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Theoretical Distributions Frequency Distribution: • • Observed Frequency Distribution Theoretical Frequency Distribution Theoretical Distributions: 1. 2. 3. 4. 5. 6. Binomial Distributions Multinomial Distributions Negative Binomial Distributions Poisson Distribution Hyper Geometric Distribution Normal Distributions A Discrete distribution is based on random variables which can assume only clearly separated values. A Continuous distribution usually results from measuring something. Discrete distributions studied include: o Binomial o Hypergeometric o Poisson. Continuous distributions include: o Uniform o Normal o Others Discrete and continuous distributions Binomial Distribution: “It is a discreet probability distribution which expresses the probability of one set of dichotomous alternative, ie. Success (p) and failure (q).” Binomial Distribution: • • Also Known as Bernoulli Distribution Given By Swiss Mathematician Jacob Bernoulli Assumptions: 1. 2. 3. 4. 5. An experiment is performed under same condition for a fixed no. of trials. In each trial, there are only two possible outcomes of experiment i.e.. Success or Failure. The probability of Success is denoted by “p”and it remains constant from trial to trial. The probability of Failure is denoted by “q” which is equal to (1-p) The trials are statistically independent i.e. the outcome of any trial do not effect the outcome of subsequent trials. The Binomial Distribution: P(r) = nCr q n-r p r Where p = probability of success q = Probability of failure (1-p) n = no. of trials r = no. of success in ‘n’ trials Constants of Binomial Distribution: Mean = np Standard Deviation = npq Measures of Skew ness = 1 = (q-p)2 / npq Measures of Kurtosis = 2 = 3 + (1-6pq/npq) Examples: 1. A coin is tossed 6 times. What is the probability of obtaining 4 or more heads. 2. Assuming that half the population is vegetarian so that the chance of an individual being vegetarian is ½ and assuming that 100 investigators make a sample of 10 individuals to see whether they are vegetarian. How many investigators would you expect to report that three or less were vegetarian. 3. If p = 0.4 and n = 50, Calculate the constants of Binomial Distribution. Seven coins are tossed and the number of heads were noted. This experiment was repeated 128 times and the following observations were made. No. of head 0 1 2 3 4 5 6 7 Throws 7 6 19 35 30 23 7 1 Fit a binomial distribution assuming the coin to be unbiased. Also find its mean and standard deviation. Solution: Given: N = 128 n=7 p=½ q = (1-1/2)= ½ For fitting a Binomial distribution: No. of Heads Expected Frequencies [ N x (nCr q n-r p r) 0 128 x 7C0 q 7-0 p 0 = 1 1 128 x 7C1 q 7-1 p 1 = 7 2 128 x 7C2 q 7-2 p 2 = 21 3 128 x 7C3 q 7-3 p 3 = 35 4 128 x 7C4 q 7-4 p 4 = 35 5 128 x 7C5 q 7-5 p 5 = 21 6 128 x 7C6 q 7-6 p 6 = 7 7 128 x 7C7q 7-7 p 7 = 1 Poisson Distribution: • Discrete Probability Distribution • Given By French Mathematician Simeon Denis Poison • It is expected where chance of an individual event being success is very small. • It is a distribution used to describe the behavior of rare events i.e. no. of printing errors in books, serious floods. It is defined as: P(r) = e-m mr / r ! Where r = 0, 1, 2……… e = 2.7183 m = mean of Poisson Distribution = np If we want to know the expected no. of occurrences for different success, we have to multiply each term by N Which is the total no. of observations Constants of Poisson Distribution: Mean = np Standard Deviation = np 1 = 1/ mean 2 = 3 + 1/Mean “ In General, the Poisson Distribution explains the behavior of those Discrete variables where the probability of occurrence of the event is small and the total no. of possible cases is sufficiently large.” Examples: 1. The mean of Poisson Distribution is 2.25. Find other constants. 2. Suppose on an average, one house in thousand in a certain district has a fire during a year. If there are two thousand houses in that district, what is the probability that exactly five houses will have a fire during the year. (Take e-2 = 0.1352) 3. Ten percent of the tools produced in a certain manufacturing process turn out to be defective. Find the probability that in a sample of 10 tools chosen at random, exactly two will be defective by using: • The Binomial Distribution • The Poisson Distribution Normal Probability Distribution: • An important Continuous Probability Distribution • The Graphical shape of normal Distribution called normal curve, is bell shaped smooth symmetrical curve. • The normal curve or distribution is a theoretical model which may be used to describe the frequency distribution of vast variety of continuous variables. • A continuous random variable is said to be normally distributed if: P(x) = 1 2 e –1/2 (x- / ) 2 r a l i t r b u i o n : = 0 , 2 = 1 Characteristics of a Normal Distribution 0 . 