chap7 - Business Statistics (BUSA 3101) Dr. Lari H. Arjomand

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Unformatted text preview: Business Statistics (BUSA 3101) Dr. Lari H. Arjomand [email protected] Slide 1 Please…Not during this lecture… Slide 2 Someone told me this is an important chapter! an Chapter 7 Continuous Probability Distributions s s s f (x) Uniform Probability Distribution Normal Probability Distribution Exponential Probability Distribution (Optional) f (x) Uniform f (x) Exponential Normal x x x Slide 4 Continuous Probability Distributions s A continuous random variable can assume any value in an interval on the real line or in a collection of intervals. s It is not possible to talk about the probability of the continuous random variable assuming a particular value. s Instead, we talk about the probability of the random variable assuming a value within a given interval. Slide 5 Data Types Data Numerical (Quantitative) Discrete Categorical (Qualitative) Continuous Slide 6 Continuous Random Variable Examples Experiment Random Variable Possible Values Weigh 100 people Weight 45.1, 78, ... Measure part life Hours 900, 875.9, ... Ask food spending Spending 54.12, 42, ... Measure time between Inter-arrival 0, 1.3, 2.78, ... arrivals time Slide 7 Continuous Probability Distribution Models In this Continuous Probability Distribution Chapter Uniform Normal Exponential Other Slide 8 Continuous Probability Distributions s 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. Uniform f (x) x x 1 2 f (x) x Normal x x 1 x 2 Slide 9 Uniform Probability Distribution s A random variable is uniformly distributed whenever the probability is proportional to the interval’s length. s The uniform probability density function is: f ((x) = 1/(b – a) for a < x < b f x) = 1/(b – a) for a < x < b f ((x) = 0 elsewhere f x) = 0 elsewhere where: a = smallest value the variable can assume b = largest value the variable can assume Slide 10 Uniform Probability Distribution s Expected Value (mean) of x E(x) = (a + b)/2 E(x) = (a + b)/2 s Variance of x 2 Var(x) = (b ­ a))2/12 Var(x) = (b ­ a /12 Slide 11 Uniform Probability Distribution s Example: Slater's Buffet Slater customers are charged for the amount of salad they take. Sampling suggests that the amount of salad taken is uniformly distributed between 5 ounces and 15 ounces. f ((x) = 1/(b – a) for a < x < b f x) = 1/(b – a) for a < x < b f ((x) = 0 elsewhere f x) = 0 elsewhere Slide 12 Uniform Probability Distribution s Uniform Probability Density Function ff((x) = 1/10 for 5 < x < 15 x) = 1/10 for 5 < x < 15 f ((x) = 0 elsewhere f x) = 0 elsewhere where: x = salad plate filling weight Slide 13 Uniform Probability Distribution s Expected Value of x E(x) = (a + b)/2 E(x) = (a + b)/2 = (5 + 15)/2 = (5 + 15)/2 = 10 = 10 s Variance and standard deviation of x Var(x) = (b ­ a))22/12 Var(x) = (b ­ a /12 = (15 – 5)22/12 = (15 – 5) /12 = 8.33 = 8.33 S = 2.887 Slide 14 Uniform Probability Distribution s Uniform Probability Distribution for Salad Plate Filling Weight f( x ) 1/10 5=a 10 15=b Salad Weight (oz.) Area = ( height) x (base) x Slide 15 Uniform Probability Distribution What is the probability that a customer will take between 12 and 15 ounces of salad? f( x ) P(12 < x < 15) = 1/10(3) = .3 1/10 5 10 12 15 Salad Weight (oz.) Area = ( height) x (base) x Slide 16 Using SWStat+ to find P(12 < x < 15) = ? Slide 17 Using SWStat+ (Click on Show Plot ) Slide 