1. 1 Chapter 6 The Standard Deviation and the Normal Model

2. 2 68-95-99.7 rule Mean and Standard Deviation (numerical) Histogram (graphical) 68-95-99.7 rule

3. 3 The 68-95-99.7 rule; applies only to mound-shaped data

4. 4 68-95-99.7 rule: 68% within 1 stan. dev. of the mean 68% 34% y-s y y+s

5. 5 68-95-99.7 rule: 95% within 2 stan. dev. of the mean 95% 47.5% y-2s y y+2s

6. 6 Example: textbook costs 286291307308315316327 328340342346347348348 349354355355360361364 367369371373377380381 382385385387390390397 398409409410418422424 425426428433434437440 480

7. 7 Example: textbook costs (cont.) 286291307308315316327328 340342346347348348349354 355355360361364367369371 373377380381382385385387 390390397398409409410418 422424425426428433434437 440480

8. 8 Example: textbook costs (cont.) 286291307308315316327328 340342346347348348349354 355355360361364367369371 373377380381382385385387 390390397398409409410418 422424425426428433434437 440480

9. 9 Example: textbook costs (cont.) 286291307308315316327328 340342346347348348349354 355355360361364367369371 373377380381382385385387 390390397398409409410418 422424425426428433434437 440480

10. 10 The best estimate of the standard deviation of the men’s weights displayed in this dotplot is 1. 10 2. 15 3. 20 4. 40

12. Changing Units of Measurement Shifting data and rescaling data, and how shifting and rescaling data affect graphical and numerical summaries of data.

13. Shifting and rescaling: linear transformations zOriginal data x 1, x 2,... x n zLinear transformation: x * = a + bx, (intercept a, slope b) x x*x* 0 a Shifts data by a Changes scale

14. Linear Transformations x* = a+ b x Examples: Changing 1.from feet (x) to inches (x*): x*=12x 2.from dollars (x) to cents (x*): x*=100x 3.from degrees celsius (x) to degrees fahrenheit (x*): x* = 32 + (9/5)x 4.from ACT (x) to SAT (x*): x*=150+40x 5.from inches (x) to centimeters (x*): x* = 2.54x 0 12 0 100 32 9/5 150 40 0 2.54

15. Shifting data only: b = 1 x* = a + x  Adding the same value a to each value in the data set:  changes the mean, median, Q 1 and Q 3 by a  The standard deviation, IQR and variance are NOT CHANGED. yEverything shifts together. ySpread of the items does not change.

16. Shifting data only: b = 1 x* = a + x (cont.) zweights of 80 men age 19 to 24 of average height (5'8" to 5'10") x = 82.36 kg z NIH recommends maximum healthy weight of 74 kg. To compare their weights to the recommended maximum, subtract 74 kg from each weight; x* = x – 74 (a=-74, b=1) z x* = x – 74 = 8.36 kg 1.No change in shape 2.No change in spread 3.Shift by 74

17. Shifting and Rescaling data: x* = a + bx, b > 0 Original x data: x 1, x 2, x 3,..., x n Summary statistics: mean x median m 1 st quartile Q 1 3 rd quartile Q 3 stand dev s variance s 2 IQR x* data: x* = a + bx x 1 *, x 2 *, x 3 *,..., x n * Summary statistics: new mean x* = a + bx new median m* = a+bm new 1 st quart Q 1 *= a+bQ 1 new 3 rd quart Q 3 * = a+bQ 3 new stand dev s* = b  s new variance s* 2 = b 2  s 2 new IQR* = b  IQR

18. Rescaling data: x* = a + bx, b > 0 (cont.) zweights of 80 men age 19 to 24, of average height (5'8" to 5'10") zx = 82.36 kg zmin=54.30 kg zmax=161.50 kg zrange=107.20 kg zs = 18.35 kg z Change from kilograms to pounds: x* = 2.2x (a = 0, b = 2.2) z x* = 2.2(82.36)=181.19 pounds z min* = 2.2(54.30)=119.46 pounds z max* = 2.2(161.50)=355.3 pounds z range*= 2.2(107.20)=235.84 pounds z s* = 18.35 * 2.2 = 40.37 pounds

19. Example of x* = a + bx 4 student heights in inches (x data) 62, 64, 74, 72 x = 68 inches s = 5.89 inches Suppose we want centimeters instead: x * = 2.54x (a = 0, b = 2.54) 4 student heights in centimeters: 157.48 = 2.54(62) 162.56 = 2.54(64) 187.96 = 2.54(74) 182.88 = 2.54(72) x * = 172.72 centimeters s * = 14.9606 centimeters Note that x * = 2.54x = 2.54(68)=172.2 s * = 2.54s = 2.54(5.89)=14.9606 not necessary! UNC method Go directly to this. NCSU method

20. Example of x* = a + bx x data: Percent returns from 4 investments during 2003: 5%, 4%, 3%, 6% x = 4.5% s = 1.29% Inflation during 2003: 2% x* data: Inflation-adjusted returns. x* = x – 2% (a=-2, b=1) x* data: 3% = 5% - 2% 2% = 4% - 2% 1% = 3% - 2% 4% = 6% - 2% x* = 10%/4 = 2.5% s* = s = 1.29% x* = x – 2% = 4.5% –2% s* = s = 1.29% (note! that s* ≠ s – 2%) !! not necessary! Go directly to this

