S & P Unit-III PDF

Noida Institute of Engineering and Technology, Greater Noida Statistics & Probability BAS0303 Unit: III Probability distribution Dr. Anil Agarwal...

Noida Institute of Engineering and Technology, Greater Noida Statistics & Probability BAS0303 Unit: III Probability distribution Dr. Anil Agarwal Associate Professor B.Tech-3rd Sem Dept. of Mathematics (DS/AIML/AI) Dr. Anil Agarwal Unit III 1 12/13/2024  Sequence of Content (1) Name of Subject with code, Course and Subject teacher. (2) Brief Introduction of Faculty. (3) Evaluation Scheme. (4) Subject Syllabus. (5) Branch Wise Application. (6) Course Objective. (7) Course Outcomes(COs). (8) Program Outcomes(POs). (9) COs and POs Mapping. (10)Program Specific Outcomes(PSOs) (11) COs and PSOs Mapping. (12)Program Educational Objectives(PEOs). 12/13/2024 Dr. Anil Agarwal Unit-III 2  Sequence of Content (13)Result Analysis. (14) End Semester Question Paper Templates. (15) Prequisite /Recap. (16) Brief Introduction about the Subject. (17) Unit Content. (18) Unit Objective. (19) Topic Objective/Topic Outcome. (20) Lecture related to topic. (21) Daily Quiz. (22) Weekly Assignment. (23) Topic Links. 12/13/2024 Dr. Anil Agarwal Unit-III 3  Sequence of Content (24) MCQ(End of Unit). (25) Glossary questions. (26) Old question Papers(Sessional + University). (27) Expected Questions For External Examination. (28) Recap of Unit. 12/13/2024 Dr. Anil Agarwal Unit-III 4  Faculty Introduction Name : Dr. Anil Agarwal Designation: Associate professor Department: Mathematics Teaching Experience: 22years Ph.D. : Agra University, Agra. 12/13/2024 Unit-III 5  Evaluation Scheme Sl. Subject Periods Evaluation Scheme End No. Codes Semester Credi t Subject Name Total L T P CT TA TOTAL PS TE PE WEEKS COMPULSORY INDUCTION PROGRAM 1 BAS0303 Statistics and Probability 3 1 0 30 20 50 100 150 4 2 ACSE0306 Discrete Structures 3 0 0 30 20 50 100 150 3 3 ACSE0305 Computer Organization & 3 0 0 30 20 50 100 150 3 Architecture 4 ACSE0302 Object Oriented Techniques 3 0 0 30 20 50 100 150 3 using Java 5 ACSE0301 Data Structures 3 1 0 30 20 50 100 150 4 6 ACSAI0301 Introduction to Artificial 3 0 0 30 20 50 100 150 3 Intelligence 7 ACSE0352 Object Oriented Techniques 0 0 2 25 25 50 1 using Java Lab 8 ACSE0351 Data Structures Lab 0 0 2 25 25 50 1 9 ACSAI0351 Introduction to Artificial 0 0 2 25 25 50 1 Intelligence Lab 10 ACSE0359 Internship Assessment-I 0 0 2 50 50 1 ANC0301/ Cyber Security*/ 11 Environmental Science * ANC0302 (Non Credit) 2 0 0 30 20 50 50 100 0 12 MOOCs** (For