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### Probability Distributions for Continuous Variables

• 1. PROBABILITY DISTRIBUTIONS Part II: Probability Distributions for Continuous Variables 1
• 2. Continuous variables can take any value in a specified interval falling within their plausible ranges. -The diameter of a fine metal rod may take a value of 40, 40.25, 40.75 or 41 millimeter - Human weight may take values of 120 lb, 155 lb, or 165.8 lb What is a Probability Distribution for Continuous Variables? A probability distribution for a continuous variable is largely similar to a relative frequency distribution of a large amount of data representing all possible outcomes of values of a continuous variable. Examples: - Uniform distribution - Normal distribution x P(x) 80 80.5 90 90.5 91 2
• 3. What is a Probability Distribution for Continuous Variables? Student grades (%) Mid-point (x) Number of students (frequency, f) Relative frequency (RF) 20 to < 30 25 16 0.0032 30 to < 40 35 20 0.004 40 to < 50 45 98 0.0196 50 to < 60 55 256 0.0512 60 to < 70 65 1490 0.298 70 to < 80 75 1675 0.335 80 to < 90 85 1111 0.2222 90 to < 100 95 334 0.0668 N = 5000 Sum = 1 Example: Suppose 5000 students took a course on statistics in a college over the last 5 years. The categories of grades and corresponding frequencies are as shown in the Table below. Construct a probability distribution of Student’s grade. 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 20 to < 30 30 to < 40 40 to < 50 50 to < 60 60 to < 70 70 to < 80 80 to < 90 90 to < 100 P(x) Grade Classes A = 1 3
• 4. Example: Suppose 5000 students took a course on statistics in a college over the last 5 years. The categories of grades and corresponding frequencies are as shown in the Table below. Calculate the mean and variance of students’ grades. Student Grades (%) Mid- Point (x) Number of students- (frequency, f) Relative Frequency p(x) x.p(x) (x-m)2 p(x). (x-m)2 20 to < 30 25 16 0.0032 0.08 2366.43 7.57 30 to < 40 35 20 0.004 0.14 1493.51 5.97 40 to < 50 45 98 0.0196 0.88 820.59 16.08 50 to < 60 55 256 0.0512 2.82 347.67 17.8 60 to < 70 65 1490 0.298 19.37 74.75 22.28 70 to < 80 75 1675 0.335 25.13 1.83 0.61 80 to < 90 85 1111 0.2222 18.89 128.91 28.64 90 to < 100 95 334 0.0668 6.35 455.99 30.46 N = 5000 Sum = 1 Mean = 73.65 Variance = 129.43  )(xxPm   222 )()](.)[( mm xPxxpx Standard Deviation = SQRT(129.43) 4
• 5. Working problem 6.1: The table below represents different categories of property tax of a large population of houses in New Jersey. - Plot the probability distribution - Calculate the expected value - Calculate the variance 5
• 6. What is a Uniform Probability Distribution for Continuous Variables? This is the simplest type of probability distribution for continuous variables and it can be used to model both discrete and continuous variables. It is rectangular in shape as a result of the fact that different data classes exhibit the same frequency or relative frequency. P(x) x P(x) x Continuous Uniform Distribution Discrete Uniform Distribution 120 125 130 135 140 6
• 7. What is a Uniform Probability Distribution for Continuous Variables? Examples: P(x) x •The time to fly via a commercial airliner from Newark airport to Atlanta, Georgia, ranges from 120 minutes to 140 minutes. If you monitor the fly time for many commercial flights it will follow more or less a uniform distribution •The time students take to finish one-hour standard test may range from 50 minutes to 60 minutes. Equal numbers of students complete the test over the 4 minutes intervals within this range, 50, 54, 56, 58, and 60. The finishing time of the test can be approximated by a uniform distribution • Time for pizza delivery by a certain restaurant to a certain region in town may range from 20 minutes to 30 minutes from the time the delivery man leaves the store. 7
• 8.  The time to fly via a commercial airliner from one airport to another, say from Raleigh, North Carolina to Atlanta, Georgia. This time may range from 55 minutes to 65 minutes. If you monitor the fly time for many commercial flights it will follow more or less a uniform distribution. The time a student takes to finish one-hour standard test may range from 50 minutes to 60 minutes. Approximately, equal numbers of students complete the test over the 4 minutes intervals within this range, 48, 52, 56, and 60. Thus, the finishing time of the test can be approximated by a uniform distribution Examples of variables following a Uniform Distribution: The time to deliver a pizza to a certain location in town may range from 20 minutes to 30 minutes from the time the delivery person leaves the store. This can be approximated by a uniform distribution The waiting time for a school bus may range from 20 minutes to 30 minutes. Within this period, waiting time can be approximated by a uniform distribution. 8
• 9. a b ab  1 P(x) x What is a Uniform Probability Distribution for Continuous Variables? elsewherexandbxaif ab xP 0, 1 )(    • Key Parameters min value ‘a’ and max value ‘b’ • The height of the distribution is always ab  1   12 2 2 ab ba      m 9
