Probability & Statistics Chapter 6. Normal Distribution

Transcription

1 I. Graphs of Normal Probability Distributions Normal Distribution Studied by French mathematician Abraham de Moivre and German mathematician Carl Friedrich Gauss. Gauss work was so important that the normal distribution is sometimes called Gaussian. Normal Curve The graph of a normal distribution. It is also called a bell-shaped curve. Important Properties of a Normal Curve: 1) The curve is bell-shaped with the highest point over the mean. 2) It is symmetrical about a vertical line through. 3) The curve approaches the horizontal axis but never touches or crosses it. 4) The transition points between cupping upward and downward occur above and. Example 1 Each of the curves below fail to be a normal curve. Give reasons why these curves are not normal curves.

2 Answers. (b) (c) (d) A normal curve gets closer and closer to the horizontal axis, but it never touches it or crosses it. A normal curve must be symmetrical. This curve is not. A normal curve is bell-shaped with one peak. Because this curve has two peaks, it is not normal. The tails of a normal curve must get closer and closer to the x axis. In this curve the tails are going away from the x axis. Example 2 The points A, B, and C are indicated on the normal curve. One of these points is, one is, and one is 2. Which point corresponds to the mean? What is the value of? (b) Which point corresponds to? Use the values of and to compute. (c) Which point corresponds to 2?

3 Example 2 Answers. The mean is under the peak of the normal curve. The point B corresponds to the mean so 10. (b) The point C where the curve changes from cupped down to cupped up is one standard deviation from the mean. The point C is. Since 12 and 10, 2. (c) Since 10 and 10, we see that (2) 6 Point A corresponds to 2. Example 3 Use the following curve to answer the questions. Do these distributions have the same mean? If so, what is it? (b) One of the curves corresponds to a normal distribution with 3 and the other to one with 1. Which curve has which? Example 3 Answers. The means are the same, since both graphs have the high point over (b) Curve A has 1, and curve B has 3. (Since curve B is more spread out, it has the larger value.)

4 The total area under any normal curve studied in this book will always be 1. Empirical Rule: For a distribution that is symmetrical and bell-shaped (in particular, for a normal distribution): Approximately 68.2% of the data values will lie within one standard deviation on each side of the mean. Approximately 95.4% of the data values will lie within two standard deviations on each side of the mean. Approximately 99.7% (or almost all) of the data values will be within three standard deviations on each side of the mean. Empirical Rule applies only to normal or symmetrical, bell-shaped distributions, whereas Chebyshev s theorem applies to all distributions.

5 Example 4 The yearly wheat yield per acre on a particular farm is normally distributed with mean 35 bushels and standard deviation 8 bushels. Shade the area under the curve that represents the probability that an acre will yield between 19 and 35 bushels. (b) Is the area the same as the area between 2 and? (c) Use the graph to find the percentage of area over the interval between 19 and 35. (d) What is the probability that the yield will be between 19 and 35 bushels per acre? Answers. See book. (b) Yes, since 35 and (8) 19 (c) The area between the values 2 and is 47.7% of the total area. (d) It is 47.7% of the total area, which is 1. Therefore the probability is that the yield will be between 19 and 35 bushels.

6 Control Charts Useful when we are examining data over a period of equally spaced time intervals or in some sequential order. A random variable x is said to be in statistical control if it can be described by the same probability distribution when it is observed at successive points in time. Control charts combine graphic and numerical descriptions of data with probability distributions. Control Charts were invented in the 1920 s by Walter Shewhart at bell Telephone Laboratories. A control chart is a warning device, it is not absolutely necessary that our assumptions and probability calculations be precisely correct.

7 Example 5 Over the next 15-day period, let s suppose that housekeeping reported the number of rooms not made up by 3:30 p.m. to the director. The data in the table shows the results. Next 15-day Report of Rooms Not Made Up by 3:30 p.m. Day x = number of rooms We assume that we are still working with the symmetrical, bell-shaped distribution of x values, with mean 19. 3and Compute the control limits of 2 and 3. Draw a control chart showing the solid line at the mean and the control limits and plot the data for the 15- day period. (b) Interpret the control chart of part a. Answers. See book. (b) Days 5 to 13 are above We have nine consecutive days on one side of the mean. This is a warning signal! It would appear that the mean is slowly drifting up beyond the target value of The chart indicates that housekeeping is out of control. Ms Tamara should take corrective measures at her next staff meeting.

8 II. Standard Units and Areas Under the Standard Normal Distribution. The z value or z score tells us the number of standard deviations the original measurement is from the mean. The z value is in standard units. z x Example 6 A student has computed that it takes an average (mean) of 17 minutes with a standard deviation of 3 minutes to drive from home, park the car, and walk to an early morning class. One day it took the student 21 minutes to get to class. How many standard deviations from the average is that? Is the z value positive or negative? Explain why it should be either positive or negative. (b) Another day it took only 12 minutes for the student to get to class. What is this measurement in standard units? Is the z value positive or negative? Why should it be positive or negative? (c) Another day it took 17 minutes for the student to go from home to class. What is the z value? Why should you expect this answer?

9 Solving Probability & Statistics z x for x yields the following result: x z Example 7 Marulla s z score on a college entrance exam is 1.3. If the raw scores have a mean of 480 and a standard deviation of 70 points, what is her raw score? Answer Here we are given z,, and. We need to find the raw score x corresponding to the z score 1.3. x z x 1.3(70) 480 x 571 When the original distribution of x values is normal, then the corresponding z values have a normal distribution as well. The standard normal distribution is a normal distribution with mean 0and standard deviation 1. Any normal distribution of x values can be converted to the standard normal distribution by converting all x values to their corresponding z values. Why convert from x to z and back? There are extensive tab les that show the area under the standard normal curve for almost any interval along the z axis. The areas are important because they are equal to the probability that the measurement of an item selected at random falls in this interval.