4 Normal curve is symmetrical . 3 0 . 2 0 . 1 f ( x 0 Theoretically, curve extends to infinity . 0 - 5 a Mean, median, and mode are equal x Properties of Normal Distribution: • The normal curve is ‘bell shaped’ and symmetrical in its appearance. • If the curves were folded along its vertical axis, the two halves would coincide. • There is only one maximum point of normal curve which occurs at mean. • Since there is one maximum point, the normal curve is unimodal. • Because of Symmetry: Mean = Median = Mode. • Curve is asymptotic to X- axis. • Total area under normal curve and above the horizontal axis is 1.00 Area under normal curve is distributed as: 1. Mean 1 covers, 68.27% area 2. Mean 2 covers, 95.45% area 3. Mean 3 covers, 99.73% area Area Relationship: Distance from Mean % of Total Area 0.5 19.46% 1.0 34.134% 1.96 47.5% 2.00 47.725% 2.5758 49.5% 3.00 49.865% The various hypothesis are generally tested at 5% level Or 1% level (ie. Taking account 95% and 99% of total area Of the normal curve. Area under normal Curve: It is possible to transform any normal random variable X With mean and variance 2 to a new normal random Variable Z with mean 0 and variance 1. This variable is known as Standard Normal Variable (SNV). The transformation of X to Z is given as Z=X- / Steps: • • Use transformation as above for converting the given normal random variable X to SNV Z. Use the table to calculate area or probability in the table areas under the standard normal curve. z X = 22000 - 20000 2000 = 1.00 The bi-monthly starting salaries of recent MBA graduates follows the normal distribution with a mean of Rs.20,000 and a standard deviation of Rs.2000. What is the z-value for a salary of Rs.22,000? MBA Appendix: Area Under Normal Curve Table of Area z 0 1 2 3 4 5 6 7 8 9 0.0 .0000 .0040 .0080 .0120 .0160 .0199 .0239 .0279 .0319 .0359 0.1 .0398 .0438 .0478 .0517 .0557 .0596 .0636 .0675 .0714 .0753 0.2 .0793 .0832 .0871 .0910 .0948 .0987 .1026 .1064 .1103 .1141 1.0 .3413 .3438 .3461 .3485 .3508 .3531 .3554 .3577 .3599 .3621 2.7 .4965 .4966 .4967 .4968 .4969 .4970 .4971 .4972 .4973 .4974 2.8 .4974 .4975 .4976 .4977 .4977 .4978 .4979 .4979 .4980 .4981 Examples: 1. X is a normal variable with mean = 25 and standard deviation = 5. Find the values of Z1 and Z2, such that P(20<X<30) = P(Z1<Z<Z2) 2. Z is a standard normal variable, use table to determine the following: • • • • P(0<Z<1.2) P(-1.2<Z<0) P(-2<Z<2) P(-<Z<1) Example: The average on a statistics test was 78 with a standard deviation of 8. If the test scores are normally distributed, find the probability that a student receives a test score less than 90. μ = 78 σ=8 z x - μ = 90 - 78 σ 8 = 1.5 P(x < 90) μ =78 90 μ =0 ? 1.5 x z The probability that a student receives a test score less than 90 is 0.8531 P(x < 90) = P(z < 1.5) = 0.5 + .3531 = 0.8531 Example: The average on a statistics test was 78 with a standard deviation of 8. If the test scores are normally distributed, find the probability that a student receives a test score greater than than 85. μ = 78 σ=8 P(x > 85) μ =78 85 μ =0 0.88 ? z = x - μ = 85 - 78 σ 8 = 0.875 0.88 x z The probability that a student receives a test score greater than 85 is 0.1894. P(x > 85) = P(z > 0.88) = .5 P(z < 0.88) = .5 0.3106 = 0.1894 Example: The average on a statistics test was 78 with a standard deviation of 8. If the test scores are normally distributed, find the probability that a student receives a test score between 60 and 80. z = x - μ = 60 - 78 = -2.25 P(60 < x < 80) μ = 78 σ=8 60 σ 8 z 2 x - μ = 80 - 78 σ 8 1 μ =78 80 2.25 μ =0 0.25 ? ? x z = 0.25 The probability that a student receives a test score between 60 and 80 is 0.5865. P(60 < x < 80) = P(2.25 < z < 0.25) = .4878 +.0987 = 0.5865 3. Two thousand students appear in an examination. Distribution of marks is assumed to be normal with mean = 30 and standard deviation = 6.25. How many students are expected to get • Between 20 and 40 marks. • Less than 35 marks. 4. Assuming mean height of soldiers to be 68.22 inches with a variance of 10.8 inches. How many soldiers in a regiment of 1000, would you expect to be over 6 feet tall? 5. The local authorities in a certain district install 10000 electric lamps in the streets of the city. If these lamps have an average life of 1000 burning hours with standard deviation of 200 hours. Assuming normality, what number of lamps might be expected to fall • in the first 800 burning hours • between 800 and 1200 burning hours