15 0.10 0.05 1
( ) ( ) ( ) ( ) ( ) ( ) ( ) The distribution function is
( ) ( ) 1 ( ) ( ) ( ) 2 ( ) ( ) ( ) ( ) 3 ( ) ( ) ( ) ( ) ( ) 4 ( ) ( ) ( ) ( ) 5 ( ) ( ) ( ) ( ) 6 ( ) ( ) ( ) ( ) 7 ( ) ( ) ( ) ( ) Another method: 1 2 3 4 5 6 7 ( ) 0.10 0.15 0.25 0.20 0.15 0.10 0.05 ( ): 0.10 0.25 0.50 0.70 0.85 0.95 1 Problem: 6
A random variable has the following probability distribution function as:
( )
i. Find the value of
ii. Find ( ) ( ) ( ) iii. Find the distribution function of .
Answer:
0 1 2 3 4 5 6 7 8
( ) a 3a 5a 7a 9a 11a 13a 15a 17a ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
I. We know that the total probability ∑ ( )
( ) ( ) ( ) 0 1 0 1 2 3 4 5 6 7 8 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) II. ( ) ( ) ( ) ( ) ( ) ( ) ( * ( ) ( ) ( ) ( ) ( ) The distribution function is
( ) ( ): Problem: 7
When a dice is thrown, denotes the number that turns up. Find ( ) ( ) ( ) and Variance of .
1 2 3 4 5 6 ( ) 1/6 1/6 1/6 1/6 1/6 1/6 Answer: ( ) ∑ ∑ ( * ( * ( * ( * ( * ( * ( ) ( ) ∑ ∑ ( * ( * ( * ( * ( * ( * ( ) ( ) ( ) ( ) , ( ) ( ) ( )- ( * ( ) ( ) ( ) , ( ( * ( ) Problem: 8
A random variable takes the values and such that ( ) ( ) ( ) ( ). Find the probability distribution of and cumulative distribution function of .
Answer:
( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 2 3 4 ( ) ( ) ( ) ( ) ( )
I. We know that the total probability ∑ ( ) ( ) ( ) ( ) ( ) The distribution function is
1 2 3 4 ( ) ( * ( * ( * ( ): Problem: 9
A random variable has the following probability distribution function as:
0 1 2 3 4 5 6 7
i. Find the value of ii. Find ( ) ( ) ( ) iii. Find , ⁄ - Answer: 0 1 2 3 4 5 6 7 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) We know that the total probability
∑ ( ) ( ) ( ) ( ) [ ] 0 1 2 3 4 5 6 7 ( ) I. ( ) ( ) , ( ) ( )- ( * ( ) II. ( ) ( )
( * ( ) III. , ⁄ - ( * ( ) ( ) , - , ⁄ - ( ) ( ) ( ) ( ) ( ) , - ( ) , ( ) ( )- , ( ) ( ) ( )- ( * , ⁄ - . / . / Problem: 10
The probability function of a random variable is defined as ( ) ( ) ( ) , where and ( ) if . Find
. ( ⁄ )
0 1 2 ( )
I. We know that the total probability ∑ ( ) ( ) ( ) ( ) 0 1 2 ( ) ( * ( * ( * ( *
II. Distribution function of
0 1 2 ( ) III. ( ) ( ) ( ) ( ) ( ) , ⁄ - ( ) ( )
( ) ( ) . / . /
Continuous random variable (CRV)
A random variable is said to be continuous, if it takes all possible values between certain intervals.
Probability density function (PDF)
For a continuous random variable , the probability density function is a function of such that
( ) ∫ ( )
( ) ∫ ( )
Cumulative distribution function (CDF)
( ) ( ) ∫ ( ) Mean ( ) ∫ ( ) Variance ( ) ( ) , ( )- * ( ) ∫ ( ) + Note: ( ) ( ) ( ) ( ) ( ) , ( ) - Problem: 11
( ) 2 Answer: The PDF of a C.R.V is . ∫ ( ) ∫ ( ) ( , - ) ( ( )) ( ) ∫ ( ) given ( ) is a PDF. Problem: 12 ( ) , ( ) ( ) Answer: The PDF of a C.R.V is . ∫ ( ) ∫ Limits: When When ∫ ( ) ∫
( ) [ ∫ ] , - ∫ ( )
( ) is a probability density function of .
