MAT 155 Statistical Analysis Dr. Claude Moore Cape Fear Community College Chapter 6 Normal Probability Distributions 6 1 Review and Preview 6 2 The Standard Normal Distribution 6 3 Applications of Normal Distributions 6 4 Sampling Distributions and Estimators 6 5 The Central Limit Theorem 6 6 Normal as Approximation to Binomial 6 7 Assessing Normality See page 9 of the following lesson for information relative to this section. S3.D1.MAT 155 Session 3 Chapter 6 155Session 3 Chapter 6 Lesson ( Package file ) Session 3 contains information about Chapter 6 Normal Probability Distributions. In addition to notes, this contains instructions and demonstrations including videos on how to use the TI 83/84 calculator, Table A 2, Statdisk, and Excel to solve problems. S3.D2.MAT 155 Using Technology for Chapter 6 155Chapter 6 Technology ( Package file ) This lesson illustrates how to use technology (TI 83/84 calculator, Statdisk, and Excel) to solve problems from Chapter 6 Normal Probability Distribution. The short video shows how to use the normal as an approximation to the binomial http://cfcc.edu/faculty/cmoore/nb1/nb1.htm Key Concept This section presents a method for using a normal distribution as an approximation to the binomial probability distribution. If the conditions of np 5 and nq 5 are both satisfied, then probabilities from a binomial probability distribution can be approximated well by using a normal distribution with mean μ = np and standard deviation Review Binomial Probability Distribution 1. The procedure must have a fixed number of trials. 2. The trials must be independent. 3. Each trial must have all outcomes classified into two categories (commonly, success and failure). 4. The probability of success remains the same in all trials. Solve by binomial probability formula, Table A 1, or technology. 1
Approximation of a Binomial Distribution with a Normal Distribution np 5 nq 5 then µ = np and σ = npq and the random variable has Procedure for Using a Normal Distribution to Approximate a Binomial Distribution 1. Verify that both np 5 and nq 5. If not, you must use software, a calculator, a table, or calculations using the binomial probability formula. 2. Find the values of the parameters µ and σ by calculating µ = np and σ = npq. 3. Identify the discrete whole number x that is relevant to the binomial probability problem. Focus on this value temporarily. Procedure for Using a Normal Distribution to Approximate a Binomial Distribution 4. Draw a normal distribution centered about µ, then draw a vertical strip area centered over x. Mark the left side of the strip with the number equal to x 0.5, and mark the right side with the number equal to x + 0.5. Consider the entire area of the entire strip to represent the probability of the discrete whole number itself. Procedure for Using a Normal Distribution to Approximate a Binomial Distribution 6. Using x 0.5 or x + 0.5 in place of x, find the area of the shaded region: find the z score; use that z score to find the area to the left of the adjusted value of x; use that cumulative area to identify the shaded area corresponding to the desired probability. 5. Determine whether the value of x itself is included in the probability. Determine whether you want the probability of at least x, at most x, more than x, fewer than x, or exactly x. Shade the area to the right or left of the strip; also shade the interior of the strip if and only if x itself is to be included. This total shaded region corresponds to the probability being sought. 2
Example Number of Men Among Passengers Finding the Probability of At Least 122 Men Among 213 Passengers Definition Continuity Correction Factor of 0.5 When we use the normal distribution (which is a continuous probability distribution) as an approximation to the binomial distribution (which is discrete), a continuity correction is made to a discrete whole number x in the binomial distribution by representing the discrete whole number x by the interval from x 0.5 to x + 0.5 (that is, adding and subtracting 0.5). Figure 6 21 x = at least 8 (includes 8 and above) Recap x = more than 8 (doesn t include 8) x = at most 8 (includes 8 and below) In this section we have discussed: Approximating a binomial distribution with a normal distribution. Procedures for using a normal distribution to approximate a binomial distribution. Continuity corrections. x = fewer than 8 (doesn t include 8) x = exactly 8 3
312/6. Probability of at least 2 traffic tickets this year. 312/8. Probability that the number of students who are absent is exactly 4. 312/10. Probability that the number of defective computer power supplies is between 12 and 16 inclusive. 312/11. Probability that the number of job applicants late for interviews is between 5 and 9 inclusive 4
Using Normal Approximation. Do the following: (a) Find the indicated binomial probability by using Table A 1 in Appendix A. (b) If np 5 and nq 5, also estimate the indicated probability by using the normal distribution as an approximation to the binomial distribution; if np < 5 or nq < 5, then state that the normal approximation is not suitable. 312/15. With n = 8 and p = 0.9, find P(at least 6). Using Normal Approximation. Do the following: (a) Find the indicated binomial probability by using Table A 1 in Appendix A. (b) If np 5 and nq 5, also estimate the indicated probability by using the normal distribution as an approximation to the binomial distribution; if np < 5 or nq < 5, then state that the normal approximation is not suitable. 312/16. With n = 15 and p = 0.4, find P(fewer than 3). 