18 Normal Probability Distribution s s The normal probability distribution is the most important distribution for describing a continuous random variable. It is widely used in statistical inference. f(X) X Mean Median Mode Slide 19 Normal Probability Distribution s It has been used in a wide variety of applications: Heights Heights of people of people Scientific Scientific measurements measurements Slide 20 Normal Probability Distribution s It has been used in a wide variety of applications: Test Test scores scores Amounts Amounts of rainfall of rainfall Slide 21 Normal Probability Distribution s Normal Probability Density Function 1 − ( x − µ )2 / 2σ 2 f (x ) = e σ 2π where: µ = mean σ = standard deviation π = 3.14159 e = 2.71828 Slide 22 Normal Probability Distribution s Characteristics 1­ The distribution is symmetric; its skewness 1­ The distribution is symmetric; its skewness measure is zero. measure is zero. x Slide 23 Normal Probability Distribution s Characteristics 2­ The entire family of normal probability 2­ The entire family of normal probability distributions is defined by its mean µ and its distributions is defined by its mean µ and its standard deviation σ . standard deviation σ . Standard Deviation σ Mean µ x Slide 24 Normal Probability Distribution s Characteristics 3­ The highest point on the normal curve is at the 3­ The highest point on the normal curve is at the mean, which is also the median and mode.. mean, which is also the median and mode x Slide 25 Normal Probability Distribution s Characteristics 4­ The mean can be any numerical value: negative, 4­ The mean can be any numerical value: negative, zero, or positive. zero, or positive. ­10 0 20 x Slide 26 Normal Probability Distribution s Characteristics 5­ The standard deviation determines the width of the 5­ The standard deviation determines the width of the curve: larger values result in wider, flatter curves. curve: larger values result in wider, flatter curves. σ = 15 σ = 25 x Slide 27 Normal Probability Distribution s Characteristics 6­ Probabilities for the normal random variable are 6­ Probabilities for the normal random variable are given by areas under the curve. The total area given by areas under the curve. The total area under the curve is 1 (.5 to the left of the mean and under the curve is 1 (.5 to the left of the mean and .5 to the right). .5 to the right). .5 .5 x Slide 28 Normal Probability Distribution s Characteristics #7 68.26% of values of a normal random variable 68.26% of values of a normal random variable +/­ 1 standard deviation are within of its mean. are within of its mean. +/­ 1 standard deviation 95.44% of values of a normal random variable 95.44% of values of a normal random variable +/­ 2 standard deviations are within of its mean. are within of its mean. +/­ 2 standard deviations 99.72% of values of a normal random variable 99.72% of values of a normal random variable +/­ 3 standard deviations are within of its mean. are within of its mean. +/­ 3 standard deviations Slide 29 Normal Probability Distribution s Characteristics #7 99.72% 95.44% 68.26% µ – 3σ µ – 1σ µ – 2σ µ µ + 3σ µ + 1σ µ + 2σ x Slide 30 Normal Probability Distribution d P( c ≤ X ≤ d ) = ∫ f ( X ) dx Probability is Probability area under curve! curve! c ? f(X) c d X Slide 31 Normal Probabilities Slide 32 Standard Normal Probability Distribution A random variable having a normal distribution A random variable having a normal