B.Tech. Hons. Degree) GRAND TOTAL 1100 24 12/13/2024 Dr. Anil Agarwal Unit-III 6  Syllabus of BAS0303 UNIT-I Descriptive measures 8 Hours Measures of central tendency – mean, median, mode, measures of dispersion – mean deviation, standard deviation, quartile deviation, variance, Moment, Skewness and kurtosis, least squares principles of curve fitting, Covariance, Correlation and Regression analysis, Correlation coefficient: Karl Pearson coefficient, rank correlation coefficient, uni-variate and multivariate linear regression, application of regression analysis, Logistic Regression, time series analysis- Trend analysis (Least square method). UNIT-II Probability and Random variable 8 Hours Probability Definition, The Law of Addition, Multiplication and Conditional Probability, Bayes’ Theorem, Random variables: discrete and continuous, probability mass function, density function, distribution function, Mathematical expectation, mean, variance. Moment generating function, characteristic function, Two dimensional random variables: probability mass function, density function, UNIT-III Probability distribution 8 Hours Probability Distribution (Continuous and discrete- Normal, Exponential, Binomial, Poisson distribution), Central Limit theorem UNIT-IV Test of Hypothesis & Statistical Inference 8 Hours Sampling and population, uni-variate and bi-variate sampling, re-sampling, errors in sampling, Sampling distributions, Hypothesis testing- p value, z test, t test (For mean), Confidence intervals, F test; Chi-square test, ANOVA: One way ANOVA, Statistical Inference, Parameter estimation, Least square estimation method, Maximum Likelihood estimation. UNIT-V Aptitude-III 8 Hours Time & Work, Pipe & Cistern, Time, Speed & Distance, Boat & Stream, Sitting Arrangement, Clock & Calendar. 12/13/2024 7  Branch wise Application Data Analysis Artificial intelligence Digital Communication: Information theory and coding. Dr. Anil Agarwal Unit III 12/13/2024 8  Course Objective The objective of this course is to familiarize the engineers with concept of Statistical techniques, probability distribution, hypothesis testing and ANOVA and numerical aptitude. It aims to show case the students with standard concepts and tools from B. Tech to deal with advanced level of mathematics and applications that would be essential for their disciplines. The student will be able to understand: The concept of Descriptive measurements. The concept of probability & Random variable. Probability distributions. The concept of hypothesis testing & Statistical inferences. The