To find CDF ( ): ( ) ( ) ∫ ( ) ∫ Limits: When ( ) ∫ ( ) ( ) Problem: 13 ( ) { Find (i). the value of (ii). Distribution function of (iii). ( ).
Answer:
I. To find the value of :
The PDF of a C.R.V is . ∫ ( ) ∫
, -
, ( ) ( )-
0 . /1 0 ( ) 1
II. To find the distribution function of :
( ) ( ) ∫ ( ) ∫ , - , ( ) ( )- 0 ( ) . /1 0 ( ) 1 ( ) 0 ( ) 1 III. To find ( ) ( ) ( ) ( ) 0 ( ) 1 0 1 ( ) Problem: 14
The distribution function of a random variable is given by ( ) ( ) Find the density function, mean and variance of .
Given ( ) ( ) ( ) ( ) , ( ) - ( ) ,( ) - , ( ) - ,( ) ( ) ( )- , ( ) - , - ( ) ( ) ∫ ( ) ∫ ∫ ∫ ( ) ( ) ( ) , ( ) ( ) ( )- [ ( ( ))] , - ( ) ( ) ∫ ( ) ∫
∫ ( ) ( ) ( ) ( ) , ( ) ( ) ( ) ( )- [ ( ( ))] , - ( ) ( ) ( ) , ( )- ( ) ( ) Problem: 15 ( ) { For what value of is ( ) a valid pdf? Also find cumulative distribution of the random variable with the above pdf.
Answer: I. Verification: The PDF of a C.R.V is . ∫ ( ) ∫ ( ) ( ) If , the given ( ) is a pdf.
II. To find the distribution function of : ( ) ( ) ∫ ( ) ∫ ( ) ( ) Problem: 16 ( ) { ( )
Find (i). the value of , (ii). ( ) (iii). What is , ⁄ - (iv). Find the distribution of ( ).
Answer:
I. To find the value :
The PDF of a C.R.V is . ∫ ( ) ∫ ∫ ( ) ( ) ( ) ( * [( * ( *] ( * ( *
II. To find the distribution function of : Case : 1 When
( ) ∫ ( )
( ) ∫ ( ) ∫ ( ) ∫ * + ( ) Case : 3 When ( ) ∫ ( ) ∫ ( ) ∫ ( ) ∫ ∫ ( ) ( ) ( ) *( ) ( *+ ( ) Case : 4 When ( ) ∫ ( ) III. ( ) ( ) ( ) ( ( ) ) ( ) , ⁄ - , , ( )
, ( ) ( ) ( ) ( ) [ ( ) ] . / Problem: 17 ( ) { what is ( ) and ( ) Answer:
I. To find the constant :
The PDF of a C.R.V is . ∫ ( ) ∫ ( ) [ ∫ ] ( ) , - ∫ ( ) II. To find ( ): ( ) ∫ ( ) ∫ ( )
( ) ( ) ( ) III. To find ( ): ( ) ∫ ( ) ∫ ( ) ( ) ( ) Problem: 18 ( ) [ ( * (⁄ *] Answer:
To find the constant :
The PDF of a C.R.V is . ∫ ( ) ∫ ( ) ( ( ) ) ( * [ ( * (⁄ *]
( * ( ) ( ) , - [ ( * (⁄ *] 0 . / . /1 0 . /1 [ ( * ( *] ∫ ( ) ∫ ( ) ( * ( * [ ( *] ∫ ( ) ∫ ( ) ( * ( ) [ ( * (⁄ *] . / . /
Problem: 19
( ) { ( ) (| | ) ( * Answer: ( ) ( ) { ( ) { ( ) ( ) { (| | ) ( ) ( ) ( ) , ( ) ( ) ( )- ( ) ( ) ( ) ( ) ( ) ( )
(| | ) ( * ( ) ( * ( ) ( ) ( * ( * ( * Mathematical expectations:
Let be a random variable with probability density function ( ). Then the mathematical expectation of is defined by ( ) and is given by
( ) ∫ ( ) ( ) ( ) ∑ ( ) ( ) Moment:
Let be a random variable with probability density function ( ). Then the rth
moment about the origin is given by ( ) ∫ ( ) ( ) ( ) ∑ ( ) ( )
Moment generating function
The Moment generating function of a random variable whose probability function ( ) is given by
( ) ∫ ( ) ( ) ( ) ∑ ( ) ( ) Problem: 20
Find the moment generating function and rth moment for the distribution whose
probability density function ( ) . Hence find the mean and variance.