• 10. Example: Suppose the random variable in question is the time to drive from Washington, DC, to New York City during normal traffic hours. Assuming that driving time is uniformly distributed from 220 minutes to 250 minutes, construct a uniform probability distribution of the driving time. Determine the mean and the standard deviation of the probability distribution. Minimum value, a = 220, Maximum value, b = 250 The height of the distribution is 1/(b-a) = 1/30 = 0.0333. 220 230 240 250 30 1 P(t) t» Mean = 235 1)220250( 30 1 A The area under the curve     min66.8 12 900 12 220250 12 235 2 250220 2 22           ab ba  m 10
• 11. Example: Using the uniform distribution of the above example, answer the following questions: •What is the probability a person may spend more than 4 hours on the road driving from Washington, DC, to New York City during normal traffic hours? •What is the probability a person will make the trip from Washington, DC, to New York City during normal traffic hours in less than 2 hours? 220 230 240 250 30 1 t» Mean = 235 P(t > 4 hours) =10(1/30)= 0.333 0.333 min (0) 11
• 12. Working problem 6.2: Your teacher is always late to the class. Let the random variable x represent the time from when the class is supposed to start until the teacher shows up. In addition, suppose that your teacher could be on time for some classes or up to 15 minutes late, with all intervals between 0 and 15 being equally likely. Construct a probability distribution for the random continuous variable, x. Determine the mean and the standard deviation? Working problem 6.3: Your teacher is always late to the class. Let the random variable x represent the time from when the class is supposed to start until the teacher shows up. In addition, suppose that your teacher could be on time for some classes or up to 15 minutes late, with all intervals between 0 and 15 being equally likely. - What is the probability that the teacher will arrive on time? - What is the probability that the teacher will arrive within 5 minutes from the start of the class? - What is the probability that the teacher will arrive in more than 10 minutes from the start of the class? 12
• 13. Working problem 6.4: Waiting period to see your eye doctor can be considered as a random variable x representing the time from signing in to the time you actually see the doctor. Further suppose that your doctor could see you as soon as you sign in (x = 0) or up to 30 minutes late (x = 30) with all intervals between 0 and 30 being equally likely. Construct a probability distribution for the random continuous variable, x. Determine the mean and the standard deviation? Working problem 6.5: Waiting period to see your eye doctor can be considered as a random variable x representing the time from signing in to the time you actually see the doctor. Further suppose that your doctor could see you as soon as you sign in (x = 0) or up to 30 minutes late (x = 30) with all intervals between 0 and 30 being equally likely. - What is the probability that your eye doctor will see you between 10 and 20 minutes? - What is the probability that your eye doctor will see you in less than 5 minutes? 13
• 14. Using Excel to simulate a Uniform Probability Distribution 2 Go to Data 3 Go to Data Analysis 1 Type a label Named “Time of Driving” 4Go to Random Number Generation 5 Press OK Example: Suppose the random variable in question is the time to drive from Washington, DC, to New York City during normal traffic hours. Assuming that driving time is uniformly distributed from 220 minutes to 250 minutes, construct a uniform probability distribution of the driving time. Determine the mean and the standard deviation of the probability distribution. 14
• 15. - Select 1 variable - Select, say 1000 random numbers - Select Uniform Distribution - You will be prompted to insert Uniform distribution parameters (220 and 250 for this example) - Specify an output right under the label 6 7 Press OK Using Excel to simulate a Uniform Probability Distribution 15
• 16. - The output of the analysis of will be a long column of 1000 random numbers representing random values of driving time from Washington, DC, to New York City during normal traffic hours Note: decimals are rounded off to one Using Excel to simulate a Uniform Probability Distribution 16
• 17. - Follow the steps for performing descriptive statistics and the steps for constructing a histogram described in Chapters 2 and 3 to obtain the Uniform frequency distribution shown here Using Excel to simulate a Uniform Probability Distribution 17
• 18. - You can convert the histogram in the previous Figure to a “Probability Distribution” By adding a Column to the frequency table label it P(X) and Calculate the probability corresponding to each frequency, this is the relative frequency (Class frequency/Total frequency) You can then copy the P(x) column and paste it on the graph, delete the previous series And change the labels on the graph. 163.0 1000 163 Using Excel to simulate a Uniform Probability Distribution 18
• 19. What is a Normal Distribution?      x exP x 22 2/ 2 1 )( m  Features of the Normal Distribution (1) Bell-Shaped (2) Defined by two parameters, m and  x P(x) Mean Mode Median  m 19
• 20. What is a Normal Distribution?      x exP x 22 2/ 2 1 )( m  Features of the Normal Distribution (1) Symmetrical (2) Area Under the Curve = 1 x P(x) Mean Mode Median  0.50.5 m 20
