# QNT561 University of Phoenix Normal Distribution Problem Discussion

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.

1. What are the “clues” that the problem can be solved with the use of the normal distribution?
2. What is the relationship between the x value (e.g 3.6) and the z score?
3. How is the probability of the event (3.6 to 5 pounds of waste) related to a specific area under the curve?
4. How do we determine the area between 2 z values?

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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 | μ = 4.43 and σ = 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 − μ σ = 3.60 − 4.43 1.32 = − 0.63 z = x − μ σ = 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 | μ = 4.43 and σ = 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