distribution with a mean of 0 and a standard deviation of 1 is with a mean of 0 and a standard deviation of 1 is said to have a standard normal probability said to have a standard normal probability distribution.. distribution Slide 33 Standard Normal Probability Distribution The letter z is used to designate the standard The letter z is used to designate the standard normal random variable. normal random variable. σ= 1 0 z Slide 34 Standard Normal Probability Distribution s Converting to the Standard Normal Distribution x−µ z= σ We can think of z as a measure of the number of standard deviations x is from µ. We use the above equation to convert normal distribution into standard normal distribution. Slide 35 Standard Normal Probability Distribution s 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. Pep Zone 5w­20 Motor Oil Slide 36 Standard Normal Probability Distribution s Example: Pep Zone The store manager is concerned that sales are being lost due to stockouts while waiting for an order. It has been determined that demand during replenishment lead­time is normally Pep distributed with a mean of 15 gallons and Zone 5w­20 a standard deviation of 6 gallons. Motor Oil The manager would like to know the probability of a stockout, P(x > 20). Slide 37 Standard Normal Probability Distribution Pep Zone s Solving for the Stockout Probability 5w­20 Motor Oil Step 1: Convert x to the standard normal distribution. Step 1: Convert x to the standard normal distribution. z = (x ­ µ)/σ z = (x ­ µ)/σ = (20 ­ 15)/6 = (20 ­ 15)/6 = .83 = .83 Step 2: Find the area under the standard normal Step 2: Find the area under the standard normal curve to the left of zz = .83 curve to the left of = .83 see next slide see next slide Slide 38 Standard Normal Probability Distribution Pep Zone 5w­20 Motor Oil Cumulative Probability Table for the Standard Normal Distribution s z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 . . . . . . . . . . . .5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224 .6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549 .7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852 .8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133 .9 .8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389 . . . . . . . . . . . P(z < .83) Slide 39 Standard Normal Probability Distribution Pep Zone s 5w­20 Motor Oil Solving for the Stockout Probability Step 3: Compute the area under the standard normal Step 3: Compute the area under the standard normal curve to the right of zz = .83 curve to the right of = .83 P((z > .83) = 1 – P((z < .83) P z > .83) = 1 – P z < .83) = 1­ .7967 = 1­ .7967 = .2033 = .2033 Probability of a stockout P(x > 20) Slide 40 Standard Normal Probability Distribution Pep Zone s 5w­20 Motor Oil Solving for the Stockout Probability Area = 1 ­ .7967 Area = .7967 P (x > 20)= .2033 0 .83 z Slide 41 Standard Normal Probability Distribution s 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. Pep Zone 5w­20 Motor Oil Slide 42 Standard Normal Probability Distribution s Example: Pep Zone The store manager is concerned that sales are being lost due to stockouts while waiting for an order. It has been determined that demand during replenishment lead­time is normally Pep distributed with a mean of 15 gallons and Zone a standard deviation of 6 gallons. 