Answer:
To find the constant : The PDF of a C.R.V is . ∫ ( ) ∫ ( * [ ∫ ] ( ) ( )
By definition of moment generating function ( ) ∫ ( ) ∫ ∫ ( ) , - * ( ) ( )+ * ( )+
[ ( )] , - ( ) ( )
To find Mean and Variance ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Mean Variance ( ) , ( )- Problem: 21
A random variable has the pdf ( ) {
. Obtain the MGF and find its first four moments about the origin. Also find the mean and variance.
Answer: By definition of MGF, we have ( ) ∫ ( ) ∫ ∫ ( ) * ( ) ( )+ ( ) ( ) ( )
First moment
( ) ( ) ( ) ( )
First moment about origin
( ) ( )
Second moment
( ) ( ) ( )
( )
Second moment about origin
( ) ( ) ( )
Third moment
( ) ( ) ( )
( )
Third moment about origin
( ) ( ) Fourth moment ( ) ( ) ( ) ( )
Fourth moment about origin
( ) ( ) ( ) ( ) Variance ( ) , ( ( * Problem: 22
Find the MGF of the random variable having the probability density function ( ) { Answer: By definition of MGF, we have ( ) ∫ ( )
∫ ∫ . / ( * Limits: When ∫ . / . / ∫ . / [ ( * ( ) ] , - ( ) ,* + * +- ( ) ( ) ( ) First moment ( ) ( ) ( ) ( )
First moment about origin
( ) ( )
Second moment
( ) ( ) ( )
( )
Second moment about origin
( ) ( ) Problem: 23 ( ) { And hence find its mean and variance.
( ) ∫ ( ) ∫ ∫ . / 0 . / . / 1 0 . / 1 0 . / 1 , - ( ) ( ) First moment ( ) ( ) ( ) ( )
First moment about origin
( ) ( ) ( )
Second moment
( ) ( ) ( )
( )
Second moment about origin
( ) ( ) ( ) ( ) ( ) Variance ( ) , ( Problem: 24 ( ) 2 And hence find its first four moments about the origin.
Answer:
By definition of moment generating function ( ) ∫ ( ) ∫ ∫ ∫ . / [ . . / . / / ( . / . / ,] [* + { . / }] , - [ . / ] 0 ( ) 1 ( ) ( ) ( ) First moment ( ) ( ) ( ) ( )
First moment about origin
( ) ( ) ( )
Second moment
( ) ( ) ( )
( )
Second moment about origin
( ) ( ) ( )
Third moment
( ) ( ) ( )
Third moment about origin
( )
Fourth moment
( ) ( ) ( )
( )
Fourth moment about origin
( ) ( ) ( )
Problem: 25
( ) { ( ) Find its moment. Hence evaluate ,( ) -.
Answer: By definition of moment , - ∫ ( ) ∫ ( ) ∫ ( ) * + [ ] [ ( )( )] , - ( )( ) , - ( )( ) , - ( )( ) ,( ) - , - , - , - , -
( * ( * , ( ) -
Problem: 26
A continuous random variable has the PDF ( ) . Find moment
generating function, mean and variance of .
