Normal Distribution

Before Starting Normal Distribution P(8<x < 12) = ? x f (x ) 5 15 12 1/10 P( < 12) = ? P( < 8) = ? x f (x ) 5 15 12 1/10 P(8<x < 12) = .7-.3 = .4

The Normal Probability Distribution Graph of the Normal Probability Density Function f (x ) x 

The Normal Curve The shape of the normal curve is often illustrated as a bell-shaped curve. The highest point on the normal curve is at the mean of the distribution. The normal curve is symmetric. The standard deviation determines the width of the curve.

The Normal Curve The total area under the curve the same as any other probability distribution is 1. The probability of the normal random variable assuming a specific value the same as any other continuous probability distribution is 0. Probabilities for the normal random variable are given by areas under the curve.

The Normal Probability Density Function where  = mean  = standard deviation  = 3.14159 e = 2.71828

The Standard Normal Probability Density Function where  = 0  = 1  = 3.14159 e = 2.71828

Given any positive value for z, the table will give us the following probability The table will give this probability Given positive z The probability that we find using the table is the probability of having a standard normal variable between 0 and the given positive z.

Given z find the probability

Given any probability between 0 and Given any probability between 0 and .5,, the table will give us the following positive z value Given this probability between 0 and .5 The table will give us this positive z

Given the probability find z find

What is the z value where probability of a standard normal variable to be greater than z is .1 10% 40%

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Standard Normal Probability Distribution (Z Distribution)

Standard Normal Probability Distribution A random variable that has a normal distribution with a mean of zero and a standard deviation of one is said to have a standard normal probability distribution. The letter z is commonly used to designate this normal random variable. The following expression convert any Normal Distribution into the Standard Normal Distribution

Example: Pep Zone Pep Zone sells auto parts and supplies including multi-grade motor oil. When the stock of this oil drops to 20 gallons, a replenishment order is placed. The store manager is concerned that sales are being lost due to stockouts while waiting for an order. It has been determined that leadtime demand is normally distributed with a mean of 15 gallons and a standard deviation of 6 gallons. In Summary; we have a N (15, 6): A normal random variable with mean of 15 and std of 6. The manager would like to know the probability of a stockout, P(x > 20).

Standard Normal Distribution z = (x -  )/ = (20 - 15)/6 = .83 Area = .5 z .83

Example: Pep Zone

The Probability of Demand Exceeding 20 .83 Area = .2967 Area = .5 Area = .2033 z The Standard Normal table shows an area of .2967 for the region between the z = 0 line and the z = .83 line above. The shaded tail area is .5 - .2967 = .2033. The probability of a stockout is .2033.

Example: Pep Zone If the manager of Pep Zone wants the probability of a stockout to be no more than .05, what should the reorder point be? Let z.05 represent the z value cutting the tail area of .05. Area = .05 Area = .5 Area = .45 z.05

Example: Pep Zone Using the Standard Normal Probability Table We now look-up the .4500 area in the Standard Normal Probability table to find the corresponding z.05 value. z.05 = 1.645 is a reasonable estimate.

Example: Pep Zone The corresponding value of x is given by x =  + z.05 = 15 + 1.645(6) = 24.87 A reorder point of 24.87 gallons will place the probability of a stockout during leadtime at .05. Perhaps Pep Zone should set the reorder point at 25 gallons to keep the probability under .05.

Example: Aptitude Test A firm has assumed that the distribution of the aptitude test of people applying for a job in this firm is normal. The following sample is available. 71 66 61 65 54 93 60 86 70 70 73 73 55 63 56 62 76 54 82 79 76 68 53 58 85 80 56 61 61 64 65 62 90 69 76 79 77 54 64 74 65 65 61 56 63 80 56 71 79 84

Example: Mean and Standard Deviation We first need to estimate mean and standard deviation

z Values What test mark has the property of having 10% of test marks being less than or equal to it To answer this question, we should first answer the following What is the standard normal value (z value), such that 10% of z values are less than or equal to it? 10%

z Values We need to use standard Normal distribution in Table 1. 10%

z Values 10% 40%

z Values 40% z = 1.28 10% z = - 1.28

z Values and x Values The standard normal value (z value), such that 10% of z values are less than or equal to it is z = -1.28 To transform this standard normal value to a similar value in our example, we use the following relationship The normal value of test marks such that 10% of random variables are less than it is 55.1.

z Values and x Values Following the same procedure, we could find z values for cases where 20%, 30%, 40%, …of random variables are less than these values. Following the same procedure, we could transform z values into x values. Lower 10% -1.28 55.1 Lower 20% -.84 59.68 Lower 30% -.52 63.01 Lower 40% -.25 65.82 Lower 50% 0 68.42 Lower 60% .25 71.02

Example : Victor Computers Victor Computers manufactures and sells a general purpose microcomputer. As part of a study to evaluate sales personnel, management wants to determine if the annual sales volume (number of units sold by a salesperson) follows a normal probability distribution. A simple random sample of 30 of the salespeople was taken and their numbers of units sold are below. 33 43 44 45 52 52 56 58 63 64 64 65 66 68 70 72 73 73 74 75 83 84 85 86 91 92 94 98 102 105 (mean = 71, standard deviation = 18.54) Partition this Normal distribution into 6 equal probability parts

Example : Victor Computers Areas = 1.00/6 = .1667 53.02 71 88.98 = 71 + .97(18.54) 63.03 78.97

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Normal Approximation of Binomial Probabilities When the number of trials, n, becomes large, evaluating the binomial probability function by hand or with a calculator is difficult. The normal probability distribution provides an easy-to-use approximation of binomial probabilities where n > 20, np > 5, and n(1 - p) > 5. Set  = np Add and subtract a continuity correction factor because a continuous distribution is being used to approximate a discrete distribution. For example, P(x = 10) is approximated by P(9.5 < x < 10.5).

The Exponential Probability Distribution Exponential Probability Density Function for x > 0,  > 0 where  = mean e = 2.71828 Cumulative Exponential Distribution Function where x0 = some specific value of x

Example: Al’s Carwash The time between arrivals of cars at Al’s Carwash follows an exponential probability distribution with a mean time between arrivals of 3 minutes. Al would like to know the probability that the time between two successive arrivals will be 2 minutes or less. P(x < 2) = 1 - 2.71828-2/3 = 1 - .5134 = .4866

Example: Al’s Carwash Graph of the Probability Density Function F (x ) .1 .3 .4 .2 1 2 3 4 5 6 7 8 9 10 P(x < 2) = area = .4866

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