concept of numerical aptitude. Dr. Anil Agarwal Unit III 12/13/2024 9  Course Outcome CO1: Understand the concept of moments, skewness, kurtosis, correlation, curve fitting and regression analysis, Time-Series analysis etc. CO2: Understand the concept of Probability and Random variables. CO3: Remember the concept of probability to evaluate probability distributions. CO4: Apply the concept of hypothesis testing and estimation of parameters. CO5: Solve the problems of Time & Work, Pipe & Cistern, Time, Speed & Distance, Boat & Stream, Sitting arrangement , Clock & Calendar etc. Dr. Anil Agarwal Unit II 12/13/2024 10  PO S.No Program Outcomes (POs) PO 1 Engineering Knowledge PO 2 Problem Analysis PO 3 Design/Development of Solutions PO 4 Conduct Investigations of Complex Problems PO 5 Modern Tool Usage PO 6 The Engineer & Society PO 7 Environment and Sustainability PO 8 Ethics PO 9 Individual & Team Work PO 10 Communication PO 11 Project Management & Finance PO 12 Lifelong Learning 12/13/2024 Dr. Anil Agarwal Unit-III 11  CO-PO Mapping(CO3) Sr. Course PO1 PO PO PO4 PO PO PO PO PO PO1 PO1 PO1 No Outcom 2 3 5 6 7 8 9 0 1 2 e 1 CO 1 3 3 3 3 1 1 2 2 CO 2 3 3 3 2 1 1 2 2 3 CO 3 3 2 3 2 1 1 1 4 CO 4 3 2 2 3 1 1 1 5 CO.5 3 3 2 2 1 1 1 2 2 *1= Low *2= Medium *3= High 12/13/2024 Dr. Anil Agarwal Unit-III 12  PSO 12/13/2024 Dr. Anil Agarwal Unit-III 13  CO-PSO Mapping(CO3) CO PSO 1 PSO 2 PSO 3 CO1 3 2 1 CO2 1 2 1 CO3 2 2 2 CO4 3 2 1 CO5 3 2 2 *1= Low *2= Medium *3= High 12/13/2024 Dr. Anil Agarwal Unit-III 14  Program Educational Objectives(PEOs) PEO-1: To have an excellent scientific and engineering breadth so as to comprehend, analyze, design and provide sustainable solutions for real-life problems using state-of-the-art technologies. PEO-2: To have a successful career in industries, to pursue higher studies or to support entrepreneurial endeavors and to face the global challenges. PEO-3: To have an effective communication skills, professional attitude, ethical values and a desire to learn specific knowledge in emerging trends, technologies for research, innovation and product development and contribution to society. PEO-4: To have life-long learning for up-skilling and re-skilling for successful professional career as engineer, scientist, entrepreneur and bureaucrat for betterment of society. Dr. Anil Agarwal Unit-III 12/13/2024 15  Result Analysis Branch Semester Sections No. of No. % Passed enrolled Passed Students Students AIML III A, B, C 199 199 100% Dr. Anil Agarwal Unit-III 12/13/2024 16  End Semester Question Paper Template Dr. Anil Agarwal Unit-III 12/13/2024 17  Prerequisite and Recap(CO3) ▪ Knowledge of Maths -I of B.Tech. ▪ Knowledge of Maths -II of B.Tech. ▪ Knowledge of Basic Statistics. 