Answer:
To find the constant : The PDF of a C.R.V is . ∫ ( ) ∫ ∫ ( ) ( ) ( ) [ ( * ( * ( * ( *] [( ) ( ( )+] ( )
By definition of moment generating function ( ) ∫ ( )
∫ ( ) [ ] ∫ ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) * ( ( ) ( )) ( ( ) ( ) ) ( ( ) ( ) ) ( ( ) ( ) )+ [( ) ( ( ( ) )+] [ ( ) ] ( ) ( ) ( ) [ ]
To find Mean and Variance
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Variance ( ) , ( )-
Problem: 27
Find the moment generating function having the PDF ( ) {
.
Answer:
By definition of moment generating function ( ) ∫ ( ) ∫ ∫ ( ) ∫ * ( ) ( )+ *( ) ( ) ( ) ( )+ *,( ) ( )- , -+ *, - ,( ) ( ) ( )-+ * + * + ( )
Binomial distribution
The probability mass function of a Binomial distribution with success in trails is given by
( )
Assumptions for Binomial distributions:
There are only two possible outcomes on each trail
The probability of success is same in each trail
There are only trails, is a constant
The trails are independent
Find the moment generating function:
By definition of MGF, we have ( ) ∑ ( ) ∑( ) ∑( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) , ( ) -
Mean of Binomial distribution
( ) ( ) ( ) ( ) ( ) Put , we get ( ) ( ) ( ) , - ( ) ( )
Variance of Binomial distribution Variance ( ) , ( )-( ) )-( ) ( ) ( ) ( ) ( )( ) ( ) Put , we get ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) , - ( ) Variance , ( ) , - , - √ Problem: 28
The mean and standard deviation of the binomial distribution are 5 and 2. Find the distribution.
Answer:
Given, Mean ( ) ( ) and √
( ) From ( ) and ( ), we have
( )
Hence the binomial distribution is
( ) ( * ( *
Problem: 29
The mean and variance of the binomial distribution are 8 and 6. Find ( ).
Answer:
Given, Mean ( ) ( ) ( ) From ( ) and ( ), we have
( )
Hence the binomial distribution is
( ) ( * ( *
( ) ( ) , ( ) ( )- * ( * ( * ( * ( * + , - Problem: 30
The probability of a bomb hitting a target is . Two bombs are enough to destroy a bridge. If six bombs are aimed at the bridge, find the probability that the bridge is destroyed?
Answer:
Let be a random variable hitting the target.
Given
The binomial distribution is
( ) ( * ( *
The bridge destroyed with two bombs are ( ):
( ) ( * ( * Problem: 31
A perfect cubic die is thrown a large number of times in sets of 8. The presence of 5 and 6 is treated as a success. In what percentage of the sets can we expect 3 successes?
Let be a random variable throwing a die.
Given
The binomial distribution is
( ) ( * ( * ( ) ( * ( *
the percentage getting 3 successes is 27.31%.
Problem: 32
In the old days, there was a probability of 0.8 of success in any attempt to make a telephone call. (This often depended on the importance of the person making the call, or the operator's curiosity!) Calculate the probability of having 7 successes in 10attempts.
Answer:
Let be a random variable denoting the attempt a telephone call.
Given
The binomial distribution is
( )
( ) ( ) ( ) ( )
( )
( ) ( )
Problem: 33
Hospital records show that of patients suffering from a certain disease, 75% die of it. What is the probability that of 6 randomly selected patients, 4 will recover?
Answer:
Let be a random variable denoting the patients suffering from a certain disease. , -
Given
The binomial distribution is
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Problem: 34
A (blindfolded) marksman finds that on the average he hits the target 4 times out of 5. If he fires 4 shots, what is the probability of (a) more than 2 hits? (b) at least 3 misses?
Answer:
Let be a random variable denoting the hit of the target , -
Given
The binomial distribution is
( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Problem: 35
4 coins were tossed simultaneously. What is the probability of getting (i) 2 heads (ii) at least 2 heads (iii) at most 2 heads.
