Bstat5 - Lessons in Business Statistics Prepared By P.K...

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Unformatted text preview: Lessons in Business Statistics Prepared By P.K. Viswanathan Chapter 5: Probability Distributions Introduction Introduction: Very often we want to describe and analyze numbers in the form of a distribution. You have seen this sharply in chapter 2 and chapter 3. Distributions can be formed by two methods; 1) from data collected and 2) with the help of standard distributions that have stood the test of time. This chapter provides the conceptual framework and applications of the three widely used probability distributions namely The Binomial Distribution, The Poisson Distribution, and The Normal Distribution. While the first two are discrete in nature, the last one is a continuous distribution. 1) What is a Probability Distribution? You rotate a pair of dice and observe the sum by totaling the numbers that turn up in both the dice. Let us designate this sum by a random variable X. When the first die shows up number 1, the second die could be 1, 2, 3, 4,5,6. Likewise when the first die shows up number 2, the second die could be 1, 2, 3, 4, 5, 6. If you continue this experiment this way until all the possibilities are exhausted, the sample space will contain 36 elements. The distribution that can be observed by us will contain a sum =2, 3,4,5,6,7,8,9,10,11,12. Please note that the starting sum will be =2 because when both the dice show up number 1, the sum =1+1 =2. Likewise, the maximum sum in the distribution will be 12 when both the dice show up a number 6. If we write the values of the random variable (sum) with associated probabilities, it will become what is called a probability distribution. Probability Distribution for the Example X=Sum P(X) 2 1/36 3 2/36 4 3/36 5 4/36 6 5/36 7 6/36 8 5/36 9 4/36 10 3/36 11 2/36 12 1/36 Please note that the probability of the sample space =1. (Adding all the probabilities above you get 1) Definition of Probability Distribution In precise terms, a probability distribution is a total listing of the various values the random variable can take along with the corresponding probability of each value. A real life example would be the pattern of distribution of the machine breakdowns in a manufacturing unit. The random variable in this example would be the various values the machine breakdowns could assume. The probability corresponding to each value of the breakdown is the relative frequency of occurrence of the breakdo wn. The probability distribution for this example is constructed by the actual breakdown pattern observed over a period of time. Statisticians use the term “observed distribution” of breakdowns. 2) Binomial Distribution The Binomial Distribution is a widely used probability distribution of a discrete random variable. It plays a major role in quality control and quality assurance function. Manufacturing units do use the binomial distribution for defective analysis. An item is considered defective if it has one or more types of defects. Reducing the number of defectives using the proportion defective control chart (p chart) is an accepted practice in manufacturing organizations. Binomial distribution is also being used in service organizations like banks, and insurance corporations to get an idea of the proportion customers who are satisfied with the service quality. Conditions for Applying Binomial Distribution (Bernoulli Process) Trials are independent and random. There are fixed number of trials (n trials). There are only two outcomes of the trial designated as success or failure. The probability of success is uniform through out the n trials Binomial Probability Function Binomial Probability Function Under the conditions of a Bernoulli process, The probability of getting x successes out of n trials is indeed the definition of a Binomial Distribution. The Binomial Probability Function is given by the following expression n x P(x) (1 p) n x p x Where P(x) is the probability of getting x successes in n trials n is the number of ways in which x successes can take place out of n trials x n! = x!(n x)! p is the probability of success, which is the same through out the n trials. p is the parameter of the Binomial distribution x can take values 0, 1, 2, ………., n Example for Binomial Distribution Example: A bank issues credit cards to customers under the scheme of Master Card. Based on the past data, the bank has found out that 60% of all accounts pay on time following the bill. If a sample of 7 accounts is selected at random from the current database, construct the Binomial Probability Distribution of accounts paying on time. Solution This problem can be structured as a Bernoulli Process where an account paying on time is taken as success and an account not paying on time is taken as failure. The random variable x represents here an account paying on time, which can take values 0,1,2,3,4,5,6,7. You need to prepare a table containing x and P(x) for all the values of x. Performing calculations using Binomial Probability Function n x P(x) (1 n for x = 0,1,2,3,4,5,6,7 is very tedious. p p) x x The best option is to use the Electronic Spread sheet of Microsoft Excel to calculate the Binomial Probabilities both for individual values and for the cumulative position. This facility is available under the option “Paste Function”. The form of the function is: =BINOMDIST(x, n, p, 0 or 1) where x is the number of successes, n is the number of trials, and p is the probability of success in each trial. The last term 0 or 1 performs a logical operation. If you enter 0, the computer returns the individual probability value; if 1 is entered, the computer gives the cumulative probability. Spreadsheet Showing the Solution Mean and Standard Deviation of the Binomial Distribution The mean of the Binomial Distribution is given by = E(x)= np The Standard Deviation is given by = np(1 p) For the example problem in the previous two slides, Mean =7 =4.2. 