1. Second graded Homework Assignment: 1.80; 1.84; 1.92; 1.110; 1.115; 1.141 (optional: 1.127) Due in Labs on Sept 14-15

2. Last Time:

3. Two Formulae for Variance! Definitional Formula: Computational Formula:

4. Comment about “Variance” The concept and definition of variance that we have so far talked about in class is often referred to as Sample Variance We will later discuss a different concept, definition (and different formula) for what is commonly referred to as Population Variance The latter is a theoretical variance, not a data variance.

5. a b Intercept Slope bx+a x Reminder: (Simple) Linear Function y=a+bx

6. If we add the constant a=5 to each observation, then we obtain new data that are shifted by 5 A shift affects the mean but not the variance New mean = Old mean + 5 New variance = Old variance

7. If we multiply each observation by the constant b=2, then we obtain a new data set that looks rather different A multiplicative constant affects both the mean and the variance New mean = Old mean x 2 New variance = Old variance x 4

8. The effect of linear transformations: Suppose that you have original data Suppose that you need to recode the data as follows:

9. The effect of linear transformations: Median of Y’s = a + b (Median of X’s) Same with other quartiles and percentiles IQR of Y’s = b (IQR of X’s)

10. The effect of linear transformations:

11. Example: Original Data: 1, 2, 4, 6, 7 Transformed Data: 4, 7, 13, 19, 22 Mean: 4 Mean: 13 = 1 + 3 (4) Median: 4 Median: 13 = 1 + 3 (4) IQR: 6.5 – 1.5 = 5 IQR: 20.5 – 5.5 = 15 = 3 (5)

12. Example: Original Data: 1, 2, 4, 6, 7 Transformed Data: 4, 7, 13, 19, 22 Variance: Y = 1 + 3 X

13. Transformation to Z-scores: Z-scores have Mean equal to zero and (Variance) Standard Deviation equal to one

14. Today: The Normal Distribution

15. Histogram of the Frequency Counts Sample Size: 60

16. Histogram of the Proportions

17. Calibrate appropriately Choose vertical unit such that the total orange area is equal to 1

18. Calibrate appropriately Choose Vertical Unit such that the total orange area is equal to 1 3/60 19/60

19. (4+19+17)/60

20. 2/3 2/3 of all Type A respondents had measurements between 55 and 69

21. Data World Theory World

22. Density Function It can be mathematically easier to work with such a curve instead of a histogram

23. Density Function The marked area is the relative frequency with which observations between a and b occur ba

24. Examples of Density Functions Median 75 th percentile 25 th percentile Area p below pth percentile Symmetric = Mean IQR

25. Examples of Density Functions Median Mean Positively Skewed (Skewed to the right)

26. Examples of Density Functions Median Mean Negatively Skewed (Skewed to the left)

27. THE NORMAL DISTRIBUTION (“Gaussian”, “bell-shaped”) Probably the single most important distribution in Statistics Symmetric

28. THE NORMAL DISTRIBUTION Notation: X ~ N( ,  ) A normal distribution is fully specified with just two ‘parameters’, the mean of the distribution (  ) and the standard deviation of the distribution (  ). Data World Theory World No formulae yet

37. Equation of the Density Function:

38. According to the Germans: “That’s right on the money”

39. Herr Gauss

40. The (theoretical) proportion of people in the population between a and b: P(a  X  b) = area between a and b under the curve defined by the density function. P(a < X < b) is equal to this area a b

41. THE NORMAL DISTRIBUTION Properties of X ~ N( ,  ) The proportion of a normally distributed X within: one standard deviation from its mean is.6826 P(  -  < X <  +  ) =.6826 two standard deviations from its mean is.9544 P(  - 2  < X <  + 2  ) =.9544 three standard deviations from its mean is.9974 P(  - 3  < X <  + 3  ) =.9974 True for any value of  and 

42. Example: The population distribution of psychometric test X is a normal distribution with mean 1.1 and standard deviation of.08: Thus, X ~ N(1.1,.08). a) P(1.1 < X) = ? b) P(1.02 < X < 1.18) = ? c) How to calculate P(1.1 < X < 1.25) ? d) How to calculate P(X > 1) ? e) How to find x such that P(X <x) =.75 ?

