## Deliverable 6 – Using Business Visuals

Question Description

Respond to the following in a minimum of 175 words: Consider the demonstration problem 6.3 which uses a normal distribution to determine the probability associated with generating between 3.6 and 5 pounds of waste per year. Discuss any one of the following concepts associated with this problem. What are the clues that the problem can be solved with the use of the normal distribution? What is the relationship between the x value (e.g 3.6) and the z score? How is the probability of the event (3.6 to 5 pounds of waste) related to a specific area under the curve? How do we determine the area between 2 z values? __________________________________________________________________________________________________DEMONSTRATION PROBLEM 6.3 Using this same waste-generation example, if a U.S. person is randomly selected, what is the probability that the person generates between 3.60 and 5.00 pounds of waste per day? We can summarize this problem as: P ( 3.60 < x < 5.00 | ? = 4.43 and ? = 1.32 = ?Figure 6.13 displays a graphical representation of the problem. Note that the area under the curve for which we are solving crosses over the mean of the distribution. Note that there are two x values in this problem (x1 = 3.60 and x2 = 5.00). The z formula can handle only one x value at a time. Thus, this problem needs to be worked out as two separate problems and the resulting probabilities added together. We begin the process by solving for each z value:z = x ? ? ? = 3.60 ? 4.43 1.32 = ? 0.63 z = x ? ? ? = 5.00 ? 4.43 1.32 = 0.43 FIGURE 6.13 Graphical Depiction of the Waste-Generation Problem with 3.60 < x < 5.00 Next, we look up each z value in the z distribution table. Since the normal distribution is symmetrical, the probability associated with z = ?0.63 is the same as the probability associated with z = 0.63. Looking up z = 0.63 in the table yields a probability of .2357. The probability associated with z = 0.43 is .1664. Using these two probability values, we can get the probability that 3.60 < x < 5.00 by summing the two areas: P ( 3.60 < x < 5.00 | ? = 4.43 and ? = 1.32 ) = .2357 + .1664 = .4021 The probability that a randomly selected person in the U.S. has between 3.60 and 5.00 pounds of waste generation per day is .4021 or 40.21%. Figure 6.14 displays the solution to this problem. FIGURE 6.14 Solution of the Waste-Generation Problem with 3.60 < x < 5.00

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