• 21. What is a Normal Distribution? Examples of random variables following a normal distribution include:  People’s income in a given nation- few earn low income, few earn high income, and the majority earns middle income.  Students’ grades in a course- few earn low grades, few earn high grades, and the majority earns middle grades.  People’s height- few are short, few are tall, and the majority has middle heights.  Education cost- some colleges charge small tuition, some charge very high tuition, and the majority charges tuition in between. 21
• 22. Example: Given the three normal distributions a, b, and C below: - Which normal curve has the greatest mean and which has the lowest mean? - Which normal curve has the greatest standard deviation and which has the lowest standard deviation? -20 0 20 40 60 80 100 120 140 160 180 200 220 A B C Solution: Distribution A: Mean ≈ 30 Distribution B: Mean ≈ 80 Distribution C: Mean ≈ 160 Distribution C is the most spread out distribution. Therefore, it has the greatest standard deviation. Distribution B is the least spread out distribution. Therefore, it has the lowest standard deviation. 22
• 23. Recall: The Empirical Rule What is a Normal Distribution? RelativeFrequency(%) Mean Mode Median m +/- 3  m +/- 2  99.74% 95.44% m +/- 1  68.26% x 23
• 25. What is a Normal Distribution? Example: Instructor Mr. Z is teaching a course of statistics in a community college. The grades of the population of students taught by the instructor over a number of years are represented by the normal distribution shown in the Figure below. Describe the pattern of this instructor’s grade. 100959085807570 f(X) X 90 • As indicated by the area under the curve for grades above 90, about 16% of Mr. Z students make an A in his course. You can also see that hardly any student in Mr. Z class Instructor Z m = 85  = 5 Percent of A students P(G>90) = 0.1587 or 16% of Students earn A Grade 25
• 26. Example: Suppose we want to compare the grades of Mr. Z with those of another instructor, Mr. W who is teaching the same course for different groups of students. The grades of the populations of students taught by the two instructors are represented by the two normal distributions shown in the next slide. Note that we are also looking at the percent of students making A grade taught by each instructor. Describe the two normal distributions and compare the grades by the two instructors. If you have a choice to take the statistics course by either instructor, which one would you chose to take the course with? What is a Normal Distribution? 26
• 27. 100959085807570 f(X) X 90 Instructor Z m = 85  = 5 Percent of A students P(G>90) = 0.1587 or 16% of Students earn A Grade f(X) X 91898785838179 90 Instructor W m = 85  = 2 Percent of A students P(G>90) = 0.0062 or < 1% of Students earn A Grade 27
• 28. 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 A C B Working Problem 6.6: Given the three normal distributions A, B, and C below: - Which normal curve has the greatest mean and which has the lowest mean? - Which normal curve has the greatest standard deviation and which has the lowest standard deviation? 28
• 29. Working Problem 6.7: The two normal distributions below describe the area (cm2) of ceramic tiles produced by two manufacturers A & B: - Compare the mean and the variability of the two distributions - Which manufacturer should you buy ceramic tiles from? 16.616.416.216.015.815.615.4 Area (cm2) m = 16  = 0.2 Manufacturer A 18.417.616.816.015.214.413.6 m = 16  = 0.8 Area (cm2) Manufacturer B 29
• 30. Review Problem 6.1: The frequency distribution given below represents the heights of 1000 students in a college. - Perform descriptive statistics to Prove whether this distribution can be approximated by a normal distribution. - Determine, the mean, the median, the mode, and the standard deviation - Given that this distribution is indeed normal, use the empirical rule to determine the heights of 68.26% of the students , the heights of 95.44% of the students , and the heights of 99.74% of the students , lower Heights (cm) upper midpoint frequency 130 < 135 133 0 135 < 140 138 4 140 < 145 143 15 145 < 150 148 37 150 < 155 153 64 155 < 160 158 125 160 < 165 163 190 165 < 170 168 196 170 < 175 173 161 175 < 180 178 97 180 < 185 183 75 185 < 190 188 28 190 < 195 193 7 195 < 200 197 1 1000 30
• 31. What is the Standard Normal Distribution? 3210-1-2-3 f(z) z Mean = m = 0 68.26% 95.44% 99.74% Transforming x to z using ./)( m xz    z ezP z 2/2 2 1 )(  Standard Deviation  = 1 A normal distribution having mean 0 and standard deviation 1 is said to be a standard normal distribution. 31
• 32. The basic properties of the standard normal distribution: •The total area under the standard normal curve is one. •The standard normal curve extends indefinitely in both directions ( ) •The standard normal curve is symmetric about 0. •Almost all the area under the standard normal curve lies between values of z of −3.4 and 3.4. •Areas under the standard normal curve can be obtained from special tables such as the ones shown in Appendix 6.A. 32