5w­20 Motor Oil The manager would like to know the probability of a stockout, P(x > 20). Students: Used Excel (SWStat) to solve this problem Slide 43 Using SWStat+ (Choose Statistics; Probability Distribution) Slide 44 Using SWStat+ (Click on Show Plot ) Slide 45 Standard Normal Probability Distribution Pep Zone 5w­20 Motor Oil s Example (Finding the X value): If the manager of Pep Zone wants the probability of a stockout to be no more than .05, what should the reorder point be? P (X = ? ) = 0.05 µ = 15 σ=6 z.05 = X=? Slide 46 Standard Normal Probability Distribution Pep Zone s 5w­20 Motor Oil Solving for the Reorder Point Area = 0.95 Area = .05 0 z.05 = z Slide 47 Standard Normal Probability Distribution Pep Zone s Solving for the Reorder Point 5w­20 Motor Oil Step 1: Find the zz­value that cuts off an area of .05 Step 1: Find the ­value that cuts off an area of .05 in the right tail of the standard normal in the right tail of the standard normal distribution. distribution. z . 1.5 1.6 1.7 1.8 1.9 . .00 . .9332 .9452 .9554 .9641 .9713 . .01 . .9345 .9463 .9564 .9649 .9719 . .02 . .9357 .9474 .9573 .9656 .9726 . .03 .04 .05 .06 .07 .08 . . . . . . .9370 .9382 .9394 .9406 .9418 .9429 .9484 .9495 .9505 .9515 .9525 .9535 .9582 .9591 .9599 .9608 .9616 .9625 .9664 .9671 .9678 .9686 .9693 .9699 .9732 .9738 .9744 .9750 .9756 .9761 We look up the complement of . . . . . the tail area (1 ­ .05 = .95) . .09 . .9441 .9545 .9633 .9706 .9767 . Slide 48 Standard Normal Probability Distribution Pep Zone s 5w­20 Motor Oil Solving for the Reorder Point Step 2: Convert zz.05 to the corresponding value of x.. Step 2: Convert .05 to the corresponding value of x x−µ z= σ x = µ + z.05σ x = µ + z.05σ = 15 + 1.645(6) = 15 + 1.645(6) = 15 + 1.645(6) = 15 + 1.645(6) = 24.87 or 25 = 24.87 or 25 Answer: The reorder point would be 25 gallons. Slide 49 Standard Normal Probability Distribution Pep Zone s Solving for the Reorder Point: Some Observation By raising the reorder point from 20 gallons to 25 gallons on hand, the probability of a stockout decreases from about .20 to .05. This is a significant decrease in the chance that Pep Zone will be out of stock and unable to meet a customer’s desire to make a purchase. 5w­20 Motor Oil Slide 50 Standard Normal Probability Distribution s Using half of the Normal Table to solve for the Reorder Point Pep Zone 5w­20 Motor Oil Area = 0.5 ­ .05 Area = .05 0.45 0 z.05 z Slide 51 Standard Normal Probability Distribution Example Continued Pep Zone 5w­20 Motor Oil The question is: P ( X = ? ) = 0.05. In another word, we need to find the value of X. The equation is: x−µ z= σ From the problem, we know that σ =6, µ = 15. The z =6 = 15. value for probability of 0.45 from the table is (1.64 + 1.65)/2 = 1.645. Thus, 1.645 = ( X – 15 )/ 6 = 24.87 or X = 25. Slide 52 Example (Continued) Using SWStat Students: Given the following find Students: P (X = ? ) = 0.05 by using Excel. (X µ = 15 σ=6 X=? Slide 53 Using SWStat+ to Find the X Value Slide 54 Using SWStat+ (Click on Show Plot) Slide 55 Standard Normal Probability Distribution More Examples f(X) Normal distributions differ by Normal mean & standard deviation. mean X Each distribution would Each require its own table. require Slide 56 Standardize the Normal Distribution Slide 57 Standardize the Normal Distribution Normal Distribution σ µ X Slide 58 Standardize the Normal Distribution X −µ Z= σ Normal Distribution Standardized Normal Distribution σ σ =1 µ X µ =0 One table! Z Slide 59 Standardizing Example Slide 60 Standardizing Example Normal Distribution σ = 