Section 10.4 Normal Distributions

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1 Section 10.4 Normal Distributions Random Variables Suppose a bank is interested in improving its services to customers. The manager decides to begin by finding the amount of time tellers spend on each transaction, rounded to the nearest minute. The times for 75 different transactions are recorded, with the results shown in the following table, where the frequencies listed in the second column are divided by 75 to find the empirical probabilities: Time Frequency Probability 1 3 3/75 = /75 = /75 = /75 = /75 = /75 = /75 = /75 = /75 = /75 =.01 The Figure below (left), on the following page, shows a histogram and frequency polygon for the data. The heights of the bars are the empirical probabilities, rather than the frequencies. The transaction times are given to the nearest minute. Theoretically, at least, they could have been timed to the nearest tenth of a minute, or hundredth of a minute, or even more precisely. Tn each case, a histogram and frequency polygon could be drawn. If the times are measured with smaller and smaller units, there are more bars in the histogram and the frequency polygon begins to look more and more like the curve in the Figure below (right) instead of a polygon. Actually, it is possible for the transaction times to take on any real-number value greater than 0. A distribution in which the outcomes can take on any real-number value within some interval is a continuous distribution. The graph of a continuous distribution is a curve. The distribution of heights (in inches) of college women is another example of a continuous distribution, since these heights include infinitely many possible measurements, such as 53, 58.5, 66.3, , and so on. The Figure on the right shows the continuous distribution of heights of college women. Here, the most frequent heights occur near the center of the interval displayed. 1

2 Normal Distributions We say that data are normal (or normally distributed) when their graph is well approximated by a bell-shaped curve. (See the Figures below.) We call the graphs of such distributions normal curves. Examples of distributions that are approximately normal are the heights of college women and cholesterol levels in adults. We use the Greek letters µ (mu) to denote the mean and (sigma) to denote the standard deviation of a normal distribution. There are many normal distributions, depending on µ and. Some of the corresponding normal curves are tall and thin, and others short and wide, as shown in the Figure above. But every normal curve has the following properties: 1. Its peak occurs directly above the mean µ. 2. The curve is symmetric about the vertical line through the mean. (That is, if you fold the graph along this line, the left half of the graph will fit exactly on the right half.) 3. The curve never touches the x-axis it extends indefinitely in both directions. 4. The area under the curve (and above the horizontal axis) is 1. (As can be shown with calculus, this is a consequence of the fact that the sum of the probabilities in any distribution is 1.) A normal distribution is completely determined by its mean µ and standard deviation. As shown in more advanced courses, its graph is the gn1ph of the function f(x) = 1 (x µ) 2 2π e 2 2, where e is the real number introduced in Section 4.1. A small standard deviation leads to a tall, narrow curve like the one in the center of the Figure above, because most of the data are close to the mean. A large standard deviation means the data are very spread out, producing a flat, wide curve like the one on the right in the Figure above. Since the area under a normal curve is 1, parts of this area can be used to determine certain probabilities. For instance, the Figure below (a) is the probability distribution of the annual rainfall in a certain region. The probability that the annual rainfall will be between 25 and 35 inches is the area under the curve from 25 to 35. The general case, shown in the Figure below (b), can be stated as follows. 2

3 To use normal curves effectively, we must be able to calculate areas under portions of them. These calculations have already been done for the normal curve with mean µ = 0 and standard deviation = 1 (which is called the standard normal curve) and are available in the Table at the back of the handouts. The following Example demonstrates how to use the Table to find such areas. Later, we shall see how the standard normal curve may be used to find areas under any normal curve. The horizontal axis of the standard normal curve is usually labeled z. Since the standard deviation of the standard normal curve is 1, the numbers along the horizontal axis (the z- values) measure the number of standard deviations above or below the mean z = 0. EXAMPLE: Find the given areas under the standard normal curve. (a) The area between z = 0 and z = 1, the shaded region in the Figure below. Solution: Find the entry 1.0 in the z-column of the Table. The entry next to it in the column is.3413, which means that the area between z = 0 and z = 1 is Since the total area under the curve is 1, the shaded area in the Figure below is 34.13% of the total area under the normal curve. (b) The area between z = 2.43 and z = 0. Solution: The Table lists only positive values of z. But the normal curve is symmetric around the mean z = 0, so the area between z = 0 and z = 2.43 is the same as the area between z = 0 and z = Find 2.4 in the z-column of the Table. The entry in the 0.03-column shows that the area is Hence, the shaded area in the Figure below is 49.25% of the total area under the curve. (c) The area between z =.88 and z = Solution: First, draw a sketch showing the desired area, as in the Figure below. From the Table, the area between z = 0 and z =.88 is Also, the area between z = 0 and z = 2.35 is As the figure shows, the total desired area can be found by adding these numbers: =.8012 The shaded area in the Figure below represents 80.12% of the total area under the normal curve. 3

4 (d) The area between z =.58 and z = Solution: First, draw a sketch showing the desired area, as in the Figure below. From the Table, the area between z = 0 and z =.58 is Also, the area between z = 0 and z = 1.94 is As the figure shows, the desired area is found by subtracting one area from the other: =.2548 The shaded area of the Figure below represents 25.48% of the total area under the normal curve. (e) The area to the right of z = Solution: The total area under a normal curve is l. Thus, the total area to the right of z = 0 is 1/2, or From the Table, the area from z = 0 to z = 2.09 is The area to the right of z = 2.09 is found by subtracting.4817 from.5000: =.0183 A total of 1.83% of the total area is to the right of 2.09 standard deviations above the mean. The Figure below (which is not to scale) shows the desired area. The key to finding areas under any normal curve is to express each number x on the horizontal axis in terms of standard deviations above or below the mean. The z-score for x is the number of standard deviations that x lies from the mean (positive if x is above the mean, negative if x is below the mean). The importance of z-scores is the following fact, whose proof is omitted. 4

5 EXAMPLE: Dixie Office Supplies finds that its sales force drives an average of 1200 miles per month per person, with a standard deviation of miles. Assume that the number of miles driven by a salesperson is closely approximated by a normal distribution. (a) Find the probability that a salesperson drives between 1200 and 1600 miles per month. Solution: Here, µ = 1200 and =, and we must find the area under the normal distribution curve between x = 1200 and x = We begin by finding the z-score for x = 1200: The z-score for x = 1600 is = = = 0 = 0 = 400 = 2.67 So the area under the curve from x = 1200 to x = 1600 is the same as the area under the standard normal curve from z = 0 to z = 2.67, as indicated in the Figure below. The Table shows that this area is Therefore, the probability that a salesperson drives between 1200 and 1600 miles per month is (b) Find the probability that a salesperson drives between 1000 and 0 miles per month. Solution: As shown in the Figure below, z-scores for both x = 1000 and x = 0 are needed. We begin by finding the z-score for x = 1000: The z-score for x = 0 is = = = 200 = 1.33 = 300 = 2.00 From the Table 2, z = l.33 leads to an area of.4082, while z = 2.00 corresponds to A total of =.8855 or 88.55%, of all drivers travel between 1000 and 0 miles per month. From this calculation, the probability that a driver travels between 1000 and 0 miles per month is

6 780 Appendix A Tables The entries in this table are the probabilities that a standard normal random variable is between 0 and z (the shaded area). z TABLE A.5 Areas of the Standard Normal Distribution 0 Z