• 33. 3210-1-2-3 f(z) z -1.50 A= 0.0668 3210-1-2-3 f(z) z 1.50 A= 0.9332 z 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 -3.4 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0002 -3.3 0.0005 0.0005 0.0005 0.0004 0.0004 0.0004 0.0004 0.0004 0.0004 0.0003 -3.2 0.0007 0.0007 0.0006 0.0006 0.0006 0.0006 0.0006 0.0005 0.0005 0.0005 .. .. .. .. .. .. .. .. .. .. .. -1.5 0.0668 0.0655 0.0643 0.0630 0.0618 0.0606 0.0594 0.0582 0.0571 0.0559 -1.4 0.0808 0.0793 0.0778 0.0764 0.0749 0.0735 0.0721 0.0708 0.0694 0.0681 .. .. .. .. .. .. .. .. .. .. .. 0.0 0.5000 0.4960 0.4920 0.4880 0.4840 0.4801 0.4761 0.4721 0.4681 0.4641 z 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359 .. .. .. .. .. .. .. .. .. .. .. 1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319 1.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545 .. .. .. .. .. .. .. .. .. .. .. 3.2 0.9993 0.9993 0.9994 0.9994 0.9994 0.9994 0.9994 0.9995 0.9995 0.9995 Construction of the Standard Normal Table Area Values 33
• 34. What is the Standard Normal Distribution?    z ezP z 2/2 2 1 )( Transforming x to z using ./)( m xz •The total area under the standard normal curve is 1 •The standard normal curve extends indefinitely in both directions  z •The standard normal curve is symmetric about 0 •Almost all the area under the standard normal curve lies between −3.4 and 3.4 •Areas under the standard normal curve can be obtained from special Tables (Appendix 6.A) •The first column of the Tables represent the z values, the first row of the Tables represent the complementary decimals, and the four-decimal-place numbers in the body of the Tables gives the area under the standard normal curve. 34
• 37. z 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.0359 0.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.0753 0.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.1141 0.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.1517 0.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.1879 0.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.2224 0.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.2549 0.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.2852 0.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.3133 0.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.3389 1.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.3621 1.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.3830 1.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.4015 1.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177 1.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.4319 1.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.4441 1.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545 1.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.4633 1.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.4706 1.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.4767 2.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817 2.1 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.4857 2.2 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.4890 2.3 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.4916 2.4 0.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.4936 2.5 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.4952 2.6 0.4953 0.4955 0.4956 0.4957 0.4959 0.4960 0.4961 0.4962 0.4963 0.4964 2.7 0.4965 0.4966 0.4967 0.4968 0.4969 0.4970 0.4971 0.4972 0.4973 0.4974 2.8 0.4974 0.4975 0.4976 0.4977 0.4977 0.4978 0.4979 0.4979 0.4980 0.4981 2.9 0.4981 0.4982 0.4982 0.4983 0.4984 0.4984 0.4985 0.4985 0.4986 0.4986 3.0 0.4987 0.4987 0.4987 0.4988 0.4988 0.4989 0.4989 0.4989 0.4990 0.4990 310-1-2-3 f(z) z 1.96 0.4750 APPENDIX 6.B Standard Normal Distribution Table )0( ztoP 37
• 38. Example: Find the area under the standard normal curve for the following z values: (a) 0 ≤ z ≤ 1.5 What is the Standard Normal Distribution? 3210-1-2-3 z 1.500.00 A = A (z = 1.5) – A (z = 0) = 0.9332 – 0.5000 = 0.4332A Areas Corresponding to z values: 0 ≤ z ≤ 1.5 z 0 0.01 0.02 0.03 0.04 0.08 0.09 0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5319 0.5359 0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5714 0.5753 … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … 1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9306 0.9319 1.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9429 0.9441 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9535 0.9545 3210-1-2-3 z 1.50 0.9332 3210-1-2-3 z 0.50 38
• 39. Example: Find the area under the standard normal curve for the following z values: (b) –0.46 ≤ z ≤ 2.30 What is the Standard Normal Distribution? 3210-1-2-3 z 2.30-0.46 A = A (z = 2.3) – A (z = -0.46) = 0.9893 – 0.3228 = 0.6665A Areas Corresponding to z values: –0.46 ≤ z ≤ 2.30 z 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.09 2.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9916 2.4 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9936 z 0 0.01 0.02 0.03 0.04 0.05 0.06 0.09 -0.5 0.3085 0.305 0.3015 0.2981 0.2946 0.2912 0.2877 0.2776 -0.4 0.3446 0.3409 0.3372 0.3336 0.33 0.3264 0.3228 0.3121 -0.3 0.3821 0.3783 0.3745 0.3707 0.3669 0.3632 0.3594 0.3483 -0.2 0.4207 0.4168 0.4129 0.409 0.4052 0.4013 0.3974 0.3859 -0.1 0.4602 0.4562 0.4522 0.4483 0.4443 0.4404 0.4364 0.4247 0 0.5 0.496 0.492 0.488 0.484 0.4801 0.4761 0.4641 3210-1-2-3 z 2.30 0.9893 3210-1-2-3 z-0.46 0.3228 39
