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Chapter 6 The Normal Distribution (The original slide set from the Bluman text has been trimmed and augmented with extra explanations and TI-84 examples. The Bluman slides have the footer and rose background.) © McGraw-Hill, Bluman, 5th ed., Chapter 6 1 6.1 Normal Distributions Many continuous variables have distributions that are bell-shaped and are called approximately normally distributed variables. The theoretical curve, called the bell curve or the Gaussian distribution, can be used to study many variables that are not normally distributed but are approximately normal. Bluman, Chapter 6 2 Normal Distributions The mathematical equation for the a normal distribution is: y e ( X )2 (2 2 ) 2 where e 2.718 3.14 population mean population standard deviation Bluman, Chapter 6 3 Normal Distributions The shape and position of the normal distribution curve depend on two parameters, the mean and the standard deviation. Each normally distributed variable has its own normal distribution curve, which depends on the values of the variable’s mean and standard deviation. Bluman, Chapter 6 4 Normal Distributions - different ones for different values of and Bluman, Chapter 6 5 Normal Distribution Properties The normal distribution curve is bell-shaped. The mean, median, and mode are equal and located at the center of the distribution. The normal distribution curve is unimodal (i.e., it has only one mode). The curve is symmetrical about the mean, which is equivalent to saying that its shape is the same on both sides of a vertical line passing through the center. Bluman, Chapter 6 6 Normal Distribution Properties The curve is continuous—i.e., there are no gaps or holes. For each value of X, here is a corresponding value of Y. The curve never touches the x axis. Theoretically, no matter how far in either direction the curve extends, it never meets the x axis—but it gets increasingly closer. Bluman, Chapter 6 7 Normal Distribution Properties They said “it never meets the x axis—but it gets increasingly closer.” Example: for the standard normal distribution where = 0, = 1, when = 5, = 0.0000015 And when = 10, = 7.7 × 10−23 8 Normal Distribution Properties The total area under the normal distribution curve is equal to 1.00 or 100%. The area under the normal curve that lies within one standard deviation of the mean is approximately 0.68 (68%). two standard deviations of the mean is approximately 0.95 (95%). three standard deviations of the mean is approximately 0.997 ( 99.7%). Bluman, Chapter 6 9 Normal Distribution Properties Bluman, Chapter 6 10 “The Empirical Rule” Bluman, Chapter 6 11 Standard Normal Distribution Since each normally distributed variable has its own mean and standard deviation, the shape and location of these curves will vary. In practical applications, one would have to have a table of areas under the curve for each variable. To simplify this, statisticians use the standard normal distribution. The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1. Bluman, Chapter 6 12 The Standard Normal Distribution is our favorite This one is the most special of all of them mean: = 0 The standard deviation: = 1 The Horizontal = ⋯ , −3, −2, −1, 0, 1, 2, 3, ⋯ Total area between curve and axis = 1.000 13 z value (Standard Value) The z value is the number of standard deviations that a particular X value is away from the mean. The formula for finding the z value is: value - mean z standard deviation z X Bluman, Chapter 6 14 x value Going the other way: If you know the z value and you need to find the x value, = ∙ + =∙+ 6 15 Area Problems KEY CONNECTION: PROBABILITY is AREA AREA is PROBABILITY HOW THIS AFFECTS YOUR LIFE: You want to answer Probability questions You find the answers by finding Areas of regions underneath the Normal Curve. 16 Three kinds of area problems “What is the area to the left of some ?” “What is the area to the right of some ?” “What is the area between 1 and 2 ?” 17 How do you find these areas? 1. 2. 