Answer:
Let be a random variable getting head or tails.
Given
The binomial distribution is
( ) ( * ( * ( ) ( ) ( * ( * ( * ( *
( ) ( ) ( ) , ( ) ( )- * ( * ( * ( * ( * + ( * ( ) ( ) ( ) ( ) ( ) ( * ( * ( * ( * ( * ( * ( * ( * ( * ( * Problem: 36
If 10% of the screws produced by an automatic machine are defective, find the
probability that out of 20 screws selected at random, there are (i) exactly 2 defective (ii) at most 3 defective (iii) at least 2 defective (iv) between 1 and 3 defective (inclusive).
Answer:
Let be a random variable denoting defective screws.
Given
The binomial distribution is
( * ( * ( ) ( ) ( * ( * ( * ( * ( ) ( ) ( ) ( ) ( ) ( ) ( * ( * ( * ( * ( * ( * ( * ( * ( ) ( ) ( ) , ( ) ( )- * ( * ( * ( * ( * + , - ( ) ( ) ( ) ( ) ( ) ( ) ( * ( * ( * ( * ( * ( *
Problem: 37
In certain town 20% samples of the population is literature and assume that 200 investigators take samples of ten individuals to see whether they are literature. How many investigators would you expect to report that 3 people or less are literatures in the samples?
Answer:
Let be a random variable denoting the literature persons.
Given and
The binomial distribution is
( ) ( * ( * ( ) ( ) ( ) ( ) ( ) ( ) ( * ( * ( * ( * ( * ( * ( * ( *
200 investigators reporting 3 people or less as literatures ( )
Problem: 38
In a large consignment of electric bulbs, 10 percent are defective. A random sample of 20 is taken for inspection. Find the probability that (1) all are good bulbs (2) at most there are 3 defective bulbs (3) exactly there are 3 defective bulbs.
Answer:
Given
The binomial distribution is
( ) ( * ( * ( ) ( ) ( * ( * ( * ( ) ( ) ( ) ( ) ( ) ( ) ( * ( * ( * ( * ( * ( * ( * ( * ( ) ( ) ( * ( *
Poisson distribution
Poisson distribution is a limiting case of binomial distribution under the following assumptions.
I. The number of trails should be indefinitely large. , - II. The probability of success for each trail is indefinitely small. III. , should be finite where is a constant.
Poisson distribution is limiting case of Binomial distribution:
The probability mass function of Binomial distributions ( ) ( ) We know the parameter
( ) ( ) ( * ( * ( )( ) ( )( ) ( )( ) ( )( ) ( ) ( * ( * . / . / . / ( * ( * . / . / . / ( * ( * . / . / . / ( * . / Taking , we get ( ) * ( * +
Probability mass function
The probability mass function of a random variable which follows Poisson distribution is given by
( )
Find the moment generating function:
By definition of MGF, we have ( ) ∑ ( ) ∑( ) ∑( ) * ( ) ( ) + * + , - ( ) ( )
Mean of Poisson distribution
( ) ( ) ( ) , - [ ( ) ] ( ) Put , we get ( ) ( ) ( )
Variance of Poisson distribution
)-( ) ( ) ( ) ( ) [ ] Put , we get ( ) [ ] [ ] , - ( ) , - Variance , ,
Recurrence relation for the Poisson distribution.
The probability mass function of a Poisson distribution is given by ( ) ( ) ( ) ( ) ( ) ( ( ) * ( * ( )( ) , ( ) - ( ) ( ) ( * ( ) Problem: 39 ( ) ( ) ( ) ( ) Answer:
( ) ( ) ( ) ( ) ( ) ( ) , - ( ) . / ( ) . /. / Problem: 40
If X is Poisson variate such that ( ) ( ) ( ) then ( )
Answer:
The probability mass function of a Poisson distribution is given by ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) [ ]
Since ( ) Also Variance ( ) ( ) , ( ) ( )- * + , - , - ( ) Problem: 41
The number of monthly breakdowns of a computer is a random variable having a Poisson distribution with mean equal to 1.8. Find the probability that this computer will function for a month (1) without any breakdown (2) with only one breakdown.