• 40. Example: Find the area under the standard normal curve for the following z values: (c) 0.80 ≤ z ≤ 2.0 What is the Standard Normal Distribution? A = A (z = 2.0) – A (z = 0.8) = 0.9772 – 0.7881 = 0.1891 3210-1-2-3 z 2.000.80 A Areas Corresponding to z values: 0.80 ≤ z ≤ 2.0 z 0 0.01 0.02 0.03 0.04 0.08 0.09 0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5319 0.5359 0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7823 0.7852 0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8106 0.8133 0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8365 0.8389 1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8599 0.8621 z 0 0.01 0.02 0.03 0.04 0.08 0.09 1.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9699 0.9706 1.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9761 0.9767 2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9812 0.9817 2.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9854 0.9857 2.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9887 0.9890 2.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9913 0.9916 310-1-2-3 z 2.00 0.9772 3210-1-2-3 z 0.80 0.7881 40
• 41. Example: Determine the value(s) of z in the following cases: (a) Area under the normal curve between 0 and z is 0.3790. What is the Standard Normal Distribution? 3210-1-2-3 z 1.170.00 0.5000 z 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621 1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.8830 1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015 1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177 1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319 1.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441 A= 0.879 0.3790 41
• 42. Example: Determine the value(s) of z in the following cases: (b) Area under the normal curve to left of z is 0.6100. What is the Standard Normal Distribution? 3210-1-2-3 f(z) z 0.28 0.6100 z 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359 0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753 0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141 0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517 0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879 0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224 0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549 0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.7852 42
• 43. Example: The mean value of course grades in a large population of students is 85, and the standard deviation is 5, assuming that the grade follows a normal distribution, determine the z values corresponding to the following grades: • Grade = x = 75 • 80 ≤ x ≤ 90 • 75 ≤ x ≤ 95 • 70 ≤ x ≤ 100 Solution: • At x = 75, z = (x-m)/ = (75-85)/5 = -2 • 80 ≤ x ≤ 90 yields (80-85)/5 ≤ z ≤ (90-85)/5, or -1 ≤ z ≤ +1 •75 ≤ x ≤ 95 yields (75-85)/5 ≤ z ≤ (95-85)/5, or -2 ≤ z ≤ +2 •70 ≤ x ≤ 100 yields (70-85)/5 ≤ z ≤ (100-85)/5, or -3 ≤ z ≤ +3 43
• 44. Example: In previous example. what percent of students made the following grades? • 80 ≤ x ≤ 90 • 75 ≤ x ≤ 95 •70 ≤ x ≤ 100 Solution: Using the empirical rule, • 80 ≤ x ≤ 90 yields (80-85)/5 ≤ z ≤ (90-85)/5, or -1 ≤ z ≤ +1 and this corresponds to 68.26% of the students’ grades. • 75 ≤ x ≤ 95 yields (75-85)/5 ≤ z ≤ (95-85)/5, or -2 ≤ z ≤ +2 and this corresponds to 95.44% of the students’ grades. • 70 ≤ x ≤ 100 yields (70-85)/5 ≤ z ≤ (100-85)/5, or -3 ≤ z ≤ +3 and this corresponds to 99.74% of the students’ grades. 44
• 45. -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 x (A) Mean = m = 2 Std. Dev. =  = 1 (B) Mean = m = 6 Std. Dev. =  = 1/2 (C) Mean = m = 12 Std. Dev. =  = 2 Working Problem 6.8: For the three normal distributions shown below, find the values of z corresponding to the values of x in the circles (A) z= ? (B) z= ? (C) z= ? 45
• 46. Working Problem 6.9: Find the area under the standard normal curve for the following z values: (a) 0 ≤ z ≤ 1.0 (b) -2 ≤ z ≤ 2 (c) z < 3 Working Problem 6.10: Find the area under the standard normal curve for the following z values: (a) -∞ ≤ z ≤ 1.3 (b) -1 ≤ z ≤ 1.3 (c) z > -3.2 Working Problem 6.11: Determine the value(s) of z in the following cases: (a)Area under the normal curve between - ∞ and z is 0.100. (b) Area under the normal curve to right of z is 0.8100. (c) Area under the normal curve to left of z is 0.7100 46
• 47. Working Problem 6.12: The three normal distributions shown below represent the grades of pre-algebra course of students obtained in three different semesters… - Describe and compare the performances of students in the three semester - What percent of students failed (<60%) in each semester m  75 m  80 m  85   5   5   5 Working Problem 6.13: The three normal distributions shown below represent the grades of pre-algebra course of students obtained in three different semesters… - Describe and compare the performances of students in the three semester - What percent of students made at least B (≥ 80%) in each semester m  75 m  80 m  85   8   6   3 47
