6 Problems following sixth meeting Beginning page 187 1, 2, 5, 7, 17, 18* Problem 18 is of a theoretical nature, but it makes you \"think out of the box\". Please remember the first exam is on the eighth meeting. Refer to the calendar so you can plan accordingly. 1.The grades on the midterm examination given in a large managerial statistics class are normally distributed with mean 75 and standard deviation 9. The instructor of this class wants to assign an A grade to the top 10% of the scores, a B grade to the next 10% of the scores, a C grade to the next 10% of the scores, a D grade to the next 10% of the scores, and an F grade to all scores below the 60th percentile of this distribution. For each possible letter grade, find the lowest accept- able score within the established range. For example, the lowest acceptable score for an A is the score at the 90th percentile of this normal distribution. Grade Distribution Grades are normally distributed with a mean of 75 and a standard deviation of 9. N(75,9) Mean= 75 \"B6\" Std= 9 \"B7\" Grades NORMINV(0.9,$B$6,$B 90th percentile is the lowest numerical grade A 87 $7) that will receive this letter grade NORMINV(0.8,$B$6,$B 80th percentile is the lowest numerical grade B 83 $7) that will receive this letter grade NORMINV(0.7,$B$6,$B 70th percentile is the lowest numerical grade C 80 $7) that will receive this letter grade NORMINV(0.6,$B$6,$B 60th percentile is the lowest numerical grade D 77 $7) that will receive this letter grade F Everything below 77 is an F 2. Suppose it is known that the distribution of purchase amounts by customers entering a popular retail store is approximately normal with mean $75 and standard deviation $20. a. What is the probability that a randomly selected customer spends less than $85 at this store? b. What is the probability that a randomly selected customer spends between $65 and $85 at this store? c. What is the probability that a randomly selected customer spends more than $45 at this store? d. Find the dollar amount such that 75% of all customers spend no more than this amount. e. Find the dollar amount such that 80% of all customers spend at least this amount. f. Find two dollar amounts, equidistant from the mean, such that 90% of all customer purchases are between these values. 5. An investor has invested in nine different investments. The dollar returns on the different investments are probabilistically independent, and each return follows a normal distribution with mean $50,000 and standard deviation $10,000. a. There is a 1% chance that the total return on the nine investments is less than what value? b. What is the probability that the investor's total return is between $400,000 and $520,000? http://www.chegg.com/homework-help/investor-invested-nine-different-investments-dollar-returnschapter-5-problem-5p-solution-9781285965529-exc 7. Suppose that the weight of a typical American male follows a normal distribution with = 180 lb and = 30 lb. Also, suppose 91.92% of all American males weigh more than I weigh. a. What fraction of American males weigh more than 225 pounds? b. How much do I weigh? c. If I weighed 20 pounds more than I do, what percentile would I be in? 17. A fast-food restaurant sells hamburgers and chicken sandwiches. On a typical weekday the demand for hamburgers is normally distributed with mean 313 and standard deviation 57; the demand for chicken sandwiches is normally distributed with mean 93 and standard deviation 22. a. How many hamburgers must the restaurant stock to be 98% sure of not running out on a given day? b. Answer part a for chicken sandwiches. c. If the restaurant stocks 400 hamburgers and 150 chicken sandwiches for a given day, what is the probability that it will run out of hamburgers or chicken sandwiches (or both) that day? Assume that the demand for hamburgers and the demand for chicken sandwiches are probabilistically independent. d. Why is the independence assumption in part c probably not realistic? Using a more realistic assumption, do you think the probability requested in part c would increase or decrease? Mean Hamburge r Chicken 313 93 Std 57 22 18. Referring to the box plots introduced in Chapter 2, the sides of the \"box\" are at the first and third quartiles, and the difference between these (the length of the box) is called the interquartile range (IQR). A mild outlier is an observation that is between 1.5 and 3 IQRs from the box, and an extreme outlier is an observation that is more than 3 IQRs from the box. a. If the data are normally distributed, what percentage of values will be mild outliers? What percentage will be extreme outliers? Why don't the answers depend on the mean and/or standard deviation of the distribution? a) 3-1.5= 1.5 IQR, 50% of observations are in 1IQR, thus 1.5= 75% will be mild outliers b) 3IQR=150% will be extreme outliers c) uses only the fact that the numbers were evenly distributed among the central value (2d Quartile) b. Check your answers in part a with simulation. Simulate a large number of normal random numbers (you can choose any mean and standard deviation), and count the number of mild and extreme outliers with appropriate formulas. Do these match, at least approximately, your answers to part a