313/20. Gender Selection The Genetics & IVF Institute developed its YSORT method to increase the probability of conceiving a boy. Among 152 women using that method, 127 had baby boys. Assuming that the method has no effect so that boys and girls are equally likely, find the probability of getting at least 127 boys among 152 babies. Does the result suggest that the YSORT method is effective? Why or why not? 313/22. Voters Lying? In a survey of 1002 people, 701 said that they voted in a recent presidential election (based on data from ICR Research Group). Voting records show that 61% of eligible voters actually did vote. Given that 61% of eligible voters actually did vote, find the probability that among 1002 randomly selected eligible voters, at least 701 actually did vote. What does the result suggest? Since the probability of getting at least 701 actual voters by chance alone is so small, the result suggests that the respondents were not truthful about their voting. Yes. Since the probability of getting at least 127 boys by chance alone is so small, it appears that the method is effective and that the genders were not being determined by chance alone. 5
313/24. Employee Hiring There is an 80% chance that a prospective employer will check the educational background of a job applicant (based on data from the Bureau of National Affairs, Inc.). For 100 randomly selected job applicants, find the probability that exactly 85 have their educational backgrounds checked. 313/26. Acceptance Sampling With the procedure called acceptance sampling, a sample of items is randomly selected and the entire batch is either rejected or accepted, depending on the results. The Telektronics Company has just manufactured a large batch of backup power supply units for computers, and 7.5% of them are defective. If the acceptance sampling plan is to randomly select 80 units and accept the whole batch if at most 4 units are defective, what is the probability that the entire batch will be accepted? Based on the result, does the Telektronics Company have quality control problems? Yes. Not only does the 7.5% rate of defectives seem too high, but also the fact that only about 26% of the batches it ships will be accepted presents a problem. 313/28. Detecting Fraud When working for the Brooklyn District Attorney, investigator Robert Burton analyzed the leading digits of amounts on checks from companies that were suspected of fraud. Among 784 checks, 479 had amounts with leading digits of 5, but checks issued in the normal course of honest transactions were expected to have 7.9% of the checks with amounts having leading digits of 5. Is there strong evidence to indicate that the check amounts are significantly different from amounts that are normally expected? Explain? 314/30. Polygraph Accuracy Polygraph experiments conducted by researchers Charles R. Honts (Boise State University) and Gordon H. Barland (Department of Defense Polygraph Institute) showed that among 57 polygraph indications of a lie, the truth was told 15 times, so the proportion of false positive results among the 57 positive results is 15/57. Assuming that the polygraph makes random guesses, determine whether 15 is an unusually low number of false positive results among the 57 positive results. Does the polygraph appear to be making random guesses? Explain. Yes. Since the probability of obtaining 479 or more such checks from normal honest transactions by chance alone is so very strong, there is strong evidence to indicate that the checks from the suspected companies do not follow the normal pattern and are likely fraudulent. No. Since 0.0003 0.05, getting 15 or fewer false positives by chance alone would be an unusual event. The polygraph does not appear to be making random guesses. 6
314/32. Passenger Load on a Boeing 767 300 An American Airlines Boeing 767 300 aircraft has 213 seats. When fully loaded with passengers, baggage, cargo, and fuel, the pilot must verify that the gross weight is below the maximum allowable limit, and the weight must be properly distributed so that the balance of the aircraft is within safe acceptable limits. When considering the weights of passengers, their weights are estimated according to Federal Aviation Administration rules. Men have a mean weight of 172 lb, whereas women have a mean weight of 143 lb, so disproportionately more male passengers might result in an unsafe overweight situation. Assume that if there are at least 122 men in a roster of 213 passengers, the load must be somehow adjusted. Assume that passengers are booked randomly, and that male passengers and female passengers are equally likely. If the aircraft is full of adults, find the probability that a Boeing 767 300 with 213 passengers has at least 122 men. Based on the result, does it appear that the load must be adjusted often? 314/32. Passenger Load on a Boeing 767 300 An American Airlines Boeing 767 300 aircraft has 213 seats. When fully loaded with passengers, baggage, cargo, and fuel, the pilot must verify that the gross weight is below the maximum allowable limit, and the weight must be properly distributed so that the balance of the aircraft is within safe acceptable limits. When considering the weights of passengers, their weights are estimated according to Federal Aviation Administration rules. Men have a mean weight of 172 lb, whereas women have a mean weight of 143 lb, so disproportionately more male passengers might result in an unsafe overweight situation. Assume that if there are at least 122 men in a roster of 213 passengers, the load must be somehow adjusted. Assume that passengers are booked randomly, and that male passengers and female passengers are equally likely. If the aircraft is full of adults, find the probability that a Boeing 767 300 with 213 passengers has at least 122 men. Based on the result, does it appear that the load must be adjusted often? No. Since the 0.0199 0.05, having at last 122 men is an unusual event. The load does not have to be adjusted very often. No. Since the 0.0199 0.05, having at last 122 men is an unusual event. The load does not have to be adjusted very often. 7