12/13/2024 Dr. Anil Agarwal Unit-III 18  Brief Introduction about the subject In first four modules, we will discuss Statistics and probability. In 5th module we will discuss aptitude part. 12/13/2024 Dr. Anil Agarwal Unit-III 19  Unit Content(CO3) Introduction of Probability distributions Binomial Distribution Poisson Distribution Normal Distribution Exponential Distribution 12/13/2024 20 Dr. Anil Agarwal Unit-III  Unit Objective(CO3) 1. A basic knowledge in probability theory. 2. The student is able to reflect developed mathematical methods in probability and statistics. 3. Understand the concept of Probability and its usage in various business applications. 4. The binomial distribution model allows us to compute the probability of observing a specified number of "successes" when the process is repeated a specific number of times 5. Poisson Distribution is a tool that helps to predict the probability of certain events from happening when you know how often the event has occurred. 6. To learn the characteristics of a typical normal curve. 7. To explore the key properties, such as the moment-generating function, mean and variance, of a normal random variable. 12/13/2024 Dr. Anil Agarwal Unit-III 21  Topic Objective (CO3) The probability distributions are very much helpful for making predictions. Estimates and predictions form an important part of research investigation. With the help of Probability distributions, we make estimates and predictions for the further analysis. 12/13/2024 Dr. Anil Agarwal Unit-III 22  Probability distributions(CO3) Theoretical probability distribution Discrete Continuous probability probability distributions distributions Binomial Normal Distribution Distribution t- Distribution Poisson Distribution F-Distribution 12/13/2024 Dr. Anil Agarwal Unit-III 23  Binomial Distribution(CO3) Binomial Probability Distribution: Probability distribution defined as follows is known as binomial Probability distribution. 𝑃 𝑋 = 𝑟 = 𝑛𝐶𝑟 𝑝𝑟 𝑞 𝑛−𝑟 ,𝑟 = 1,2 … 𝑛 Where n is no of trial which are finite ,r be the success in n trials and 𝑝 + 𝑞 = 1, p is probability of success and q is probability of failure. Assumptions For Binomial distribution: n, the number of trials is finite Each trial has only two possible outcomes usually called success and failure. All trials are independent. p and q is constant for all trials. 