3. By using a printed table that gives areas to the left of = −#. ## and = #. ## Or by using the TI-84 normalcdf(z1,z2) function. Or by using Excel’s =NORM.S.DIST(z,TRUE) function 18 Area under the Standard Normal Distribution Curve 1. To the left of any z value: Look up the z value in the table and use the area given. Bluman, Chapter 6 19 Area under the Standard Normal Distribution Curve 2. To the right of any z value: Look up the z value and subtract the area from 1. Bluman, Chapter 6 20 Area under the Standard Normal Distribution Curve 3. Between two z values: Look up both z values and subtract the corresponding areas. Bluman, Chapter 6 21 TI-84 Methods To get normalcdf() 2ND DISTR Area between two z vals. Normalcdf(zLow, zHigh) It’s on the VARS key 2:normalcdf( Beware! You do NOT want to use 1:normalpdf( in most cases in this course. Example: Area between = −1 and = +1 Agrees with the Empirical Rule (68/95/99.7) ! 22 TI-84 Methods Area to the left of z normalcdf(-99,z) Example: Area to the left of z=1.23 Area to the right of z normalcdf(z,99) Example: Area to the right of z=1.23 Observe: area to left + area to right = 1.0000000 Not a coincidence!!!! 23 Example 6-1: Area under the Curve Find the area to the left of z = 1.99. The value in the 1.9 row and the .09 column of Table E is .9767. The area is .9767. Bluman, Chapter 6 24 Example 6-1: Area under the Curve Find the area to the left of z = 1.99. They got 0.9767 using the printed table. 25 Example 6-2: Area under the Curve Find the area to right of z = -1.16. The value in the -1.1 row and the .06 column of Table E is .1230. The area is 1 - .1230 = .8770. Bluman, Chapter 6 26 Example 6-2: Area under the Curve Find the area to right of z = -1.16. They used the printed table, which only gives areas to the left, so they had to subtract, 1 -0.1230 = 0.8770. The TI-84 normalcdf() was more direct. 27 Example 6-3: Area under the Curve Find the area between z = 1.68 and z = -1.37. The values for z = 1.68 is .9535 and for z = -1.37 is .0853. The area is .9535 - .0853 = .8682. Bluman, Chapter 6 28 Example 6-3: Area under the Curve Find the area between z = 1.68 and z = -1.37. With the printed tables, they had to do two lookups and subtract the results to get .8682 The TI-84 normalcdf() was more direct. Bluman, Chapter 6 29 Problems that work backwards They give you the area You have to work backwards to find the z score. 30 Example of a backwards problem What z value divides the area under the standard normal curve so that the area to the left of that z is 0.7123? And what is the area to the right of that z score? 31 How to solve it using the printed table The area to the left is .7123. Then look for that value inside Table E. The z value is 0.56. Bluman, Chapter 6 32 Backwards problem using TI-84 invNorm( is the tool 2ND DISTR again 3:invNorm( Stands for “Inverse Normal” Tell him area to the left He responds with the z score. Example: area 0.7123 33 If they give you area to the right Example: Find z so area to the right of z is 0.7500 But Tables & TI-84 deal with areas to the left. COMPLEMENT: If area to the right is 0.7500, are to the left is 1 – 0.7500 So we seek the z that has 0.2500 to its left. How to solve it In table: Lookup 0.2500 in table. If it’s not there, take the closest value. Read out to find z again. With TI-84 invNorm( 34 Backwards Area-Between “Find the z scores that delimit the middle 80% of the area” DRAW A PICTURE!!! 0.8000 is in the middle So 1.0000 – 0.8000 = 0.2000 in two tails 0.2000 ÷ 2 = 0.1000 in each tail 35 Backwards Area-Between Using Printed Table Look deep in table for closest match to 0.1000 Read out to find the negative z on the left. Because of symmetry, the positive z on the right is the opposite of that value. Using TI-84 invNorm(.1000) Because of symmetry, the positive z on the right is the opposite of that value. Answers: = −1.28 and = +1.28 36 Confirming this with TI-84 Using normalcdf() normalcdf(-1.28,1.28) should give about area .8000 they asked for More decimals for more precision Using DRAW 2ND DRAW 1:ClrDraw 2ND DISTR Right arrow to DRAW 1:ShadeNorm(z1,z2) 37 Confirming this with TI-84 WINDOW settings It probably won’t turn out well on its own. You may need to do some thinking. What we did for this one: Using DRAW 2ND DRAW 1:ClrDraw 2ND DISTR Right arrow to DRAW 1:ShadeNorm(z1,z2) 38