10 µ =5 6.2 X Slide 61 Standardizing Example Normal Distribution X − µ 6.2 − 5 Z= = = .12 σ 10 σ = 10 µ =5 6.2 X Slide 62 Standardizing Example Normal Distribution X − µ 6.2 − 5 Z= = = .12 σ 10 Standardized Normal Distribution σ = 10 µ =5 σ =1 6.2 X µ =0 .12 Z Slide 63 Obtaining the Probability (Areas Under the Normal Curve) Standardized Normal Standardized Probability Table (Portion) Probability Z .00 .01 .02 σ =1 0.0 .0000 .0040 .0080 .0478 0.1 .0398 .0438 .0478 0.2 .0793 .0832 .0871 µ =0 .12 Z 0.3 .1179 .1217 .1255 Probabilities Slide 64 Example P(3.8 ≤ X ≤ 5) Slide 65 Example P(3.8 ≤ X ≤ 5) Normal Normal Distribution Distribution σ = 10 3.8 µ =5 X Slide 66 Example P(3.8 ≤ X ≤ 5) X − µ 3.8 − 5 Z= = = − .12 σ 10 Normal Normal Distribution Distribution σ = 10 3.8 µ =5 X Slide 67 Example P(3.8 ≤ X ≤ 5) X − µ 3.8 − 5 Z= = = − .12 σ 10 Normal Normal Distribution Distribution Standardized Standardized Normal Distribution Normal σ = 10 σ =1 .0478 3.8 µ =5 X -.12 µ =0 Z Slide 68 Example P(2.9 ≤ X ≤ 7.1) Slide 69 Example P(2.9 ≤ X ≤ 7.1) Normal Normal Distribution Distribution σ = 10 2.9 5 7.1 X Slide 70 Example Example P(2.9 ≤ X ≤ 7.1) Normal Normal Distribution Distribution X − µ 2.9 − 5 Z= = = − .21 σ 10 X − µ 7.1 − 5 Z= = = .21 σ 10 σ = 10 2.9 5 7.1 X Slide 71 Example P(2.9 ≤ X ≤ 7.1) Normal Normal Distribution Distribution X − µ 2.9 − 5 Z= = = − .21 σ 10 X − µ 7.1 − 5 Z= = = .21 σ 10 Standardized Standardized Normal Distribution Normal σ = 10 σ =1 .1664 .0832 .0832 2.9 5 7.1 X -.21 0 .21 Z Slide 72 Example (Continued) Using SWStat Students: Use Excel find: Students: Use P (2.9 <= X <= 7.1) Slide 73 Using SWStat+ Slide 74 Using SWStat+ to Plot Slide 75 Example P(X ≥ 8) Slide 76 Example P(X ≥ 8) Normal Normal Distribution Distribution σ = 10 µ =5 8 X Slide 77 Example P(X ≥ 8) X −µ 8−5 Z= = = .30 σ 10 Normal Normal Distribution Distribution σ = 10 µ =5 8 X Slide 78 Example P(X ≥ 8) X −µ 8−5 Z= = = .30 σ 10 Normal Normal Distribution Distribution Standardized Standardized Normal Distribution Normal σ = 10 σ =1 .5000 .1179 µ =5 8 X µ =0 .30 .3821 .3821 Z Slide 79 Example P(7.1 ≤ X ≤ 8) Slide 80 Example P(7.1 ≤ X ≤ 8) Normal Normal Distribution Distribution σ = 10 µ =5 7.1 8 X Slide 81 Example Example P(7.1 ≤ X ≤ 8) Normal Normal Distribution Distribution X − µ 7.1 − 5 Z= = = .21 σ 10 X −µ 8−5 Z= = = .30 σ 10 σ = 10 µ =5 7.1 8 X Slide 82 Example P(7.1 ≤ X ≤ 8) Normal Normal Distribution Distribution X − µ 7.1 − 5 Z= = = .21 σ 10 X −µ 8−5 Z= = = .30 σ 10 Standardized Standardized Normal Distribution Normal σ = 10 σ =1 .1179 .0347 .0347 .0832 µ =5 7.1 8 X µ =0 .21 .30 Z Slide 83 Some Areas under the Standard Normal Curve Find the following, and graph each case: s P (­1 ≤ Z ≤ 1) = ? s P ( ­3 ≤ Z ≤ 3) = ? s P ( ­2 ≤ Z ≤ 2) = ? s P ( ­2.53 ≤ Z ≤ 2.53) = ? s P (Z ≥ ­1 ) = ? s P ( 0 ≤ Z ≤ 1) = ? s P (Z ≥ 1 ) = ? s Slide 84 Some Areas under the Standard Normal Curve Slide 85 P ( -1 ≤ Z ≤ 1) = ? Using SWStat+ P ( -2 ≤ Z ≤ 2) = ? -2 Slide 86 Using SWStat+ to Find P ( ­2.53 ≤ Z ≤ 2.53) = ? Slide 87 Using SWStat+ To Graph P ( ­2.53 ≤ Z ≤ 2.53) =0.988 Slide 88 Calculating P(z ≥ ­1) Slide 89 Using SWStat+ Calculating P(z ≥ ­1) Slide 90 Calculating P(z ≥ 1) Slide 91 Using SWStat + Calculating P(z ≥ 1) Slide 92 Normal Distribution Thinking Challenge You work in Quality Control for GE. Light bulb life has a normal distribution with mean = 2000 hours & standard deviation = 200 hours. What’s the probability that a bulb