Answer:
Let be a random variable denoting the monthly breakdowns of a computer . Given Mean
The probability mass function of a Poisson distribution is given by ( ) ( ) ( )
( ) ( ) Problem: 42
1% of jobs arriving at a computer system need to wait until weekends for scheduling, owing to core-size limitations. Find the probability that among a sample of 200 jobs there are no jobs that have to wait until weekends.
Answer:
Let be a random variable denoting no. of jobs that have to wait until weekends. Given waiting time is 1% (Success)
The probability mass function of a Poisson distribution is given by ( ) ( ) ( ) Problem: 43
3% of the electric bulbs are manufactured by a company are defective, find the probability that in a sample of 100 bulbs exactly 5 bulbs are defective.
Answer:
Let be a random variable denoting the defective bulbs. Given defective bulbs are 3% (Success)
The probability mass function of a Poisson distribution is given by ( ) ( ) ( ) Problem: 44
A manufacture of pins knows that 2% of his products are defective. If he sells pins in boxes of 100 and guarantees that not more than 4 pins will be defective, what is the probability that a box fail to meet the guaranteed quality?
Answer:
Let be a random variable denoting the defective pins. Given defective pins are 2% (Success)
The probability mass function of a Poisson distribution is given by ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( * ( ) ( ) ( ) Problem: 45
It is known that form past experience in a certain plant there are on the average 4 industrial accidents. Find the probability that in a given year there will be less than 4 accidents.
Answer:
Let be a random variable denoting the industrial accidents. ( )
The probability mass function of a Poisson distribution is given by ( ) ( ) ( ) ( ) ( ) ( ) ( ) [ ] Problem: 46
An insurance company found that only of the population is involved in a certain type of accidents each year. If its 1000 policy holders were randomly selected from
the population, what is the probability that not more than two of its clients are involved in such an accident next year?
Answer:
Let be a random variable denoting the clients involved in a certain type of accidents. Given the persons involved in such accidents 0.01% (Success)
The probability mass function of a Poisson distribution is given by ( ) ( * ( ) ( ) ( ) ( ) [ ] Problem: 47
Out of 1000 balls 50 are red and the rest of them are white, if 60 balls are picked at random, what is the probability of picking up (i) 3 red balls (ii) not more than 3 red balls in the sample. Assume Poisson distribution for the number of red balls picked in the sample.
Answer:
Let be a random variable denoting the number of red balls in the sample. Here success = picking red ball from the sample.
No. of red balls = 50
The probability mass function of a Poisson distribution is given by ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) [ ] Problem: 48
If electricity power failures occur according to a Poisson distribution with an average of 3 failures every twenty weeks, calculate the probability that there will not be more than one failure during a particular week.
Let be a random variable denoting the electricity power failures. Given that an average of 3 failures every twenty weeks.
The probability mass function of a Poisson distribution is given by ( ) ( ) ( ) ( ) ( ) ( ) Problem: 49
Vehicles pass through a junction on a busy road at an average rate of 300 per hour. Find the probability that none passes in a given minute.
Answer:
Let be a random variable denoting the vehicles pass on road. Given that an average of 300 vehicles passing in an hour.
The probability mass function of a Poisson distribution is given by ( ) ( ) ( ) ( )
Continuous distributions Uniform distribution
A random variable is said to have a continuous uniform distribution if its probability density function is given by
( ) {
Moment generating function
By definition of MGF, we have ( ) ∫ ( ) ∫ ( ) [ ∫ ] ( ) ( ) ( )
Mean of Uniform distribution
( ) ∫ ( ) ∫ ( ) ( ) * ( )( ) +
( )
Variance of Uniform distribution
( ) ∫ ( ) ∫ ( ) ( ) , ( ) - ( ) ( ) , ( ) - ( ) ( ) ( ) ( ),( ) ( ) ( ) Variance ( ) , ( ( * ( ) ( ) Problem: 50
A random variable is uniformly distributed over ( ). Find ( ) ( ) ( ) and ( ).