43. STANDARD NORMAL DISTRIBUTION Z ~ N( 0, 1)  Know everything about Z ~ N(0,1): Table in your book (inside cover) tabulates values P(Z<z) z

44. STANDARD NORMAL DISTRIBUTION Z ~ N( 0, 1)  Know everything about Z ~ N(0,1): Table in your book (inside cover) tabulates values P(Z<z) (note the table goes over two pages)  Note: you can think of values z of Z ~ N(0,1) as “z many standard deviations from the mean” z

45. AMAZING PROPERTY OF NORMAL DISTRIBUTIONS If X is normally distributed then a+bX (b>0) is also normally distributed. More precisely: X ~ N( ,  )  (a+bX) ~ N(a+b , b  ) Note: This type of relationship is not necessarily true for other distributions

46. AMAZING PROPERTY OF NORMAL DISTRIBUTIONS Consequence: Any normally distributed quantity can be standardized: This enables us to use Z~N(0,1) (which we know everything about) whenever information about X ~ N( ,  ) is needed.

47. Example: The population distribution of psychometric test X is a normal distribution with mean 1.1 and standard deviation of.08: Thus, X ~ N(1.1,.08). a) P(1.1 < X) = b) P(1.02 < X < 1.18) = c) How to calculate P(1.1 < X < 1.25) ? d) How to calculate P(X > 1) ? e) How to find x such that P(X <x) =.75 ?

48. Example: The population distribution of psychometric test X is a normal distribution with mean 1.1 and standard deviation of.08: Thus, X ~ N(1.1,.08). a) P(1.1 < X) = Mean is 1.1 X~N(1.1,.08)

49. Example: The population distribution of psychometric test X is a normal distribution with mean 1.1 and standard deviation of.08: Thus, X ~ N(1.1,.08). b)P(1.02 < X < 1.18) Mean is 1.1 X~N(1.1,.08)

50. c) How to calculate P(1.1 < X < 1.25) ? 1. Draw a picture! 2. Standardize: 3. Draw a picture again! 4. Find the probability (using the table) P(0<Z<1.875) =? Example: The population distribution of psychometric test X is a normal distribution with mean 1.1 and standard deviation of.08: Thus, X ~ N(1.1,.08). Z~N(0,1) X~N(1.1,.08) 1.1 0

51. c) How to calculate P(0 < Z < 1.875) ? Example: The population distribution of psychometric test X is a normal distribution with mean 1.1 and standard deviation of.08: Thus, X ~ N(1.1,.08). Z~N(0,1) 0 0

52. d) How to calculate P(X > 1) ? 1. Draw a picture! 2. Standardize: 3. Draw a picture again! 4. Find the probability (using the table) P(Z > -1.25) =? Example: The population distribution of psychometric test X is a normal distribution with mean 1.1 and standard deviation of.08: Thus, X ~ N(1.1,.08). X~N(1.1,.08) 1.1 Z~N(0,1) 0

53. d) How to calculate P(Z > -1.25) ? Example: The population distribution of psychometric test X is a normal distribution with mean 1.1 and standard deviation of.08: Thus, X ~ N(1.1,.08). Z~N(0,1) 0 0

54. e) How to find x such that P(X < x) =.75? 1. Draw a picture! 2. Standardize: 3. Draw a picture again! 4. Find z (using the table) P(Z <z) 5. Solve for x using Example: The population distribution of psychometric test X is a normal distribution with mean 1.1 and standard deviation of.08: Thus, X ~ N(1.1,.08). 1.1 X~N(1.1,.08) 0 Z~N(0,1)