will last between 2000 and 2400 hours? less than 1470 hours? Slide 93 Solution P(2000 ≤ X ≤ 2400) X − µ 2400 − 2000 Z= = = 2.0 σ 200 Normal Distribution Standardized Normal Distribution σ = 200 σ =1 .4772 .4772 µ = 2000 2400 X µ =0 2.0 Z Slide 94 Using SWStat+ to Find P(2000 ≤ X ≤ 2400) Slide 95 Solution P(X ≤ 1470) X − µ 1470 − 2000 Z= = = − 2.65 σ 200 Normal Normal Distribution Distribution Standardized Standardized Normal Distribution Normal σ = 200 σ =1 .5000 .4960 .0040 .0040 1470 µ = 2000 X -2.65 µ =0 Z Slide 96 Using SWStat+ to Find P( X ≤ 1470) Slide 97 Finding Z Values for Known Probabilities Slide 98 Finding Z Values for Known Probabilities What is Z given What P(Z) = .1217? P(Z) .1217 µ =0 σ =1 ? Z Slide 99 Finding Z Values for Known Probabilities Standardized Normal Standardized Probability Table (Portion) Probability What is Z given What P(Z) = .1217? P(Z) .1217 σ =1 Z .00 .01 0.2 0.0 .0000 .0040 .0080 0.1 .0398 .0438 .0478 µ =0 ? Z 0.2 .0793 .0832 .0871 0.3 .1179 .1217 .1255 Slide 100 Finding Z Values for Known Probabilities Standardized Normal Standardized Probability Table (Portion) Probability What is Z given What P(Z) = .1217? P(Z) .1217 σ =1 Z .00 .01 0.2 0.0 .0000 .0040 .0080 0.1 .0398 .0438 .0478 µ =0 .31 Z 0.2 .0793 .0832 .0871 0.3 .1179 .1217 .1255 Slide 101 Finding X Values for Known Probabilities Slide 102 Finding X Values for Known Probabilities Normal Distribution σ = 10 .1217 .1217 µ =5 ? X Slide 103 Finding X Values for Known Probabilities Normal Distribution Standardized Normal Distribution σ = 10 σ =1 .1217 .1217 µ =5 ? X .1217 .1217 µ =0 .31 Z Slide 104 Finding X Values for Known Probabilities Normal Distribution Standardized Normal Distribution σ = 10 σ =1 .1217 .1217 µ =5 ? X .1217 .1217 µ =0 .31 Z X = µ + Z ⋅ σ = 5 + ( .31) (10 ) = 8. 1 Slide 105 Using SWStat+ To Find X Values for Known Probabilities 0.6217=0.5+0.1217 Slide 106 Finding X Values for Known Probabilities Another Example Dr. L.A. has determined that the final averages in his statistics course is normally distributed with a mean of 72 and a standard deviation of 5. He decides to assign his grades for his current course such that the top 15% of the students receive an A. What is the lowest average a student must receive to earn an A? Let X be the lowest average. We need to find X such that P(X ≥ ?) = 0.15. The corresponding z value is 1.04 (approximately). Thus we have (X ­ 72)/5 = 1.04, from which X = 77.2. (See graph) Slide 107 0.35 0.35 0.15 0.15 1.04 1.04 z Slide 108 Using SWStat+ To Find X Values for Known Probabilities 0.85=0.5+0.35 Slide 109 Using SWStat+ to Plot Slide 110 Another Problem Use SWStat+ s X is a normally distributed random variable with a mean of 50 and a standard deviation of 5. Use Excel to calculate the following: • 1. P(x ≤ 45) • 2. P(45 ≤ x ≤ 55) • 3. P(x ≥ 55) • 4. X value with 0.20 in the lower tail • 5. X value with 0.01 in the upper tail STUDENT Slide 111 Another Problem (Solution) s s s s s 1. P(x ≤ 45) = 0.16 =NORMDIST(45,50,5,TRUE) 2. P(45 ≤ x ≤ 55) = 0.68 =NORMDIST(55,50,5,TRUE)­NORMDIST(45,50,5,TRUE) 3. P(x ≥ 55) = 0.16 =1­NORMDIST(55,50,5,TRUE) 4. x value with 0.20 in the lower tail = 45.79 =NORMINV(0.2,50,5) 5. x value with 0.01 in the upper tail = 61.63 =NORMINV(0.99,50,5) Slide 112 End of Chapter 7 Slide 113 ...
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This document was uploaded on 11/25/2011.

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