Answer:
The probability density function of a uniform distribution is ( ) { ( ) { ( ) ∫ ( ) ∫ , ( ) ( ) ∫ ( ) ∫ , ( ) ( ) ∫ ( ) ∫
, ( )
( )
Problem: 51
Trains arrive at a station at 15 minutes intervals starting at 4 a.m. If a passenger arrives at a station at a time that is uniformly distributed between 9.00 and 9.30, find the probability that he has to wait for the train for (1) less than 6 minutes (2) more than 10 minutes.
Answer:
Let be a random variable that a passenger waits for train. Here is uniformly distributed in the interval ( ).
The probability density function is ( ) { ( ) ( ) ( ) ∫ ( ) ∫ ( ) ∫ ∫ *, - , + *( ) ( )+ ( ) ( ) ( ) 𝑻𝟏 𝑻𝟐 𝑻𝟑 𝟗 𝟎𝟎 𝟗 𝟏𝟓 𝟗 𝟑𝟎 𝟗 𝟎𝟓 𝟗 𝟏𝟎 𝟗 𝟐𝟎 𝟗 𝟐𝟓
∫ ( ) ∫ ( ) ∫ ∫ *, - , - + *( ) ( )+ Problem: 52
The number of computers sold daily at a computer shop is uniformly distributed with minimum of 2000 PC and a maximum of 5000 PC. (1). Find the probability that daily sales will fall between 2500 and 3000 PC (2). What is the probability that the computer shop will sell at least 4000 PC? (3). what is the probability that the computer shop will exactly sell 2500 PC?
Answer:
Let be a random variable that a daily sales of PC in a computer shop. Here is uniformly distributed in the interval 2000 PC to 5000 PC. The probability density function is
( ) { ( ) ( ) ∫ ( ) ∫ *, - + ( ) ( ) ( ) 𝟗 𝟎𝟓 𝑻𝟏 𝑻𝟐 𝑻𝟑 𝟗 𝟎𝟎 𝟗 𝟏𝟎 𝟗 𝟏𝟓 𝟗 𝟑𝟎 𝟗 𝟐𝟎 𝟗 𝟐𝟓
∫ ( ) ∫ *, - + ( ) ( ) ( ) Normal Distribution
The normal distribution is also referred to as Gaussian or Gauss distribution. The distribution is widely used in natural and social sciences. It is made relevant by the Central Limit Theorem, which states that the averages obtained from independent, identically distributed random variables with almost the same normal distributions, regardless of the type of distributions they are sampled from.
Example:
Life of items subjected to wear and tear like currency notes.
Weekly sales of an item in a store. Height and weight of children at birth
Probability density function
A continuous random variable is said to follow normal distribution with Mean and standard deviation , if its density function is defined as
( )
√
( )
Properties of a normal distribution
The mean, mode and median are all equal.
The curve is symmetric at the center (i.e. around the mean, μ).
Exactly half of the values are to the left of center and exactly half the values are to the right. The total area under the curve is 1.
The Standard Normal Model
A standard normal model is a normal distribution with a mean of 1 and a standard deviation of 1.
Note:
( )
Standard Normal distribution
The Normal distribution with mean , is called the Standard Normal distribution.
( )
√
Problem: 53
A radar unit is used to measure speeds of cars on a motorway. The speeds are normally distributed with a mean of 90 km/hr and a standard deviation of 10 km/hr. What is the probability that a car picked at random is travelling at more than 100 km/hr?
Answer:
Let be the normal variable representing the oxygen consumption. Given ⁄ and S.D ⁄
( ) ( ) ( ) Problem: 54
The time taken to assemble a car in a certain plant is a random variable having a normal distribution of 20 hours and a standard deviation of 2 hours. What is the probability that a car can be assembled at this plant in a period of time a) less than 19.5 hours? b) between 20 and 22 hours?