• 48. The za Notation Example: Find the za values for the following cases: • a = 0.01 • a = 0. 05 • a = 0.10 (a) a = 0.01 or 1% 1 – a = 0.99 or 99% (a) za = z0.01 = 2.33 One-Sided (b) a = 0.05 or 5% 1 – a = 0.95 or 95% za = z0.05 = 1.64 One-Sided (c) a = 0.10 or 10% 1 – a = 0.9 or 90% za = z0.1 = 1.28 One-Sided 3210-1-2-3 z A = 1- a = 0.90 or 90% a = 0.10 za = 1.28 3210-1-2-3 z a = 0.05 za = 1.64 A = 1- a = 0.95 or 95% 3210-1-2-3 z za =2.33 a = 0.01A = 1- a = 0.99 or 99% 48
• 49. Example: Find the za/2 values for the following cases: • a/2 = 0.005 • a/2 = 0. 025 • a/2 = 0.05 (a) a = 0.01 or 1% 1 – 2 a/2 = 0.99 or 99% za/2 = z0.01/2 = z0.005 = +/- 2.58 Two-Sided (b) a = 0.05 or 5% 1 – 2 a/2 = 0.95 or 95% za/2 = z0.05/2 = z0.025 = +/- 1.96 Two-Sided (c) a = 0.10 or 10% 1 – 2 a/2 = 0.90 or 90% za/2 = z0.1/2 = z0.05 = +/- 1.64 Two-Sided 3210-1-2-3 z A = 1- a = 0.99 or 99% za/2 = +2.58za/2 = -2.58 a/2 = 0.005 a/2 = 0.005 3210-1-2-3 z A = 1- a = 0.95 or 95% za/2 = +1.96za/2 = -1.96 a/2 = 0.025a/2 = 0.025 3210-1-2-3 z A = 1- a = 0.90 or 90% za/2 = +1.64za/2 = -1.64 a/2 = 0.05a/2 = 0.05 49
• 50. Applications of the standard normal distribution In most applications dealing with the standard normal distribution require the following basic steps: Step 1: Sketch the normal curve associated with the variable to describe the problem in question Step 2: Shade the region of interest and mark its delimiting x-value(s). Step 3: Calculate the z-score(s) corresponding to the x values: . : . Step 4: Use the standard normal distribution table to find the area under the standard normal curve delimited by the z-score(s). Step 5: Express the findings in terms of x values. 50
• 51. Example: The Thermosense Company produces digital thermometers that have a 0oC midpoint, which is the reading expected at the freezing point of water. In actual testing of a large number of thermometers, the temperature at freezing points fluctuates around the 0oC from negative values (below 0oC) to positive values (above 0oC). The temperature follows a normal distribution with a mean value of 0oC and a standard deviation of 1oC. If one thermometer is randomly selected, find the probability that, at the freezing point of water, the reading is more than -1.2oC. What is the probability that, at the freezing point of water, the reading is less than 0.5oC. Solution: With a mean value m = 0 and a standard deviation  = 1, this is a standard normal distribution. Using the normal table in Appendix 6-A, we can find the area P(-∞ ≤ z ≤ -1.2), which is 0.1151 as shown below. This yields value of P(z > -1.2) of 0.8849 (or 1- 0.1151). This answer means that the chance that the reading at the freezing point of water will be more than -1.2oC is 0.8849 or about 88.5%. z 0 0.01 0.02 0.03 0.04 0.05 -3.3 0.0005 0.0005 0.0005 0.0004 0.0004 0.0004 -1.4 … … … … … … -1.3 0.0968 0.0951 0.0934 0.0918 0.0901 0.0885 -1.2 0.1151 0.1131 0.1112 0.1093 0.1075 0.1056 -1.1 0.1357 0.1335 0.1314 0.1292 0.1271 0.1251 -1.0 … … … … … … 3210-1-2-3 z -1.20 0.8849 0.1151 51
• 52. What is the probability that, at the freezing point of water, the reading is less than 0.5oC. Solution: The probability that, at the freezing point of water, the reading is less than 0.5oC can be obtained using the normal table in Appendix 6-A, we can find the area P(z < 0.5), which is 0.6915 as shown below. This answer means that the chance that the reading at the freezing point of water will be less than 0.5oC is 0.6915 or about 69.15%. z 0 0.01 0.02 0.03 0.0 0.5000 0.5040 0.5080 0.5120 … … … … … 0.5 0.6915 0.6950 0.6985 0.7019 0.6 0.7257 0.7291 0.7324 0.7357 3210-1-2-3 z 0.50 0.6915 0.3085 52
• 53. Working Problem 6.14: The property tax of houses in a large City in the State of New Jersey is normally distributed. The normal curve shown below represents this distribution. What is the mean value of property tax? Estimate the standard deviation of this normal distribution. 6,0005,5005,0004,5004,0003,5003,000 f(X) x 53
• 54. Example: A sewing mill pays workers by the quantity of garments they make. The average annual pay per worker is \$18,000 and the standard deviation is \$4000. Find the probability that a worker selected randomly earns between \$13,000 and \$20,000. Solution: Following the procedure in the above example, the probability that a worker selected randomly earns between \$13,000 and \$20,000 is calculated using the steps shown in Figure 6.20. As can be seen in this Figure, this probability is 0.5859. This result also implies that about 58.6% of the workers earn wages between \$13,000 and \$20,000. 30,00026,00022,00018,00014,00010,0006,000 f(x) x m = \$18,000  = \$4000 3210-1-2-3 0.50-1.25 30,00026,00022,00018,00014,00010,0006,000 f(x) x 20,00013,000 A= ? 0.5859 Finding the probability a worker selected earns between \$13,000 and \$20,000 54
• 55. Working Problem 6.15: The weekly gross income of restaurant assistant managers follows a normal distribution with a mean of \$1,000 and a standard deviation of \$100. The variation in the weekly income of assistant managers is a result of managers getting a commission in addition to their weekly salary. - What are the z values for the income of assistant managers earning between \$900 and \$1,100 weekly? What is the percent of assistant managers earning this range of income? - What are the z values for the income of assistant managers earning between \$800 and \$1,200 weekly? What is the percent of assistant managers earning this range of income? - What are the z values for the income of assistant managers earning between \$700 and \$1,100 weekly? What is the percent of assistant managers earning this range of income? - What is the z value for the income of assistant managers earning less than \$860 weekly? What is the percent of assistant managers earning this range of income? - What is the z value for the income of assistant managers earning more than \$1,050 weekly? What is the percent of assistant managers earning this range of income? 55
