1. The Normal Distribution “the bell curve” Some slides downloaded from www.registart.co.uk

2. The Most Important Distribution in Statistics! Describes the characteristics of many real- world data sets: – test scores for large groups of students – actual sizes (length, width) of jeans at Kohl’s – eyesight of all 20-year-olds in Kissimmee – actual lifetimes of 1000 AA batteries – testosterone level of all male students at GHS – length of middle finger of 250 students

3. Characteristics Symmetric, bell-shaped curve. X can take any value (continuous RV) Shape of curve depends on 2 parameters: – Center of dist n is population mean  – Spread is determined by std deviation  Most values fall around the mean, but a few values are much smaller and a few are much larger (equal chance).

4. Probability Density Function (PDF) “X is distributed normally with mean μ and standard deviation σ”

5. Shape Depends on Mean, Std. Dev

6. As a Histogram (Area of rectangle = probability)

7. Decrease interval size...

8. Decrease interval size more…. Almost a nice continuous curve

9. Probability = Area under Curve Curve describes probability of getting a range of values – e.g., P(X > 60), P(X < 30), P(20 < X < 50) Area under whole curve = 1 Probability of getting specific number is 0, e.g. P(X=60) = 0 – so P(x < 60) is the same as P(x ≤ 60)

10. Probability that X is less than a #

11. Probability that X is more than a #

12. Probability that X is between 2 #s

13. Standard Deviation Graph (H&H p.730)

14. Draw with GDC Set window – X from μ - 3σ to μ + 3σ (99.7% of all values) – Y from 0 to Draw – 2 nd PRGM (DRAW)  CLRDRAW (#1) – 2 nd VARS (DISTR)  DRAW  ShadeNorm(lower limit, upper limit, [μ, σ]) if omit [μ, σ]  0, 1

15. Draw with GDC (con’t) For normally distributed X with mean 15, std dev 2: P(8 ≤ X ≤ 12):use ShadeNorm(8, 12, 15, 2) P(X ≥ 17):use ShadeNorm(17, E99, 15, 2) P(X ≤ 16):use ShadeNorm(-E99, 16, 15, 2) use E99 in place of ∞, -E99 in place of -∞

16. Calculate with GDC For normally dist’d X with mean 71.5, std dev 3.8: 2 nd VARS (DISTR)  #2 normalcdf(lower limit, upper limit, [μ, σ]) if omit [μ, σ]  0, 1 P(62.1 ≤ X ≤ 68.7):use normalcdf(62.1, 68.7, 71.5, 3.8) P(X ≥ 89.0):use normalcdf(89.0, E99, 71.5, 3.8) P(X ≤ 42.5):use normalcdf(-E99, 42.5, 71.5, 3.8) Note: P(62.1 ≤ X ≤ 68.7) is just P(X ≤ 68.7) - P(X ≤ 62.1)

17. Standard Normal Distribution (Z-Dist n ) To make a table of values for X, need to know both μ and σ – One table for each combination of μ and σ –  LOTS of tables!!! Make a new random variable Z = (X – μ)/σ Z is called the standard normal distribution Need only one table of values for Z, since μ = 0 and σ = 1 always Symmetric, so P(Z k)

18. The Standard Normal Distribution  (Z)

19. Z-Values (“Z-Scores”) Value of Z is just the # of standard deviations from the mean: – Z = -2 corresponds to X = μ - 2σ – Z = -1 corresponds to X = μ - σ – Z = 0 corresponds to X = μ – Z = 1 corresponds to X = μ + σ – Z = 2 corresponds to X = μ + 2σ Etc.

20. (Insert graph of preceding slide)

21. Z-Values with GDC P(-1.5 ≤ Z ≤ 2.1)  normalcdf(-1.5, 2.1) (Omitting μ and σ means μ = 0 and σ = 1) If starting with X-values (μ ≠ 0 and/or σ ≠ 1), don’t forget to convert to Z, then back to X

22. The Standard Normal Distribution The probabilities are given by the area under the curve P(Z<-1.6)

23. The Standard Normal Distribution The probabilities are given by the area under the curve P(Z< -1.6) =0.0548 By symmetry: P (Z 1.6) P(Z < -1.6) = 1 - P(Z < 1.6)

24. Z-Values from Tables Table in formula packet “Area under the standard normal curve (topic 6.11)” Gives probability that Z is less than (actually < or ≤) a specified value Table is for positive values of Z, only Before using table, convert X-values to Z

25. Reading Table of Z-Values (INSERT picture of table), with animations showing how to read z to 2 decimal places Highlight Z-values on top and on left, highlight cross-indexed area

26. Example : Using Table of Z-Values For X with mean μ = 26, std dev σ = 1.4, find P(X < 27.1) Z = (X – μ)/σ = (27.1 – 26)/1.4 = 0.786  use Z = 0.79 P(X < 27.1) = P(Z < 0.79) = 0.7852 compare to answer from normalcdf(-E99, 27.1, 26, 1.4)  P(X < 27.1) = 0.7840 (slightly different because we rounded Z) P(X < 27.106) = P(Z < 0.79) (no rounding) = 0.7852 (to 4 d.p.’s)

27. Extending the Table Table from formula packet only works for: – P(Z < z) – Positive Z-values What to do if you want P(Z > z), or if Z is a negative value? Think of the graph and which areas you should shade…

28. Calculating P(Z > z) from Table Use the fact that the total area under the bell curve equals 1 P(Z z) = 1 (remember P(Z = z) = 0)  P(Z > z) = 1 – P(Z < z)

29. Example: P(Z > z) from Table Find P(Z > 2.58) P(Z > 2.58) = 1 – P(Z < 2.58) From table, P(Z < 2.58) = 0.9951  P(Z > 2.58) = 1 – 0.9951 = 0.0049