0.6 4.2(1 0.60) = 1.30 Standard Deviation = 3) Poisson Distribution Poisson Distribution is another discrete distribution which also plays a major role in quality control in the context of reducing the number of defects per standard unit such as number of defects per item, number of defects per transformer produced, number of defects per 100 m2 of cloth, etc. Other real life examples would include 1) The number of cars arriving at a highway check post per hour; 2) The number of customers visiting a bank per hour during peak business period. Poisson Process The probability of getting exactly one success in a continuous interval such as length, area, time and the like is constant The probability of a success in any one interval is independent of the probability of success occurring in any other interval. The probability of getting more than one success in an interval is 0. Poisson Probability Function Poisson Distribution Formula x e λλ P(x) x! where P(x) = Probability of x successes given an idea of λ λ Average number of successes = e = 2.71828(based on natural logarithm) x = successes per unit which can take values 0, 1, 2, 3,………….. λis the Parameter of the Poisson Distribution. Mean of the Poisson Distribution is = λ Standard Deviation of the Poisson Distribution is = λ Example If on an average, 6 customers arrive every two minutes at a bank during the busy hours of working, a) what is the probability that exactly four customers arrive in a given minute? b) What is the probability that more than three customers will arrive in a given minute? Solution λ 3 ( 6 customers every two minutes implies 3 customers per = minute) a) You are required to find out P(x =4). Substituting in the formula the value of x =4, you can get the answer after e λ x simplifying the expression P(x) λ x! However, you are strongly encouraged to use the Microsoft Excel. Standard Poisson table or manual calculation cannot match “Excel”. Use the Paste Function option and then click Poisson. Follow the prompts, you get the answer. Just like the Binomial Distribution, here also the logical operator 0 gives the individual probabilities, and 1 gives the cumulative probabilities. The output from Excel is given in the next slide for the values of x ranging from 0 to 10 Spreadsheet showing Solution Look at the table and you will find that corresponding to x =4 in column B, the probability P(x=4) is 0.168031 found in column C. You want P(x > 3) = 1- P(x < = 3) = 1 -0.647232 = 0.352768 ( Please note that P(x< = 3) is found in Column D corresponding to the value of x =3 in column B. You could see 0.647232 in row 8 and column D. 4) Normal Distribution The Normal Distribution is the most widely used continuous distribution. It occupies a unique place in the field of statistics. In fact, the entire quality control function that employs the statistical techniques heavily will come to a grinding halt without the use of the normal distribution. The control charts for reducing and stabilizing variation relies on the normal distribution. Process capability studies to meet the customer specifications needs the normal distribution. The whole subject matterinferential statistics is based on the normal distribution. In all management functions including the human side, the observed frequency distributions encountered are all fairly close to the normal distribution when the sample size is reasonably large. Properties of Normal Distribution The normal distribution is a continuous distribution looking like a bell. Statisticians use the expression “Bell Shaped Distribution”. It is a beautiful distribution in which the mean, the median, and the mode are all equal to one another. It is symmetrical about its mean. If the tails of the normal distribution are extended, they will run parallel to the horizontal axis without actually touching it. (asymptotic to the x-axis) The normal distribution has two parameters namely the mean and the standard deviation Normal Probability Density Function In the usual notation, the probability density function of the normal distribution is given below: f ( x) 1 e σ 2π (x ) 2 μ 2 2σ x is a continuous normal random variable with the property x meaning x can take all real numbers in the interval x . You need not worry about this complicated function because all calculations can be done without the help of this expression. What is important is that you should understand the normal distribution and you should be able to apply it to practical problems. For this purpose, the knowledge of the actual probability density function is not required. If x follows a normal distribution with mean and standard deviation it is a , general practice among st atisticians to symbolize this expression as x~ N (). , Property of Symmetry Property of Symmetry-Continues Property of Symmetry-Continues Standard Normal Distribution For practical problems, which require the help of the normal distribution for solutions, it is not possible to compute the probability in a straightforward manner because each of those normal variables may have different units of measurements. One problem may involve the unit meter, another may involve kilograms, and so on. Therefore you need a probability distribution that can tackle any unit of measurement. Fortunately, you have the Standard Normal Distribution defined as follows: x μ z σ z is called the standard normal variable. Please note that z is a pure number independent of the unit of measurement. The random variable z follows a normal distribution with mean=0 and standard deviation =1. You first convert the original variable in a given problem into z. It is only a transformation of scale. The probability table for z is available for computing the necessary probabilities for a given situation. The elegant way of getting the probability value based on z is to take advantage of the Microsoft Excel. Example Problem The mean weight of a morning breakfast cereal pack is 0.295 kg with a standard deviation of 0.025 kg. The random variable weight of the pack follows a normal distribution. a) b) c) What is the probability that the pack weighs less than 0.280 kg? What is the probability that the pack weighs more than 0.350 kg? What is the probability that the pack weighs between 0.260 kg to 0.340 kg? Solution-a) x μ z = (0.280-0.295)/0.025 = -0.6. Click “Paste Function” of Microsoft Excel, σ then click the “statistical” option, then click the standard normal distribution option and enter the z value. You get the answer. Excel accepts directly both the negative and positive values of z. Excel always returns the cumulative probability value. When z is negative, the answer is direct. When z is positive, the answer is =1- the probability value returned by Excel. The answer for part a) of the question = 0.2743(Direct from Excel since z is negative). This says that 27.43 % of the packs weigh less than 0.280 kg. Solution-b) x μ z = (0.350- 0.295)/0.025 =2.2. Excel returns a value of 0.9861. Since z is σ positive, the required probability is = 1-0.9861 =0.0139. This means that 1.39% of the packs weigh more than 0.350 kg. Solution-c) For this part, you have to first get the cumulative probability up to 0.340 kg and then x μ subtract the cumulative probability up to 0.260. z = (0.340- 0.295)/0.025 σ x μ =1.8(up to 0.340). z = (0.260 -0.295)/0.025 = - 1.4(up to 0.260). These two σ probabilities from Excel are 0.9641 and 0.0808 respectively. The answer is = 0.96410.0808 = 0.8833. This means that 88.33% of the packs weigh between 0.260 kg and 0.340 kg. ...
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This note was uploaded on 02/24/2012 for the course BUSINESS 281 taught by Professor Gray during the Spring '12 term at Florida State College.

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