• 56. Example: According to the controversial 2002 book titled “IQ and the Wealth of Nations” by Richard Lynn, and Tatu Vanhanen (Praeger/Greenwood Publication, Westport, Connecticut, London, 2002), The average IQ (Intelligence Quotient) test score of the world was 88 and the standard deviation was 12. Using the following criteria typically describes one’s intelligence with respect to IQ score: Applications of the Standard Normal Distribution IQ Description 130+ Very superior 120-129 Superior 110-119 High average 90-109 Average 80-89 Low average 70-79 Borderline Below 70 Extremely low IQ Criteria (http://iq-test.learninginfo.org/iq04.htm) • Determine the percent of people in the world that their intelligences are considered above average or better (IQ ≥ 110)? • Determine the percent of people in the world that their intelligences are considered average (IQ = 90-109)? 56
• 57. Example: m = 88  = 12 Applications of the Standard Normal Distribution IQ Description 130+ Very superior 120-129 Superior 110-119 High average 90-109 Average 80-89 Low average 70-79 Borderline Below 70 Extremely low IQ Criteria (http://iq-test.learninginfo.org/iq04.htm) • Determine the percent of people in the world that their intelligences are considered above average or better (IQ ≥ 110)? 124112100766452 f( IQ) IQ m = 88  = 12 (a) 12411210088766452 f(IQ) IQ 110 (b) 3210-1-2-3 f(z) z z = 1.83 Area of interest = 0.0336(c) 0.9664 1.83 57
• 58. Example: m = 88  = 12 Applications of the Standard Normal Distribution IQ Description 130+ Very superior 120-129 Superior 110-119 High average 90-109 Average 80-89 Low average 70-79 Borderline Below 70 Extremely low IQ Criteria (http://iq-test.learninginfo.org/iq04.htm) • Determine the percent of people in the world that their intelligences are considered average (IQ = 90-109)? 124112100766452 f( IQ) IQ m = 88 (a)  = 12 10990 12411210088766452 f(IQ) IQ A=? (b) 3210-1-2-3 f(z) z 1.750.17 Area of interest = 0.3920 (c) 58
• 59. Using Microsoft Excel® to find the Area under the Normal Distribution Curve 1 2 3 • Go to Excel Spreadsheet • Click fx on the button bar • Select Statistical from the “Or select a category” drop down list box •Select NORMDIST from the “Select a function” list • Click OK Example: Determine the area to the left (cumulative area) of a certain value of x, say 109 in a normal distribution with mean value of 88 and standard deviation of 12. 59
• 60. 12411210088766452 f(X) X 109 0.95994 4 •Type 109 in the X text box •Click in the Mean text box and type 88 •Click in the Standard deviation text box and type 12 •Click in the Cumulative text box and type TRUE •You should be able to see the value of the area as illustrated by the circle shown here, or you can Click Ok and the value will be presented 60
• 61. Using Microsoft Excel® to Generate Random Numbers following a Normal Distribution 2 3 m  Example: Sales Revenues m = 100,000  = 15000 Example: In the analysis of the net income of a textile company, it was found that the total sales revenues per week of the company follows a normal distribution with a mean value of \$100,000, and a standard deviation of \$15,000. Generate a normal distribution for the sales revenues per week using Microsoft Excel® data analysis. 61
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### Editor's Notes

1. In chapter 5, we introduced probability distribution and divided them into two main types: probability distributions for discrete variables and probability distributions for continuous variables. We also discussed probability distributions for discrete variables using common distributions such as the binomial and the Poisson distributions. In this chapter, we turn our attention to probability distributions for continuous variables. In this regard, we remind students that a continuous variable can take any value in a specified interval falling within its plausible range. For example, the diameter of a fine metal rod may take a value of 40, 40.25, 40.75 or 41 millimeter, and human weight may take values of 120 lb, 155 lb, or 165.8 lb. These data imply continuity of the variable values and the variables are classified as continuous variables.   A probability distribution for a continuous variable is largely similar to a relative frequency distribution of a large amount of data representing all possible outcomes of values of a continuous variable. As we discussed earlier in chapter 3, a histogram or a frequency curve is constructed by dividing the data under consideration into pre-specified categories or classes of equal widths and determine the frequencies corresponding to different classes. This type of distributions is the basis for a probability distribution of continuous variables.