10 Example 8 In this exercise we will find the area under the standard normal curve from z 0 to z We will use the abbreviated Table 6-5 on P Shade the area we are to find in the figure below. (b) For z 2. 53, the units value is and the tenths value is, so we look in the column under z for the number. (c) For z 2. 53, the hundredths value is, so we look in the column headings for the number (d) The area between z 0 and z 2. 53is given by the entry in the row beginning with 2.5 and in the column headed by What is the area? Answers. (b) The units value is 2 and the tenths value is 0.5, so we look in the z column for the value 2.5. (c) The hundredths value is 0.03 (d) The area is

11 Table 5 of Appendix II gives areas under the normal curve for regions beginning at z 0 and extending to a specified positive z value. However, because the normal curve is symmetrical, we also can use the table directly to find areas beginning with a negative z value and extending to z 0. Example 9 Find the area from z to z First, we draw the picture and observe the location of the requested area. Next, we find component areas from Table 5 (Appendix II) and combine them appropriately. Look at the figure. Should we add or subtract component areas? (b) Find the area under the standard normal curve between z 0 and z (c) Find the area under the standard normal curve between z and z 0. (d) Use parts b and c to find the area under the standard normal curve between z 3.00 and z

12 Answers. Since the area extends from the left side of z 0 to the right side, we add the component areas. (b) We look under the z column of Table 5 (Appendix II) until we find 2.6; then we stay in this row and move to the right until we are in the column headed by The area from z 0 to z 2. 65is given by the entry (c) Since the area from z to z 0 is the same as that from z 0 to z 3.00, we look down the z column until we find 3.0. Then we move to the right in this row until we are in the column headed by This entry is , which is the area from z to z 0. (d) (Area from) = (area from) + (area from) (-3.00 to 2.65) (-3.00 to 0) (0 to 2.65) = The desired area is

13 Example 10 Let z be a random variable with a standard normal distribution. Find the P z probability z 1.15 P refers to the probability that z values lie to the right of Shade the corresponding area under the standard normal curve. (b) The area to the right of z 0 equals. (c) Find the area between z 0 and z (d) Use the area of parts b and c to find the area to the right of z Answers. See book. (b) The area to the right of z 0 equals (c) By Table 5 of Appendix II, the area is (d) =

14 III. Areas Under Any Normal Curve Often the original normal curve is not the standard normal curve. In most cases there will not be a table of areas available for the original normal curve. We must convert original measurements x, a and b to z values. Example 11 In the last section we talked about Sunshine Stereo cassette decks. The cassette deck life was normally distributed with a mean of 2.3 years and a standard deviation of 0.4 years. We wanted to know the probability that a cassette deck will break down during the guarantee period of 2 years. Let x represent the life of a cassette deck. The statement that the cassette deck breaks during the 2 year guarantee period means the life is less than 2 years, or x 2. Convert this to a statement about x. (b) Indicate the area to be found in the figure below. Does this area correspond to the probability that z 0. 75? (c) The shaded portion of the figure corresponds to the probability that z To compute this probability, we must do the subtraction of the areas shown below. Do that subtraction. P z 0.75 P z 0 P 0.75 z 0 (d) What is the probability that the cassette deck will break before the end of P x 2 P z 0.75.] the guarantee period? [Hint:

15 Answers. x z So x 2means z (b) The shaded area does correspond to the probability that z (c) = (d) The probability is This means the company will fix or replace about 23% of the cassette decks. See Calculator Note P. 315!

16 Example 12 Find the z value so that 3% of the area under the standard normal curve lies to the left of z. Draw a standard normal curve; shade the region so that 3% of the area lies to the left of z. (b) What portion of the area lies between this z value and 0? (c) Use Table 5 in Appendix II to find the z value so that 47% of the area under the standard normal curve is between 0 and z. (d) Use the symmetry of the standard normal curve to find the z value so that 3% of the area lies to the left of z. (That is, find the z value indicated in the figure in part a.) Answers. (b) Since half the area lies to the left of zero, then 50% - 3% = 47% lies between the z value and 0. (c) The area value of Table 5 nearest to is The corresponding z value is (d) 3% of the area lies to the left of z =

17 IV. Normal Approximation to the Binomial Distribution n = number of trials r = number of successes p = probability of success q = probability of failure = 1 p np and npq Example 13 From many years of observation a biologist knows the probability is only 0.65 that any given Arctic tern will survive the migration from its summer nesting area to its winter feeding grounds. A random sample of 500 Arctic terns were banded at their summer nesting area. Use the normal approximation to the binomial and the following steps to find the probability that between 310 and 340 of the banded Artic terns will survive the migration. Let r be the number of surviving terns. The approximate P 310 r 340, we use the normal cure with and (b) P 310 r 340 is approximate equal to P x where x is a variable from the normal distribution described in part a. (c) Convert the condition x to a condition in standard units. (d) P 310 r 340 P x P 1.45 z 1.45 (e) Will the normal distribution make a good approximation to the binomial for this problem?

18 Answers. np 500(0.65) 325 npq 500(0.65)(0.35) (b) Since 310 is the left endpoint, we subtract 0.5, and since 340 is the right endpoint, we add 0.5. Consequently, P 310 r 340 P x (c) P 1.45 z P 0 z Which is approximately the probability we were seeking. (d) np 500(0.65) 325 np 500(0.35) 175 Are both greater than 5, the normal distribution will be a good approximation to the binomial.