21. Example zOriginal data x: Jim Bob’s jumbo watermelons from his garden have the following weights (lbs): 23, 34, 38, 44, 48, 55, 55, 68, 72, 75 s = 17.12; Q 1 =37, Q 3 =69; IQR = 69 – 37 = 32 zMelons over 50 lbs are priced differently; the amount each melon is over (or under) 50 lbs is: zx* = x  50 (x* = a + bx, a=-50, b=1) -27, -16, -12, -6, -2, 5, 5, 18, 22, 25 s* = 17.12; Q* 1 = 37 - 50 =-13, Q* 3 = 69 - 50 = 19 IQR* = 19 – (-13) = 32 NOTE: s* = s, IQR*= IQR

22. SUMMARY: Linear Transformations x* = a + bx z Linear transformations do not affect the shape of the distribution of the data -for example, if the original data is right- skewed, the transformed data is right-skewed

23. SUMMARY: Shifting and Rescaling data, x* = a + bx, b > 0

24. 24 Z-scores: Standardized Data Values Measures the distance of a number from the mean in units of the standard deviation

25. 25 z-score corresponding to y

26. 26 n Exam 1: y 1 = 88, s 1 = 6; exam 1 score: 91 Exam 2: y 2 = 88, s 2 = 10; exam 2 score: 92 Which score is better?

27. 27 Comparing SAT and ACT Scores n SAT Math: Eleanor’s score 680 SAT mean =500 sd=100 n ACT Math: Gerald’s score 27 ACT mean=18 sd=6 n Eleanor’s z-score: z=(680-500)/100=1.8 n Gerald’s z-score: z=(27-18)/6=1.5 n Eleanor’s score is better.

28. Z-scores: a special linear transformation a + bx Example. At a community college, if a student takes x credit hours the tuition is x* = $250 + $35x. The credit hours taken by students in an Intro Stats class have mean x = 15.7 hrs and standard deviation s = 2.7 hrs. Question 1. A student’s tuition charge is $941.25. What is the z-score of this tuition? x* = $250+$35(15.7) = $799.50; s* = $35(2.7) = $94.50

29. Z-scores: a special linear transformation a + bx (cont.) Example. At a community college, if a student takes x credit hours the tuition is x* = $250 + $35x. The credit hours taken by students in an Intro Stats class have mean x = 15.7 hrs and standard deviation s = 2.7 hrs. Question 2. Roger is a student in the Intro Stats class who has a course load of x = 13 credit hours. The z-score is z = (13 – 15.7)/2.7 = -2.7/2.7 = -1. What is the z-score of Roger’s tuition? Roger’s tuition is x* = $250 + $35(13) = $705 Since x* = $250+$35(15.7) = $799.50; s* = $35(2.7) = $94.50 This is why z-scores are so useful!! The linear transformation did not change the z-score!

30. 30 Z-scores add to zero Student/Institutional Support to Athletic Depts For the 9 Public ACC Schools: 2013 ($ millions) SchoolSupporty - ybarZ-score Maryland15.56.41.79 UVA13.14.01.12 Louisville10.91.80.50 UNC9.20.10.03 VaTech7.9-1.2-0.34 FSU7.9-1.2-0.34 GaTech7.1-2.0-0.56 NCSU6.5-2.6-0.73 Clemson3.8-5.3-1.47 Mean=9.1000, s=3.5697 Sum = 0

31. 31 Average IQ by Browser Nationally: Mean IQ=100 sd = 15 Story was exposed as a hoax

32. 32 NORMAL PROBABILITY MODELS The Most Important Model for Data in Statistics

33. 33 X 8 369120  A family of bell-shaped curves that differ only in their means and standard deviations. µ = the mean  = the standard deviation µ = 3 and  = 1

34. 34 Normal Probability Models The mean, denoted ,  can be any number n The standard deviation  can be any nonnegative number n The total area under every normal model curve is 1 n There are infinitely many normal distributions

35. 35 Total area =1; symmetric around µ

36. 36 The effects of  and  How does the standard deviation affect the shape of f(x)?  = 2  =3  =4  = 10  = 11  = 12 How does the expected value affect the location of f(x)?

37. 37 X 369120 X 369 0  µ = 3 and  = 1  µ = 6 and  = 1

38. 38 X 8 369120  X 8 369 0  µ = 6 and  = 2 µ = 6 and  = 1

39. 39 area under the density curve between 6 and 8. 36912 µ = 6 and  = 2 0 X

40. 40 area under the density curve between 6 and 8

41. 41 Standardizing Suppose X~N(  Form a new random variable by subtracting the mean  from X and dividing by the standard deviation  : (X  n This process is called standardizing the random variable X.

42. 42 Standardizing (cont.) (X  is also a normal random variable; we will denote it by Z: Z = (X   has mean 0 and standard deviation 1: E(Z) =  = 0; SD(Z) =   n The probability distribution of Z is called the standard normal distribution.