2. A frequency distribution of the data in the Table and the Figure shown here. We can also calculate the relative frequency for each class, as discussed in Chapter 3, and construct a relative frequency distribution as shown in the Figure. Since 5000 students can be considered as a finite population, one can consider the relative frequency distribution as a probability distribution of student grade. Note that both the frequency distribution and the probability distribution exhibit the same shape with the difference being the values of the vertical axis as explained in Chapter 3.
3. The simplest type of probability distributions for continuous variables is the uniform distribution. Ideally, the uniform distribution is rectangular in shape as a result of the fact that different data classes exhibit the same frequency or relative frequency. The distribution is fully defined by two parameters: the minimum and the maximum value of a variable.
4. Comparison between the distributions of grades of the two instructors shown here reveals the following key points:   The grades of the two instructors follow a normal distribution The two distributions have the same mean value (85) but different values of standard deviation (5, 2, respectively) The majority of students taken the course with both instructors (the mode) earn a B grade or 85. Grades given by instructor Z seems to cover a wider range by virtue of the higher standard deviation of 5 than those given by instructor W (standard deviation of only 2). Using the empirical rule, the two instructors hardly fail any student, but Instructor Z gives more A grade (16%) than Instructor W (less than 1%).
5. The idea of a standard normal distribution came about as a result of the need for a more simplified normal function that can be easily integrated to obtain the area under the normal curve. This simplification is achieved by moving the center of the normal distribution to the origin (at x = 0), via transforming the variable x to a new variable z, defined by . Common names of the z statistics are the z score, the standard normal deviate, the standard normal value, or the normal deviate. Note that the empirical rule, discussed in Chapter 3, can be used for the standard normal distribution with 68.26 percent of the observations falling within plus and minus one of the mean; 95.44 percent of the observations falling within plus and minus two of the mean; and 99.74 percent falling within plus and minus three of the mean.
6. The construction of the standard normal table is illustrated in this slide. The first column of the table represents the z values, the first row of the tables represent the complementary decimals of z values, and the four-decimal-place numbers in the body of the tables gives the area under the standard normal curve from - ∞ to z.
7. This is another form of the table in which the area presented is from the origin 0 to some z value is shown in Appendix 6.B (textbook).
8. The za notation is an important tool that students will become very familiar with when we discuss inferential statistics (Chapters 7 through 10). This notation indicates the z-score that has an area alpha under the standard normal distribution curve to the right of z. At this point, few examples will familiarize the student with the zalpha notation using values of alpha that are commonly utilized in inferential statistics analysis.
9. In practice, the standard normal distribution is used in numerous applications. Examples of these applications are presented in the next few slides. We should point out that most applications dealing with the standard normal distribution require the basic steps described here.
10. Step 1: Go to Excel Spreadsheet. Step 3: Click fx on the button bar. Step 3: Select Statistical from the “Or select a category” drop down list box. Step 4: Select NORMDIST from the “Select a function” list. Step 5: Click OK. Step 6: Type in the x value 109 in the X text box. Step 7: Click in the Mean text box and type in the mean value 88. Step 8: Click in the Standard deviation text box and type in the standard deviation value 12. Step 9: Click in the Cumulative text box and type TRUE.
11. Example : In the analysis of the net income of a textile company, it was found that the total sales revenues per week of the company follows a normal distribution with a mean value of \$100,000, and a standard deviation of \$15,000. Generate a normal distribution for the sales revenues per week using Microsoft Excel® data analysis. The normal distributions for sales revenues per week can be generated using Excel® Data Analysis (Random Number Generation) as discussed above. The only difference is that when you are asked to select a distribution, you select the normal distribution and insert the values of the parameters mu and sigma associated with the variable in question as shown here. The other steps used are exactly similar to those discussed for the uniform distribution. See next slides.
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