30. Example: P(X > x) For X with mean μ = 26 and std dev σ = 1.4, find P(X > 26.8) Z = (X – μ)/σ = (26.8 – 26)/1.4 = 0.571  use Z = 0.57 P(X > 26.8)  P(Z > 0.57) = 1 - P(Z < 0.57) = 1 - 0.7157 = 0.2843 compare to P(X > 26.8) using normalcdf(26.8, E99, 26, 1.4): P(X > 26.8) = 0.2839 (again, difference due to rounding)

31. Using Table with Negative Z’s Use the fact that the bell curve is symmetric! (insert graph) P(Z z) = 1 – P(Z < z) P(Z > -z) = P(Z < z)

32. Example: Using Table with Z < 0 Given normally dist’d X with μ = 54.4, σ = 6.7, find P(X < 49.8) Z = (X – μ)/σ = (49.8 – 54.4)/6.7 = -0.687  use Z = -0.69 P(Z 0.69) = 1 – P(Z < 0.69) = 1 – 0.7549 = 0.2451 Compare: normalcdf(-E99, 49.8, 54.4, 6.7) = 0.2462

33. Using Table for P(a < X < b) Subtract the areas P(a < X < b) = P(X < b) – P(X < a) INSERT pictures

34. Example: P(a < X < b) from Table Given normally dist’d X with μ = 54.4, σ = 6.7, find P(45.0 < X < 49.8) Z 1 = (X – μ)/σ = (45.0 – 54.4)/6.7 = -1.40 Z 2 = (X – μ)/σ = (49.8 – 54.4)/6.7 = -0.69 P(-1.40 < Z < -0.69) = P(Z < -0.69) – P(Z < -1.40) = P(Z > 0.69) – P(Z > 1.40) = [1 – P(Z < 0.69)] – [1 – P(Z < 1.40)] = [1 – 0.7549] – [1 – 0.9192] = 0.1643 Compare: normalcdf(45.0, 49.8, 54.4, 6.7) = 0.1659

35. Inverse Normal Probabilities Now we work backwards: – know the probability – want to find corresponding value of X (or Z) Examples of questions: – Find k such that P(X ≤ k) = 95.4% – If P(-0.10 < X < b) = 0.357 (i.e., 35.7%), find b – Find μ so that P(X > 0.771) = 80.8% Could use trial and error, but there’s a better way

36. Inverse Normal Probabilities by GDC Use 2 nd VARS (DISTR)  #3 invNorm(area, [μ, σ]) μ and σ are optional If omitted, then: μ = 0 σ = 1(omit when using Z-score, not X)

37. Example: Inv. Normal Prob. by GDC X is normally distributed with μ = 70, σ = 10. Find k such that P(X ≤ k) = 0.954 (i.e., 95.4%) 2 nd VARS (DISTR)  invNorm(0.954, 70, 10) = 86.8  k = 86.8 Check: normalcdf(-E99, 86.8, 70, 10) = 0.954 

38. Inverse Normal Probabilities by Table Table in formula packet (2 pages) “Inverse Normal Probabilities (topic 6.11)” Gives probability that Z is less than (actually < or ≤) a specified value Table is for probabilities between 0.5 and 0.999, and only for positive values of Z Before using table, convert X-values to Z

39. Reading Inverse Probability Table (INSERT picture of table), with animations showing how to read z to 2 decimal places Highlight probabilities on top and on left, highlight cross-indexed Z-score

40. Examples: Using Inverse Table Find k such that P(Z < k) = 0.695 p = 0.695  read Z = 0.5101  k = 0.5101 Check: normalcdf(-E99, 0.5101) = 0.695  (omit μ, σ) Find k such that P(Z > k) = 0.128 P(Z k) = 1 – 0.128 = 0.872 p = 0.872  read Z = 1.1359  k = 1.1359 Check: normalcdf(1.1359, E99) = 0.128  (omit μ, σ)

41. Example: Using Inverse Table for X X is dist’d normally with μ = 24.6, σ = 0.8 For what value of k is P(X < k) = 0.602? p = 0.602  read Z = 0.2585 Z = (X – μ)/σ  X = Z  σ + μ = (0.2585)(0.8) + 24.6 = 24.8 Check: normalcdf(-E99, 24.8, 24.6, 0.8) = 0.599 (difference due to rounding X)

42. Extending the Inverse Table Table from formula packet only works for 0.5 0 What to do if p < 0.5? – use P(Z k) = 1 What to do if P(Z > k)? – use P(Z > k) = P(Z < -k) (symmetry)

43. Example: Inverse Table for p < 0.5 For what value of k is P(Z < k) = 0.210? P(Z k)  P(Z > k) = 1 – 0.210 = 0.79 (which is > 0.5, so we can use the table now) By symmetry, P(Z > k) = P(Z < -k) (the table requires P(Z < k)) p = 0.79  read Z = 0.8064  -k = 0.8064  k = -0.8064 Check: normalcdf(-E99, -0.8064) = 0.210 

44. Example: Inverse Table, P(a < X < b) Insert example…

45. Example: Inverse Table, μ = ? X is dist’d normally with σ = 1.75 but unknown μ. Find μ if P(X > 4.92) = 0.4. P(X > 4.92) = 0.4  P(Z > k) = 0.4  1 - P(Z < k) = 0.4  P(Z < k) = 1 - 0.4 = 0.6  read Z = 0.2534 Z = (X – μ)/σ  0.2534 = (4.92 – μ)/1.75 μ = 4.92 – (0.2534)(1.75) = 4.48 Check: normalcdf(4.92, E99, 4.48, 24.6, 1.75) = 0.401  (difference due to rounding X)