12/13/2024 Dr. Anil Agarwal Unit-III 24  Cont…(CO3) Recurrence or recursion formula: 𝑛! 𝑃 𝑟 = 𝑛𝐶𝑟 𝑝𝑟 𝑞 𝑛−𝑟 = 𝑝𝑟 𝑞 𝑛−𝑟 ….(1) 𝑟!(𝑛−𝑟)! Equation (1) denote binomial distribution. 𝑛 𝑛! 𝑃 𝑟 + 1 = 𝐶𝑟+1 𝑝𝑟+1 𝑞 𝑛−𝑟−1 = 𝑝𝑟+1 𝑞 𝑛−𝑟−1 ….(2) (𝑟+1)!(𝑛−𝑟−1)! By equation (1) and (2) 𝑃(𝑟 + 1) (𝑛 − 𝑟)! 𝑟! 𝑝 = × × 𝑃(𝑟) (𝑛 − 𝑟 − 1)! (𝑟 + 1)! 𝑞 (𝑛−𝑟) 𝑝 𝑃 𝑟+1 =.𝑃 𝑟 …..(3) (𝑟+1) 𝑞 Equation (3) is known as Recurrence Formula. 12/13/2024 Dr. Anil Agarwal Unit-III 25  Cont…(CO3) Mean Of Binomial distribution: 𝒏 𝝁 = ෍ 𝒓𝑷(𝒓) 𝒓=𝟎 𝑛 For Binomial distribution 𝑃 𝑟 = 𝐶𝑟 𝑝𝑟 𝑞 𝑛−𝑟 𝑛 𝜇 = ෍ 𝑟 𝑛𝐶𝑟 𝑝𝑟 𝑞 𝑛−𝑟 𝑟=0 By expanding we have ⟹ 0 + 1. 𝑛𝐶1 𝑝𝑞 𝑛−1 + 2. 𝑛𝐶2 𝑝2 𝑞 𝑛−2 + ⋯ + 𝑛. 𝑛𝐶𝑛 𝑝𝑛 ⟹ 𝑛𝑝 𝑛−1𝐶0 𝑞 𝑛−1 + 𝑛−1𝐶1 𝑝𝑞 𝑛−2 + ⋯ + 𝑛−1𝐶𝑛−1 𝑝𝑛−1 ⟹ 𝑛𝑝 𝑞 + 𝑝 𝑛−1 = 𝑛𝑝 Hence mean of binomial distribution is np. 12/13/2024 Dr. Anil Agarwal Unit-III 26  Cont…(CO3) Variance of Binomial Distribution: 𝑛 𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 𝜎 2 = ෍ 𝑟 2 𝑃 𝑟 − 𝜇2 𝑟=0 𝑛 = ෍[𝑟 + 𝑟 𝑟 − 1 ]𝑃 𝑟 − 𝜇2 𝑟=0 𝑛 𝑛 = ෍ 𝑟𝑃 𝑟 + ෍ 𝑟(𝑟 − 1)𝑃 𝑟 − 𝜇2 𝑟=0 𝑟=0 𝑛 = 𝑛𝑝 + ෍ 𝑟(𝑟 − 1)𝑃 𝑟 − 𝑛2 𝑝2 𝑟=0 = 𝑛𝑝 + 𝑛 𝑛 − 1 𝑝2 − 𝑛2 𝑝2 = 𝑛𝑝 1 − 𝑝 = 𝑛𝑝𝑞 12/13/2024 Dr. Anil Agarwal Unit-III 27  Cont…(CO3) Hence the Variance of binomial distribution is 𝒏𝒑𝒒and Standard deviation is 𝑛𝑝𝑞. Moment generating function of binomial Distribution: i. About origin 𝑛 𝑀𝑥 𝑡 = 𝐸 𝑒 𝑥𝑡 = ෍ 𝑛𝐶𝑥 (𝑝𝑒 𝑡 )𝑥 𝑞 𝑛−𝑥 = (𝑞 + 𝑝𝑒 𝑡 )𝑛 𝑥=0 ii. About mean 𝑀𝑥−𝑛𝑝 𝑡 = 𝐸 𝑒 𝑡(𝑥−𝑛𝑝) = 𝑒 −𝑛𝑝𝑡 𝐸 𝑒 𝑥𝑡 = 𝑒 −𝑛𝑝𝑡 𝑀𝑥 𝑡 = 𝑒 −𝑛𝑝𝑡 (𝑞 + 𝑝𝑒 𝑡 )𝑛 = (𝑞𝑒 −𝑝𝑡 + 𝑝𝑒 𝑞𝑡 )𝑛 12/13/2024 Dr. Anil Agarwal Unit-III 28  Cont…(CO3) Applications of Binomial Distribution: 1. In problem concerning no. of defectives in sample production line. 2. In estimation of reliability of systems. 3. No. of rounds fired from a gun hitting a target. 4. In radar detection. Q1. If 10% of bolts are produced by a machine are defective , determine the probability that out of 10 bolts chosen at random i. 1 ii. None iii. At most 2 bolts will be defective 12/13/2024 Dr. Anil Agarwal Unit-III 29  Problems based on Binomial Distribution(CO3) Solution: let p and q are the probability of defective and non defective bolts respectively. 10 1 1 9 𝑝= = ,𝑞 =1− = and n=10 (no of bolts chosen) 100 10 10 10 The Probability of r defective bolts out of n bolt chosen at random is given by 𝑃 𝑟 = 𝑛𝐶𝑟 𝑝𝑟 𝑞 𝑛−𝑟 i. Here r=1, 1 10−1 1 9 𝑃 1 = 10𝐶1 = 0.3874 10 10 ii. Here r=0 0 10−0 1 9 𝑃 0 = 10𝐶0 = 0.3486 10 10 12/13/2024 Dr. Anil Agarwal Unit-III 30  Problems based on Binomial distribution(CO3) iii.Prob.that at most 2 bolts will be defective =𝑃(≤ 2) = 𝑃(0) + 𝑃(1) + 𝑃(2) 2 10−2 10 1 9 𝑃 2 = 𝐶2 10 10 1 =45 0.43046 = 0.1937 100 From(4).Required Probability=𝑃 0 + 𝑃 1 + 𝑃 2 = 0.3486 + 0.3874 + 0.1937 = 0.9297 Q2. Out of 800 families with 4 children each, how many families would be expected to have i. 2 boys and 2 girls ii. At least one boy iii no girl iv. Atmost two girls? Assume equal probability for boys and girls. 