43. 43 Standardizing (cont.) n If X has mean  and stand. dev. , standardizing a particular value of x tells how many standard deviations x is above or below the mean . n Exam 1:  =80,  =10; exam 1 score: 92 Exam 2:  =80,  =8; exam 2 score: 90 Which score is better?

44. 44 X 8 369120 µ = 6 and  = 2 Z 0123-2-3.5 µ = 0 and  = 1 (X-6)/2

45. 45 Z = standard normal random variable  = 0 and  = 1 Z 0123-2-3.5 Standard Normal Model.5

46. 46 Important Properties of Z #1. The standard normal curve is symmetric around the mean 0 #2.The total area under the curve is 1; so (from #1) the area to the left of 0 is 1/2, and the area to the right of 0 is 1/2

47. 47 Finding Normal Percentiles by Hand (cont.) n Table Z is the standard Normal table. We have to convert our data to z-scores before using the table. n The figure shows us how to find the area to the left when we have a z-score of 1.80:

48. 48 Areas Under the Z Curve: Using the Table Proportion of area above the interval from 0 to 1 =.8413 -.5 =.3413 0 1 Z.1587.3413.50

49. 49 Standard normal areas have been calculated and are provided in table Z. The tabulated area correspond to the area between Z= -  and some z 0 Z = z 0 Area between -  and z 0 z0.000.010.020.030.040.050.060.070.080.09 0.00.50000.50400.50800.51200.51600.51990.52390.52790.53190.5359 0.10.53980.54380.54780.55170.55570.55960.56360.56750.57140.5753 0.20.57930.58320.58710.59100.59480.59870.60260.60640.61030.6141 10.84130.84380.84610.84850.85080.85310.85540.85770.85990.8621 1.20.88490.88690.88880.89070.89250.89440.89620.89800.89970.9015

50. 50 n Example – begin with a normal model with mean 60 and stand dev 8 In this example z 0 = 1.25 0.8944 z0.000.010.020.030.040.050.060.070.080.09 0.00.50000.50400.50800.51200.51600.51990.52390.52790.53190.5359 0.10.53980.54380.54780.55170.55570.55960.56360.56750.57140.5753 0.20.57930.58320.58710.59100.59480.59870.60260.60640.61030.6141 10.84130.84380.84610.84850.85080.85310.85540.85770.85990.8621 1.20.88490.88690.88880.89070.89250.89440.89620.89800.89970.9015

51. 51 Example n Area between 0 and 1.27) = 1.270 z Area=.3980.8980-.5=.3980

52. 52 Example Area to the right of.55 = A 1 = 1 - A 2 = 1 -.7088 =.2912 0.55 A2A2

53. 53 Example n Area between -2.24 and 0 = Area=.4875.5 -.0125 =.4875 z -2.24 0 Area=.0125

54. 54 Example Area to the left of -1.85=.0322

55. 55 Example A1A1 A2A2 02.73 z -1.18 n Area between -1.18 and 2.73 = A - A 1 n =.9968 -.1190 n =.8778.1190.9968 A1A1 A

56. 56 Area between -1 and +1 =.8413 -.1587 =.6826.8413.1587.6826 Example

57. 57 Example Is k positive or negative? Direction of inequality; magnitude of probability Look up.2514 in body of table; corresponding entry is -.67 -.67

58. 58 Example

59. 59 Example.9901.1230.8671

60. 60 N(275, 43); find k so that area to the left is.9846

61. 61 Area to the left of z = 2.16 =.9846 0 2.16 Z.1587.4846 Area=.5.9846

62. 62 Example Regulate blue dye for mixing paint; machine can be set to discharge an average of  ml./can of paint. Amount discharged: N( ,.4 ml). If more than 6 ml. discharged into paint can, shade of blue is unacceptable. Determine the setting  so that only 1% of the cans of paint will be unacceptable

63. 63 Solution

64. 64 Solution (cont.)

65. 65 A random variable X with mean  and standard deviation  is normally distributed if its probability density function is given by Normal Distributions

66. 66 The Shape of Normal Distributions Normal distributions are bell shaped, and symmetrical around   Why symmetrical? Let  = 100. Suppose x = 110. Now suppose x = 90 11090

67. 67 Are You Normal? Normal Probability Plots Checking your data to determine if a normal model is appropriate

68. 68 Are You Normal? Normal Probability Plots n When you actually have your own data, you must check to see whether a Normal model is reasonable. n Looking at a histogram of the data is a good way to check that the underlying distribution is roughly unimodal and symmetric.

69. 69 n A more specialized graphical display that can help you decide whether a Normal model is appropriate is the Normal probability plot. n If the distribution of the data is roughly Normal, the Normal probability plot approximates a diagonal straight line. Deviations from a straight line indicate that the distribution is not Normal. Are You Normal? Normal Probability Plots (cont)

70. 70 n Nearly Normal data have a histogram and a Normal probability plot that look somewhat like this example: Are You Normal? Normal Probability Plots (cont)

71. 71 n A skewed distribution might have a histogram and Normal probability plot like this: Are You Normal? Normal Probability Plots (cont)