Answer:
Let be the normal variable denoting car assemble. Given and ( ) ( ) ( ) ( ) ( ) 𝑃(𝑍 ) 𝑃(𝑍 )
( ) ( ) ( ) ( ) Problem: 55
In a test on 2000 electric bulbs, it was found that the life of Philips bulbs was normally distributed with an average of 2100 hours and S.D. of 60 hours. Estimate the number of bulbs likely to burn for (i) more than 2150 hours, (ii) less than 1950 hours.
Answer:
Let be the normal random variable representing the life time of an electric bulbs Given and ( ) ( ) ( ) ( ) ( ) For 2000 bulbs ( ) ( ) ( ) ( ) ( ) ( ) 𝑃( 𝑍 ) 𝑃(𝑍 ) 𝑃(𝑍 )
For 2000 bulbs ( )
Problem: 56
Assume that the reduction of a person’s oxygen consumption during a period of Transcendental Meditation (T.M) is a continuous random variable X normally distributed with mean 37.6 cc/mim and S.D 4.6 cc/min. Determine the probability that during a period of T.M. a person’s oxygen consumption will be reduced by
i. at least 44.5 cc/min i. at most 35.0 cc/min
ii. anywhere from 30.0 to 40.0 cc/mim.
Answer:
Let be the normal variable representing the oxygen consumption. Given and ( ⁄ ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) Problem: 57
The marks obtained by a number of students in a certain subject are assumed to be normally distributed with mean 65 and standard deviation 5. If 3 students are selected at random from this group, what is the probability that two of them will have marks over 70 and atleast one of them will have marks over 70?
Solution:
Let be the variable denoting the marks of the students. Given and ( ) ( ) ( ) ( ) ( )
If 3 students are selected at random. Success scoring more than 70
The binomial distribution is ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) , ( ) ( ) - Problem: 58
Given that is distributed normally. If ( ) and ( ) , find the mean and standard deviation of the distribution.
Answer: ( ) ( * ( * ( * ( * ( * ( ) ( ) ( ) ( * ( * ( * ( * ( * ( ) ( ) Solving equations ( ) and ( ) , we get
Problem: 59
Let and be independent normal variates with mean 45 and 44 and standard deviation 2 and 15 respectively. What is the probability that randomly chosen values of and differ by 1.5 or more?
Answer: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) , ( ) ( ) ( )- √ ,( ) -: , - ,( ) - ( ) ( )
Two Marks 1. Define random variable.
Answer:
A real valued function defined on the outcome of a probability experiment is called random variables.
2. Define discrete random variable.
Answer:
A random variable whose set of possible values is either finite or countably infinite is called discrete random variable.
3. Define continuous random variable.
Answer:
A random variable is said to be continuous, if it takes all possible values between certain intervals.
4. and are independent random variables with variance 2 and 3. Find the variance of .
Answer:
Given that ( ) and ( ) .
We know that ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ⁄ ) Answer: ( * ( ) ( ) , - , ⁄ - ( ) ( ) ( ) ( )
( ) ∫ ( ) ∫ ( ) ( ) ( ) ∫ ( ) ∫ ( ) ( ) , ⁄ - . / . /
6. Assume that X is a continuous random variable with the probability density
( ) { ( ) ( ) Answer: ( ) ∫ ( ) ∫ ( ) ( ) *( ) ( )+
7. A continuous random variable X has probability density function
( ) { ( ) Answer: ( ) ∫ ( ) ∫ ( )
8. Find , if a continuous random variable has the density function
( ) Answer: The PDF of a C.R.V is . ∫ ( ) ∫ , - , ( ) ( )- 0 . /1 0 ( ) 1 ( ) ,
Find the PDF and mean of
Answer: ( ) ( ) , { ( * ( ) , - ∫ ( ) ∫
Limits: When ∫ ∫ [ ( * ( *] , ( )- ( ) { ( )
Find the PDF of and expected value of .
Answer: ( ) ( ) { ( ) { ( ) ( ) , - ∫ ( ) ∫ ( )