12/13/2024 Dr. Anil Agarwal Unit-III 31  Cont…(CO3) Solution: Probability for boys and girls are equal 1 𝑝 = 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 ℎ𝑎𝑣𝑖𝑛𝑔 𝑎 𝑏𝑜𝑦 = , 2 1 𝑞 = 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 ℎ𝑎𝑣𝑖𝑛𝑔 𝑎 𝑔𝑖𝑟𝑙 = n=4 N=800 2 i. The expected number of families having 2 boys and 2 girls 2 2 1 1 = 800 4𝐶2 = 300 2 2 ii. The expected number of families having at least one boy 3 1 2 2 1 3 1 1 1 1 1 1 = 800 ൥ 4𝐶1 + 4𝐶2 + 4𝐶3 2 2 2 2 2 2 4 1 + 4𝐶4 ൩ = 750 2 12/13/2024 Dr. Anil Agarwal Unit-III 32  Cont…(CO3) iii. The expected number of families having no girl i.e. having 4 boys 4 1 4 = 800 𝐶4 = 50 2 iv. The expected number of families having almost two girls i.e. having at least 2 boys 2 2 1 3 4 4 1 1 4 1 1 4 1 = 800 𝐶2 + 𝐶3 + 𝐶4 = 550 2 2 2 2 2 12/13/2024 Dr. Anil Agarwal Unit-III 33  Daily Quiz(CO3) 1. Four persons in a group of 20 are graduates. If 4 persons are selected at random from 20, find the probability that all 4 are graduates. Ans: 0.0016 2. The Prob. that a bulb produced by a factory will fuse after use of 150 days is 0.05. Find the probability that out of 5 such bulbs at least one bulb will fuse after use of 150 days of use. 19 5 Ans: 1 − 20 12/13/2024 Dr. Anil Agarwal Unit-III 34  Poisson Distribution(CO3) Poisson distribution: Probability distribution defined as follows is known as Poisson Probability distribution. 𝑒 −𝜆 𝜆𝑟 𝑃 𝑋=𝑟 = , 𝑟 = 0,1,2,3 …. 𝑟! Where 𝜆 𝑓𝑖𝑛𝑖𝑡𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 = 𝑛𝑝. Recurrence formula for Poisson Distribution: 𝑒 −𝜆 𝜆𝑟 Poisson distribution𝑃 𝑟 = ……… 1 𝑟! 𝑒 −𝜆 𝜆𝑟+1 𝑃 𝑟+1 = ……..(2) (𝑟+1)! 𝑃(𝑟 + 1) 𝜆𝑟! 𝜆 = = 𝑃(𝑟) (𝑟 + 1)! 𝑟 + 1 12/13/2024 Dr. Anil Agarwal Unit-III 35  Cont…(CO3) 𝜆 𝑃 𝑟+1 = 𝑃 𝑟 , 𝑟 = 01,2,3 … … 𝑟+1 This is called the recurrence or recursion formula for Poisson distribution. Mean of the Poisson distribution: Mean 𝜇 = σ∞ 𝑟=0 𝑟𝑃 𝑟 ∞ 𝑒 −𝜆 𝜆𝑟 = ෍ 𝑟. 𝑟! 𝑟=0 ∞ 𝜆𝑟 = 𝑒 −𝜆 ෍ (𝑟 − 1)! 𝑟=1 12/13/2024 Dr. Anil Agarwal Unit-III 36  Cont…(CO3) 2 3 𝜆 𝜆 = 𝑒 −𝜆 𝜆 + + + ⋯. 1! 2! 𝜆 𝜆2 = 𝜆𝑒 −𝜆 1 + + + ⋯. 1! 2! 𝜆 𝜆2 = 𝜆𝑒 −𝜆 1 + + + ⋯. 1! 2! = 𝜆𝑒 −𝜆 𝑒 𝜆 = 𝜆 Mean = 𝜆 for Poisson distribution. Variance of Poisson Distribution: ∞ 𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 𝜎 2 = ෍ 𝑟 2 𝑃 𝑟 − 𝜇2 𝑟=0 12/13/2024 Dr. Anil Agarwal Unit-III 37  Cont…(CO3) ∞ 𝑒 −𝜆 𝜆𝑟 = ෍ 𝑟2 − 𝜇2 𝑟! 