Statistical techniques in business ecohomics chap007

7- Chapter Seven McGraw- © 2005 The McGraw-Hill Companies, Inc., All 7- Chapter Seven Continuous Probability Distributions GOALS When you have completed this chapter, you will be ONE able to: Understand the difference between discrete and continuous distributions TWO Compute the mean and the standard deviation for a uniform distribution THREE Compute probabilities using the uniform distribution FOUR List the characteristics of the normal probability distribution Goals 7- Chapter Seven continued Continuous Probability GOALS Distributions When you have completed this chapter, you will be able to: FIVE Define and calculate z values SIX Determine the probability an observation will lie between two points using the standard normal distribution SEVEN Determine the probability an observation will be above or below a given value using the standard normal distribution EIGHT Use the normal distribution to approximate the binomial probability distribution Goals 7- 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 7- The Uniform distribution Is rectangular in shape Is defined by minimum and maximum values Has a mean computed as follows: Has a standard deviation computed as follows: σ= f(x) x µ = a+b where a and b are the minimum and maximum values (b-a)2 12 The uniform distribution 7- Calculates its height as P(x) = if a < x < b and elsewhere (b-a) Calculates its area as Area = height* base =(b-a) *(b-a) The uniform distribution 7- Suppose the time that you wait on the telephone for a live representative of your phone company to discuss your problem with you is uniformly distributed between and 25 minutes What is the mean wait time? µ =a + b 5+25 = = 15 What is the standard deviation of the wait time? (b-a)2 σ= 12 (25-5) = = 5.77 12 Example 7- What is the probability of waiting more than ten minutes? The area from 10 to 25 minutes is 15 minutes Thus: P(10 < wait time < 25) = height*base = (25-5) *15 = 75 What is the probability of waiting between 15 and 20 minutes? The area from 15 P(15 < wait time < 20) = height*base to 20 minutes is = *5 = 25 minutes Thus: (25-5) Example continued 7- The Normal probability distribution is bell-shaped and has a single peak at the center of the distribution Is symmetrical about the mean is asymptotic That is the curve gets closer and closer to the X-axis but never actually touches it its mean, µ , to determine its location and its standard deviation, σ , to determine its dispersion Has 7- 10 r a l i t r b u i o n : µ = , σ2 = Characteristics of a Normal Distribution Normal curve is symmetrical f ( x Theoretically, curve extends to infinity - a Mean, median, and mode are equal x 7- 18 7- 19 What percent of the population use between 18 and 26 gallons per day? z = z= X −µ σ X −µ σ 26 − 20 = = 1.20 18 − 20 = = −0.40 EXAMPLE continued 7- 20 The area The area associated with a associated with a z-value of –0.40 z-value of 1.20 is is 1554 Adding these 3849 areas, the result is 5403 We conclude that 54.03 percent of the residents use between 18 and 26 gallons of water per day EXAMPLE continued 7- 21 Professor Mann has determined that the scores in his statistics course are approximately normally distributed with a mean of 72 and a standard deviation of He announces to the class that the top 15 percent of the scores will earn an A What is the lowest score a student can earn and still receive an A? EXAMPLE 7- 22 To begin let X be the score that separates an A from a B If 15 percent of the students score more than X, then 35 percent must score between the mean of 72 and X The z-value associated corresponding to 35 percent is about 1.04 EXAMPLE continued 7- 23 We let z equal 1.04 and solve the standard normal equation for X The result is the score that separates students that earned an A from those that earned a B Those with a score of 77.2 or more earn an A X − 72 1.04 = X = 72 + 1.04(5) = 72 + 5.2 = 77.2 EXAMPLE continued 7- 24 The normal distribution (a continuous distribution) yields a good approximation of the binomial distribution (a discrete distribution) for large values of n The normal probability distribution is generally a good approximation to the binomial probability distribution when nπ and n(1- π ) are both greater than The Normal Approximation to the Binomial 7- 25 Recall for the binomial experiment: oThere are only two mutually exclusive outcomes (success or failure) on each trial oA binomial distribution results from counting the number of successes oEach trial is independent oThe probability is fixed from trial to trial, and the number of trials n is also fixed The Normal Approximation continued 7- 26 Continuity Correction Factor The value subtracted or added, depending on the problem, to a selected value when a binomial probability distribution (a discrete probability distribution) is being approximated by a continuous probability distribution (the normal distribution) Continuity Correction Factor 7- 27 How to Apply the Correction Factor: For the probability that more than X occur, use the area above (X+.5) For the probability that fewer than X occur, use the area below (X-.5) For the probability at least X occur, use the area above (X-.5) For the probability that X or fewer occur, use the area below (X+.5) Continuity Correction Factor 7- 28 A recent study by a marketing research firm showed that 15% of American households owned a video camera For a sample of 200 homes, how many of the homes would you expect to have video cameras? µ = nπ = (.15)(200) = 30 This is the mean of a binomial distribution EXAMPLE 7- 29 What is the variance? σ = nπ (1 − π ) = (30)(1−.15) = 255 What is the standard deviation? σ = 25.5 = 5.0498 EXAMPLE continued 7- 30 What is the probability that less than 40 homes in the sample have video cameras? We use the correction factor (X-.5) for fewer than, so X-.5 is 39.5 The value of z is 1.88 X − µ 39.5 − 30.0 z= = = 1.88 σ 5.0498 EXAMPLE continued 7- 31 From Appendix D the area between and 1.88 on the z scale is 4699 So the area to the left of 1.88 is 5000 + 4699 = 9699 The likelihood that less than 40 of the 200 homes have a video camera is about 97% EXAMPLE continued 7- 32 ... values A Continuous distribution usually results from measuring something Discrete distributions studied include: o Binomial o Hypergeometric o Poisson Continuous distributions include: o Uniform... be able to: FIVE Define and calculate z values SIX Determine the probability an observation will lie between two points using the standard normal distribution SEVEN Determine the probability... more than ten minutes? The area from 10 to 25 minutes is 15 minutes Thus: P(10 < wait time < 25) = height*base = (25-5) *15 = 75 What is the probability of waiting between 15 and 20 minutes? The