MGMT305-Lec3

# MGMT305-Lec3 - Chapter 6(6.2 Normal Probability...

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Unformatted text preview: Chapter 6 (6.2) Normal Probability Distribution s s Continuous Probability Distributions Normal Probability Distribution f( x ) µ x Slide 1 Continuous Probability Distributions s s s s A continuous random variable can assume any value in an interval on the real line or in a collection of intervals. Here, we will not talk about the probability of the random variable assuming a particular value. Instead, we talk about the probability of the random variable assuming a value within a given interval. The probability of the random variable assuming a value within some given interval from x1 to x2 is defined to be the area under the graph of the probability density function between x1 and x2. Slide 2 Example s Let X be the height of a student (measured exactly). • • How much is P (X = 65.3”)? What is P (62” < X < 70”)? Slide 3 Normal Probability Distribution s Graph of the Normal Probability Density Function f( x ) µ x Slide 4 Normal Probability Distribution s Normal Probability Density Function 1 − ( x − µ )2 / 2σ2 f ( x) = e 2 πσ where: µ = mean σ = standard deviation π = 3.14159 e = 2.71828 Slide 5 Normal Probability Distribution s Characteristics of the Normal Probability Distribution • The shape of the normal curve is often illustrated as a bell­shaped curve. • Two parameters, µ (mean) and σ (standard deviation), determine the location and shape of the distribution. • The highest point on the normal curve is at the mean, which is also the median and mode. • The mean can be any numerical value: negative, zero, or positive. … continued Slide 6 Normal Probability Distribution s Characteristics of the Normal Probability Distribution • The normal curve is symmetric. • The standard deviation determines the width of the curve: larger values result in wider, flatter curves. • The total area under the curve is 1 (.5 to the left of the mean and .5 to the right). • Probabilities for the normal random variable are given by areas under the curve. Slide 7 Normal Probability Distribution s % of Values in Some Commonly Used Intervals • 68.26% of values of a normal random variable are within +/­ 1 standard deviation of its mean. • 95.44% of values of a normal random variable are within +/­ 2 standard deviations of its mean. • 99.72% of values of a normal random variable are within +/­ 3 standard deviations of its mean. Slide 8 Normal Probability Distribution s % of Values in Some Commonly Used Intervals 99.72% 95.44% 68.26% μ-3σ μ-2σ μ-σ μ μ+σ μ+2σ μ+3σ s Where do the values of the random variable X lie? Is it on the x­axis or the region above the x­axis? Slide 9 Standard Normal Probability Distribution s s 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. Converting to the Standard Normal Distribution x−µ z= σ s Standard Normal Probability Density Function 1 − z2 / 2 f ( z) = e , 2π s notation N (0,1) x = µ + zσ , x is z standard deviations away from µ . Slide 10 Using the Standard Normal Distribution Table s Example 1. Find the following probabilities, where Z is the standard normal random variable. a. P(­.82 ≤ Z ≤ 1.36) a. P(.47 ≤ Z ≤ 2.12) Slide 11 Using the Standard Normal Distribution Table s Example 2. If Z is a standard normal random variable, find the value z0 for which: a. a. a. P(­ z0 ≤ Z ≤ z0) = .75 P(Z ≥ z0) = .123 P(Z ≥ z0) = .67 Slide 12 Probabilities for any Normal Random Variable Suppose we want to compute P(x1≤ X ≤ x2) s STEPS: 1. Subtract µ from each term, then P(x1≤ X ≤ x2) = P(x1 ­ µ ≤ X ­ µ ≤ x2 ­ µ ) s 2. Divide each term by σ , then P(x1≤ X ≤ x2) = P((x1 ­ µ )/σ ≤ (X ­µ )/σ ≤ (x2 ­ µ )/σ ) Let Z = (X ­ µ )/σ and z1 = (x1 ­ µ )/σ , z2 = (x2 ­ µ )/σ Then P(x1 ≤ X ≤ x2) = P(z1 ≤ Z ≤ z2) 3. Use standard normal table to compute P(z1 ≤ Z ≤ z2 ). Slide 13 Example: Pep Zone Pep Zone sells auto parts and supplies including a popular multi­grade 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. The manager would like to know the probability of a stockout, P(X > 20). a. What is the value of P(X > 20)? Slide 14 Example: Pep Zone s Standard Normal Probability Distribution The Standard Normal table shows an area of .2967 for the region between the z = 0 and z = .83 lines below. The shaded tail area is .5 ­ .2967 = .2033. The probability of a stock­out is .2033. Area = .2967 Area = .5 ­ .2967 = .2033 Area = .5 0 .83 z Slide 15 Example: Pep Zone s Using the Standard Normal Probability Table .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 z .0 .0000 .0040 .0080 .0120 .0160 .0199 .0239 .0279 .0319 .0359 .1 .0398 .0438 .0478 .0517 .0557 .0596 .0636 .0675 .0714 .0753 .2 .0793 .0832 .0871 .0910 .0948 .0987 .1026 .1064 .1103 .1141 .3 .1179 .1217 .1255 .1293 .1331 .1368 .1406 .1443 .1480 .1517 .4 .5 .6 .7 .8 .9 .1554 .1591 .1628 .1664 .1700 .1736 .1772 .1808 .1844 .1879 .1915 .1950 .1985 .2019 .2054 .2088 .2123 .2157 .2190 .2224 .2257 .2291 .2324 .2357 .2389 .2422 .2454 .2486 .2518 .2549 .2580 .2612 .2642 .2673 .2704 .2734 .2764 .2794 .2823 .2852 .2881 .2910 .2939 .2967 .2995 .3023 .3051 .3078 .3106 .3133 .3159 .3186 .3212 .3238 .3264 .3289 .3315 .3340 .3365 .3389 Slide 16 Example: Pep Zone b. If the manager of Pep Zone wants the probability of a stockout to be no more than .05, what should the reorder point be? Area = .05 Area = .5 Area = .45 z.05 0 Let z.05 represent the z value cutting the .05 tail area. Slide 17 Example: Pep Zone s Using the Standard Normal Probability Table We now look­up the .4500 area in the Standard Normal Probability table to find the corresponding z.05 value. z . .00 . .01 . .02 . .03 . .04 . .05 . .06 . .07 . .08 . .09 . 1.5 .4332 .4345 .4357 .4370 .4382 .4394 .4406 .4418 .4429 .4441 1.6 .4452 .4463 .4474 .4484 .4495 .4505 .4515 .4525 .4535 .4545 1.7 .4554 .4564 .4573 .4582 .4591 .4599 .4608 .4616 .4625 .4633 1.8 .4641 .4649 .4656 .4664 .4671 .4678 .4686 .4693 .4699 .4706 1.9 .4713 .4719 .4726 .4732 .4738 .4744 .4750 .4756 .4761 .4767 . . . . . . . . . . . z.05 = 1.645 is a reasonable estimate. Slide 18 Example: Pep Zone s Original Normal Random Variable The corresponding value of x is given by x = µ + z.05σ A reorder point of ? gallons will place the A reorder point of ? gallons will place the probability of a stockout during leadtime at .05. Perhaps Pep Zone should set the reorder point at ? gallons to keep the probability under .05. Slide 19 ...
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## This note was uploaded on 12/04/2009 for the course MGMT 305 taught by Professor Priya during the Summer '08 term at Purdue.

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