𝑟=0 ∞ 𝑟 2 𝜆𝑟 = 𝑒 −𝜆 ෍ − 𝜆2 𝑟! 𝑟=0 2 1 2 2 2 3 −𝜆 1 𝜆 2 𝜆 3 𝜆 =𝑒 + + + ⋯ − 𝜆2 1! 2! 3! 2𝜆 3𝜆 2 = 𝜆𝑒 −𝜆 1 + + + ⋯ − 𝜆2 1! 2! (1 + 1)𝜆 (1 + 2)𝜆2 = 𝜆𝑒 −𝜆 1 + + + ⋯ − 𝜆2 1! 2! 2 2 𝜆 𝜆 𝜆 2𝜆 = 𝜆𝑒 −𝜆 1 + + + ⋯ + + +⋯ − 𝜆2 1! 2! 1! 2! 12/13/2024 Dr. Anil Agarwal Unit-III 38  Cont…(CO3) 2 𝜆 𝜆 = 𝜆𝑒 −𝜆 𝑒 𝜆 + 𝜆 1 + + + ⋯ − 𝜆2 1! 2! = 𝜆𝑒 −𝜆 𝑒 𝜆 + 𝜆𝑒 𝜆 − 𝜆2 = 𝑒 𝜆 𝑒 −𝜆 𝜆 1 + 𝜆 − 𝜆2 = 𝜆 1 + 𝜆 − 𝜆2 =𝜆 Hence , the Variance of the Poisson distribution is also 𝜆. Applications of Poisson Distribution: i. Arrival pattern of the defective vehicles in a workshop. ii. Patients in hospitals. iii. Telephone calls. iv. Emission of radioactive 𝛼 particles. 12/13/2024 Dr. Anil Agarwal Unit-III 39  Problem based on Poisson Distributions(CO3) Q1. If the Variance of the Poisson distribution is 2 , find the probability for r=1,2,3,4 from the recurrence relation of the Poisson Distribution. Also find 𝑃 𝑟 ≥ 4. Solution: Given that Variance = 2 = 𝜆 Recurrence relation for Poisson distribution 𝜆 2 𝑃 𝑟+1 = 𝑃 𝑟 = 𝑃 𝑟 … … (1) 𝑟+1 𝑟+1 𝑒 −𝜆 𝜆𝑟 Poisson Distribution 𝑃 𝑟 = 𝑟! 𝑒 −2 2𝑟 𝑒 −2 20 So 𝑃 𝑟 = ⟹𝑃 0 = ⟹ 𝑒 −2 = 0.1353 𝑟! 0! Now putting r=0,1,2,3 in equation (1) 12/13/2024 Dr. Anil Agarwal Unit-III 40  Cont…(CO3) 𝑃 1 = 2𝑃 0 = 2 × 0.1353 = 0.2706 2 2 𝑃 2 = 𝑃 1 = × 0.2706 = 0.2706 2 2 2 2 𝑃 3 = 𝑃 2 = × 0.2706 = 0.1804 3 3 2 1 𝑃 4 = 𝑃 3 = × 0.1804 = 0.0902 4 2 Now to calculate 𝑃 𝑟 ≥ 4 , 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑃 𝑟 ≥ 4 = 1 − [𝑃 0 + 𝑃 1 + 𝑃 2 + 𝑃 3 ] = 1 − 0.1353 + 0.2706 + 0.2706 + 0.1804 = 0.1431 Q2. Fit a Poisson distribution to the following data and calculate theoretical frequencies. 12/13/2024 Dr. Anil Agarwal Unit-III 41  Cont…(CO3) Deaths: 0 1 2 3 4 Frequencies 122 60 15 2 1 σ 𝑓𝑥 Mean of Poisson distribution λ = σ𝑓 0 × 122 + 1 × 60 + 2 × 15 + 3 × 2 + 4 × 1 = ෍ 𝑓 = N = 200 122 + 60 + 15 + 2 + 1 = 0.5 Required Poisson distribution 𝑒 −𝜆 𝜆𝑟 𝑒 −0.5 (0.5)𝑟 (0.5)𝑟 = 𝑁. = 200 = (121.306) 𝑟! 𝑟! 𝑟! 12/13/2024 Dr. Anil Agarwal Unit-III 42  Cont…(CO3) r N.P(r) Theoretical frequencies (0.5)0 0 121.306 = 121.306 121 0! (0.5)1 1 121.306 = 60.653 61 1! (0.5)2 2 121.306 = 15.163 15 2! (0.5)3 3 121.306 = 2.527 3 3! (0.5)4 4 121.306 = 0.3159 0 4! Total=200 12/13/2024 Dr. Anil Agarwal Unit-III 43  Topic objective of Normal Distribution(CO3) To define the probability density function of a normal random variable. To learn the characteristics of a typical normal curve. To explore the key properties, such as the moment- generating function, mean and variance, of a normal random variable. 12/13/2024 Dr. Anil Agarwal Unit-III 44  Normal Distribution(CO3) Normal Distribution: The general equation of the normal distribution is given by 1 1 𝑥−𝜇 2 − 𝑓 𝑥 = 𝑒 2 𝜎 𝜎 2𝜋 for Basic properties Normal distributions: i. 𝑓 𝑥 ≥ 0 ∞ ii. ‫׬‬−∞ 𝑓 𝑥 𝑑𝑥 = 1 i.e. the total area under the normal curve above x-axis is 1. iii. Normal curve is symmetrical about its mean. iv. The mean, mode and median coincide for this distribution. 12/13/2024 Dr. Anil Agarwal Unit-III 45  Cont…(CO3) Standard form of the normal distribution: The probability density function for the normal distribution in standard form is given by 1 −1𝑧 2 𝑓 𝑧 = 𝑒 2 2𝜋 𝑥−𝜇 By taking 𝑧 = , standard normal curve is formed. The total 𝜎 area under the curve is 1 and it is divided into two parts by z=0. 12/13/2024 Dr. Anil Agarwal Unit-III 46  Mean and Variance of Normal distribution(CO3) Mean of Normal Distribution: A.M. of continuous distribution f(x) is given by ∞ ‫׬‬−∞ 𝑥𝑓 𝑥 𝑑𝑥 𝐴. 𝑀 𝑥ҧ = ∞ ‫׬‬−∞ 𝑓 𝑥 𝑑𝑥 ∞ ∞ 𝑥ҧ = ‫׬‬−∞ 𝑥𝑓 𝑥 𝑑𝑥 because ‫׬‬−∞ 𝑓 𝑥 𝑑𝑥 = 1 1 𝑥−𝜇 2 ∞ 1 −2 𝜎 So 𝑥ҧ = ‫׬‬−∞ 𝑥 𝑒 𝑑𝑥 𝜎 2𝜋 𝑥−𝜇 Let = 𝑧 so that 𝑥 = 𝜇 + 𝜎𝑧 therefore dx = 𝜎𝑑𝑧 𝜎 ∞ 1 1 − 𝑧2 𝑥ҧ = න (𝜇 + 𝜎𝑧) 𝑒 2 𝜎𝑑𝑧 𝜎 2𝜋 −∞ 12/13/2024 Dr. Anil Agarwal Unit-III 47  Cont…(CO-3) ∞ ∞ 1 1 −2𝑧 2 𝜎 1 −2𝑧 2 =𝜇 න 𝑒 𝑑𝑧 + න 𝑧𝑒 𝑑𝑧 2𝜋 2𝜋 −∞ −∞ 1 ∞ 1 −2𝑧 2 Because ‫׬‬−∞ 𝑒 𝑑𝑧=1 so in above equation 2𝜋 ∞ 𝜎 1 −2𝑧 2 =𝜇+ න 𝑧𝑒 𝑑𝑧 2𝜋 −∞ ∞ 𝜎 1 −2𝑧 2 𝑧2 =𝜇+ න 𝑒 𝑑 2𝜋 2 −∞ ∞ 𝜎 1 − 𝑧2 =𝜇+ 𝑒 2 −∞ 2𝜋 12/13/2024 Dr. Anil Agarwal Unit-III 48  Cont…(CO-3) 𝑥ҧ = μ Variance of Normal distribution : ∞ = න (𝑥 − 𝑥)ҧ 2 𝑓 𝑥 𝑑𝑥 −∞ ∞ = න 𝑥 2 𝑓 𝑥 𝑑𝑥 −𝑥ҧ 2 … … … … (1) −∞ ഥ 2 1 𝑥−𝑥 ∞ 2 ∞ 2 1 − Let I = ‫׬‬−∞ 𝑥 𝑓 𝑥 𝑑𝑥 = ‫׬‬−∞ 𝑥 𝜎 2𝜋 𝑒 2 𝜎 𝑑𝑥 𝑥−𝑥ҧ = z so that 𝑥 = 𝑥ҧ + 𝜎𝑧 therefore d𝑥 = 𝜎𝑑𝑧 𝜎 12/13/2024 Dr. Anil Agarwal Unit-III 49  Cont... (CO3) ∞ 1 1 −2𝑧 2 I = න (𝑥ҧ + 𝜎𝑧)2 𝑒 𝜎𝑑𝑥 𝜎 2𝜋 −∞ ∞ ∞ 2 2 𝜎 1 −2𝑧 2 𝜎 1 −2𝑧 2 =− 𝑧𝑒 −∞ + න 𝑒 𝑑𝑥 + 𝑥ҧ 2 2𝜋 2𝜋 −∞ =0+ 𝜎2+ 𝑥ҧ 2 = 𝜎 2 +𝑥ҧ 2 In equation(1) Variance = 𝜎 2 +𝑥ҧ 2 − 𝑥ҧ 2 = 𝜎 2 s.d. of normal distribution is 𝜎. 12/13/2024 Dr. Anil Agarwal Unit-III 50  Problem based on Normal distribution(CO3) Q1. A sample of 100 dry battery cells tested to find the length of the life produced the following results : 𝑥ҧ = 12 ℎ𝑜𝑢𝑟𝑠, 𝜎 = 3 ℎ𝑜𝑢𝑟𝑠 Assuming the data to be normally distributed, what percentage of battery cells are expected to have life i. More than 15 hours ii. Less than 6 hours iii. between 10 and 14 hours Solution: x denotes the life of dry battery cells. 𝑥−𝑥ҧ 𝑥−12 And 𝑧 = = 𝜎 3 i. When x = 15 then z = 1 12/13/2024 Dr. Anil Agarwal Unit-III 51  Cont…(CO3) Therefore 𝑃 𝑥 > 15 = 𝑃(𝑧 > 1) =𝑃 0