1. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 1 THE Normal PROBABILITY DISTRIBUTION

2. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 2 When you have completed this chapter, you will be able to: 1. Explain how probabilities are assigned to a continuous random variable. 2. Explain the characteristics of a normal probability distribution. 3. Define and calculate z value corresponding to any observation on a normal distribution.

3. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 3 Use the normal probability distribution to approximate the binomial probability distribution. 5. Determine the probability a random observation is in a given interval on a normal distribution using the standard normal distribution. 4.

4. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 4 …the set of all the values in any interval is uncountable or infinite! ….we will now study the class of continuous probability distributions. Recall… Continuous Random Variable

5. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 5 Continuous … can assume any value within a specified range! … can assume any value within a specified range! e.g. - Pressure in a tire - Weight of a pork chop - Height of students in a class Quantitative Numerical Observations … can be classified as either Discrete or Continuous Characteristics Variables Also Recall that…

6. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 6 Continuous Random Variable …the Total sum of probabilities should always be 1! When dealing with a Continuous Random Variable we assume that the probability that the variable will take on any particular value is 0! Instead, Probabilities are assigned to intervals of values!

7. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 7 Continuous Probability Distributions P(a<X<b) P(a<X<b) f(x) x a b Range of Values

8. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 8 Data: Heights of adult Canadian males… a) n = 50, class interval = 2 b) n = 500, class interval = 1 c) n = 5000, class interval = 0.4 d) Probability function Relative Frequency Histogram  Probability Function Relative Frequency Histogram  Probability Function See Histograms

9. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 9 0.08 - 0.07 - 0.06 - 0.05 - 0.04 - 0.03 - 0.02 - 0.01 - 0.00 - 155 165 175 185 195 (A) n = 50 150 160 170 180 190 0.08 - 0.07 - 0.06 - 0.05 - 0.04 - 0.03 - 0.02 - 0.01 - 0.00 - (B) n = 500 0.08 - 0.07 - 0.06 - 0.05 - 0.04 - 0.03 - 0.02 - 0.01 - 0.00 - 145 155 165 175 185 195 (C) n = 5000 0.07 - 0.06 - 0.05 - 0.04 - 0.03 - 0.02 - 0.01 - 0.00 - 140 150 160 170 180 190 200 Probability Function

10. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 10 A probability function can have any shape, as long as it is non-negative (the curve is above the x-axis, and the total area under the curve is 1) …that all probability functions are not normal !

11. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 11 The arithmetic mean, median, and mode of the distribution are equal and located at the peak. The normal curve is bell-shaped Normal Probability Distribution and has a single peak at the exact centre of the distribution. …thus half the area under the curve is above the mean and half is below it.

12. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 12 Normal Probability Distribution The normal probability distribution is symmetrical about its mean. The normal probability distribution is asymptotic, i.e. the curve gets closer and closer to the x-axis but never actually touches it.

13. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 13 Mean, median, and mode are equal Normal Probability Distribution Summary Normal curve is symmetrical Theoretically, curve extends to infinity +

14. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 14 part of a “family” of curves How does the Standard Deviation values affect the shape of f(x)?    = 2  =3  =4 Normal Probability Distribution

15. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 15 How does the following Expected values affect the location of f(x)?     part of a “family” of curves Normal Probability Distribution

16. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 16 The Standard Normal Probability Distribution The Standard Normal Probability Distribution … is a normal distribution with a mean of 0 and a standard deviation of 1. Also called the Z Distribution A z-value is the distance between a selected value (designated X ) and the population mean, divided by the population S tandard D eviation, . A z-value is the distance between a selected value (designated X ) and the population mean, divided by the population S tandard D eviation, .   µ )  (X(X z Formula A z-value is, therefore, a location on the standard normal curve.

17. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 17 Total Area under the curve is 100% or 1 The Standard Normal Probability Distribution The Standard Normal Probability Distribution …since the mean is located at the peak of the curve, half the area under the curve is above the mean and Area = +0.5 half is below the mean

18. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 18 The bi-monthly starting salaries of recent MBA graduates follows the normal distribution with a mean of $3,000 and a standard deviation of $300. What is the z-value for a salary of $3,300? $300 $3,000$3,300   A z-value of 1 indicates that the value of $3,300 is one standard deviation above the mean of $3,000. The Standard Normal Probability Distribution The Standard Normal Probability Distribution z = 1.00 Show curve z Formula   µ )  (X(X z

19. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 19  z   z  $3000 $3300 A z-value of 1 indicates that the value of $3,300 is one standard deviation above the mean of $3,000.

20. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 20 The Standard Normal Probability Distribution The Standard Normal Probability Distribution The bi-monthly starting salaries of recent MBA graduates follows the normal distribution with a mean of $3,000 and a standard deviation of $300. $300 $3,000$2,550   A z-value of –1.50 indicates that the value of $2,550 is one and a half (1.5) standard deviations below the mean of $3,000. Z = -1.50 What is the z-value for a salary of $2,550? Show curve z Formula   µ )  (X(X z

21. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 21  z   z  A z-value of –1.50 indicates that the value of $2,550 is one and a half standard deviations below the mean of $3,000. $3000 $2550 The Standard Normal Probability Distribution The Standard Normal Probability Distribution

22. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 22 The Standard Normal Probability Distribution The Standard Normal Probability Distribution  … there is an infinite # of normal distributions, but only one table  … each distribution is determined by  &   …work with  &  to calculate normal probabilities, by figuring out … how many standard deviations is x away from the mean?  … Z becomes the metre stick for measuring

23. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 23

24. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 24 0 20 40 50 60 80 100  How many  is 40 away from  ? If…  = 50  = 8 If…  = 50  = 8 + 40 is 10 units(-10)away from 50 If 1 standard deviation (  ) is 8 units, then 10 units must be 1.25 SD Using z 40 is –1.25  away from 50

25. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 25 The Standard Normal Probability Distribution The Standard Normal Probability Distribution the Z Distribution A z-value is the distance between a selected value (designated X ) and the population mean µ, divided by the population S tandard D eviation, . A z-value is the distance between a selected value (designated X ) and the population mean µ, divided by the population S tandard D eviation, . … distance from the  … divide by  to get z     X z Formula 40 – 50 8 z   – 10 8 

26. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 26 The Standard Normal Probability Distribution The Standard Normal Probability Distribution the Z Distribution + z-value Right of mean - z-value Left of mean 0 20 40 50 60 80 100 The mean µ, of a standard normal distribution is 0, and S tandard D eviation  is 1. 0 Z -scale

27. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 27 The Standard Normal Probability Distribution The Standard Normal Probability Distribution Calculate P(50 < X < 60) 0 20 40 50 60 80 100 If…  = 50  = 8 If…  = 50  = 8 Transferring each value into z- scores, 01.25 Z 1 = (60-50)/8 = 1.25 P(50 < X < 60) = P(0 < < 1.25) = P(0 < z < 1.25) z- table   µ )  (X(X z

28. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 28 Normal Distribution Table To find the area between a z score of 0 and a score of 1.25 Provides the z-score to 1 decimal place Provides the second decimal place 1.20 0.05 0.3944 …thus

29. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 29 The Standard Normal Probability Distribution The Standard Normal Probability Distribution the Z Distribution 01.25 0 20 40 50 60 80 100 … 0.3944 is the area between the mean and a positive z score 1.25 0.3944 Therefore, the Probability of 50<X<60 is 39.44% New

30. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 30 …and The Standard Normal Probability Distribution The Standard Normal Probability Distribution Calculate P(40 < X < 60) If…  = 50  = 8 If…  = 50  = 8 0 1.25-1.25 Z 1 = (40-50)/8 = -1.25 Z 2 = (60-50)/8 = 1.25 Transferring each value into z- scores, 0 20 40 50 60 80 100   µ )  (X(X z

31. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 31 The Standard Normal Probability Distribution The Standard Normal Probability Distribution = for a Total area of: 0.3944 + 0.3944 or 0.7888 Z 2 = 1.25 (determined earlier) 0 20 40 50 60 80 100 0 1.25-1.25 0.3944 … because both sides are symmetrical, the left side (the area between the mean and a negative z score -1.25) must have the same area………….. 0.3944.

32. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 32 The Standard Normal Probability Distribution The Standard Normal Probability Distribution the Z Distribution Calculate P(40 < X < 60) 0 20 40 50 60 80 100 If…  = 50  = 8 If…  = 50  = 8 P(40 < X < 60) = P(40  X  60) …since P(X= 40 or 60) is zero …since P(X= 40 or 60) is zero

33. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 33 Using the Normal Distribution Table Practise Finding Areas

34. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 34 Practise using the Normal Distribution Table Look up 1.00 in Table Locate A rea on the normal curve 0 1.00 1 Be sure to always sketch the curve, insert the given values and shade the required area. z- table …the area between: Z 1 = 0 and Z 2 = 1.00 2

35. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 35 Normal Distribution Table To find the area between a z score of 0 and a score of 1.00 Provides the z-score to 1 decimal place Provides the second decimal place 1.00 0.00 0.3413 The Area between 0<Z<1.00 is 0.3413 Therefore, the Probability of 0<Z<1.00 is 34.13%

36. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 36 Practise using the Normal Distribution Table 1.64 The Area between 0<Z<1.64 is 0.4495 Therefore, the Probability of 0<Z<1.64 is 44.95% Look up 1.64 in Table Locate A rea on the normal curve 12 …the area between: Z 1 = 0 and Z 2 = 1.64 0

37. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 37 Practise using the Normal Distribution Table 0.47 The Area between 0<Z<0.47 is 0.1808 Therefore, the Probability of 0<Z<0.47 is 18.08% Look up 0.47 in Table Locate A rea on the normal curve 12 …the area between: Z 1 = 0 and Z 2 = 0.47 0

38. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 38 Practise using the Normal Distribution Table -1.64 Look up 1.64 in Table Locate A rea on the normal curve 12 …the area between: Z 1 = 0 and Z 2 = -1.64 0 The Area between -1.64<Z<0 is 0.4495 Therefore, the Probability of -1.64<Z<0 is 44.95% Because of symmetry, this area is the same as between z of 0 and positive 1.64

39. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 39 Practise using the Normal Distribution Table -2.22 Locate A rea on the normal curve 1 Look up 2.22 in Table 2 …the area between: Z 1 = 0 and Z 2 = -2.22 0 The Area between –2.22<Z<0 is 0.4868 Therefore, the Probability of –2.22<Z<0 is 48.68% 2.22

40. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 40 A z-value is a location on the standard normal curve! A z-value is a location on the standard normal curve! Therefore, a z-value(also called z-score) can have a positive or negative value ! However, a reas and p robabilities are always positive values! Practise using the Normal Distribution Table

41. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 41 Practise using the Normal Distribution Table -2.96 Look up 2.96 in Table Locate A rea on the normal curve 12 …the area between: Z 1 = 0 and Z 2 = -2.96 0 The Area between –2.96<Z<0 is 0.4985 Therefore, the Probability of –2.96<Z<0 is 49.85% 2.96

42. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 42 Practise using the Normal Distribution Table -1.65 Look up 1.65 in Table Locate A rea on the normal curve 12 …the area between: Z 1 = -1.65 and Z 2 = 1.65 0 The Area (a 1 ) between –1.65<Z<0 is 0.4505 Therefore, the required Total Area is 0.9010 1.65 The Area (a 2 ) between 0<Z<1.65 is 0.4505 Add together

43. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 43 Practise using the Normal Distribution Table Locate A rea on the normal curve 1 Look up – 2.00 then 1.00 in Table 2 …the area between: Z 1 = -2.00 and Z 2 = 1.00 The Area (a 1 ) between –2.00<Z<0 is 0.4772 Therefore, the required Total Area is 0.8185 The Area (a 2 ) between 0<Z<1.00 is 0.3413 Add together -2.00 0 1.00

44. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 44 Locate A rea on the normal curve 1 Look up – 0.44 then 1.96 in Table 2 …the area between: Z 1 = -0.44 and Z 2 = 1.96 The Area (a 1 ) between –0.44<Z<0 is 0.1700 Therefore, the required Total Area is 0.6450 The Area (a 2 ) between 0<Z<1.96 is 0.4750 Add together -0.44 0 1.96 Practise using the Normal Distribution Table

45. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 45 The Standard Normal Probability Distribution The Standard Normal Probability Distribution Total Area under the curve is 100% or 1 …since the mean is located at the peak of the curve, half the area under the curve is above the mean and Area = 0.5 half is below the mean

46. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 46 a1a1 Locate A rea on the normal curve 1 Look up 2.00 in Table 2 …the area to the Left of Z 1 = 2.00 The Area to the left of Z 1 is a 1 +.5 = 0.4772 + 0.5 Therefore, the required Total Area is 0.9772 0 2 Area = 0.5 The Area (a 1 ) between 0<Z<2.0 is 0.4772 3 Adjust as needed Practise using the Normal Distribution Table

47. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 47 a1a1 Locate A rea on the normal curve 1 Look up 1.96 in Table 2 …the area to the Left of Z 1 = 1.96 The Area to the left of Z 1 is a 1 +.5= 0.4750 + 0.5 Therefore, the required Total Area is 0.9750 1.96 The Area (a 1 ) between 0<Z<1.96 is 0.4750 3 Adjust as needed Practise using the Normal Distribution Table 0 Area = 0.5

48. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 48 Area = 0.5 Locate A rea on the normal curve 1 Look up 1.64 in Table 2 …the area to the Left of Z 1 = -1.64 The Area to the left of Z 1 is 0.5 - a 1 = 0.5 - 0.4495 Therefore, the required Total Area is 0.0505 0 -1.64 3 Adjust as needed The Area (a 1 ) between –1.64<Z<0 is 0.4495 Practise using the Normal Distribution Table a1a1

49. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 49 Area = 0.5 Locate A rea on the normal curve 1 Look up 0.95 in Table 2 0 -0.95 The Area to the left of Z 1 is 0.5 - a 1 = 0.5 - 0.3289 Therefore, the required Total Area is 0.1711 3 Adjust as needed The Area (a 1 ) between –0.95<Z<0 is 0.3289 Practise using the Normal Distribution Table …the area to the Left of Z 1 = -0.95 a1a1

50. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 50 Area = 0.5 Locate A rea on the normal curve 1 Look up 0.95 in Table 2 0 -0.95 The Area to the left of Z 1 is 0.5 + a 1 = 0.5 + 0.3289 Therefore, the required Total Area is 0.8289 3 Adjust as needed The Area (a 1 ) between –0.95<Z<0 is 0.3289 a 1 Practise using the Normal Distribution Table …the area to the Right of Z 1 = -0.95

51. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 51 Locate A rea on the normal curve 1 Look up 1.00 in Table 2 0 1.00 The Area to the left of Z 1 is 0.5 - a 1 = 0.5 - 0.3413 Therefore, the required Total Area is 0.1587 0 3 Adjust as needed The Area (a 1 ) between 0<Z<1.00 is 0.3413 Practise using the Normal Distribution Table …the area to the Right of Z 1 = 1.00 Area = 0.5 a1a1 A

52. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 52 Locate A rea on the normal curve 1 …the area between: 2 Look up 1.96 then 2.58 in Table The Area (a 1 ) between 0<Z<1.96 is 0.4750 Therefore, the required Total Area is 0.0201 The Area (a 2 ) between 0<Z<2.58 is 0.4951 Subtract 3 Adjust as needed Practise using the Normal Distribution Table 1.96 2.58 00 a2a2 a1a1 Z 1 = 1.96 and Z 2 = 2.58 Find

53. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 53 Z 1 = 0.55 and Z 2 = 1.96 Locate A rea on the normal curve 1 …the area between: 2 Look up 0.55 then 1.96 in Table 3 Adjust as needed Practise using the Normal Distribution Table 1.96.55 0 a2a2 The Area (a 1 ) between 0<Z<0.55 is 0.2088 Therefore, the required Total Area is 0.2662 The Area (a 2 ) between 0<Z<1.96 is 0.4750 Subtract a1a1

54. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.  About 68 percent of the area under the normal curve is within one standard deviation of the mean.  +/-   About 95 percent is within two standard deviations of the mean.  +/- 2   Practically all are within three standard deviations of the mean.  +/- 3  Examples Areas under the Normal Curve

55. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 55  Areas under the Normal Curve Between: 95.44% 99.74%  - 1   - 2   - 3  68.26%

56. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. The daily water usage per person in Newmarket, Ontario is normally distributed with a mean of 80 litres and a standard deviation of 10 litres. Areas under the Normal Curve About 68% of the daily water usage will lie between 70 and 90 litres About 68 percent of those living in Newmarket will use how many litres of water?

57. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 57 robability that… a person from Newmarket, selected at random, will use between 80 and 88 litres per day, is…? 10 - 80 80  0.00     (X(X z1z1 10 - 80 88  0.80  Locate A rea on the normal curve 12 Calculate the Z scores 88 80 a1a1 Z -scale 0.0 0.80    (X(X z1z1

58. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 58 robability that… a person from Newmarket, selected at random, will use between 80 and 88 litres per day, is…? The area under a normal curve between a z-value of 80 and a z-value of 88 is 0.2881, therefore, we conclude that 28.81% of the residents will use between 80 to 88 litres per day! 3 Look up 0.80 in Table A =.2881

59. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 59 robability that… a person from Newmarket, selected at random, will use between 76 and 92 litres per day, is…? 10 - 80 76  0.40  10 - 80 92  1.20  Locate A rea on the normal curve 12 Calculate the Z scores 92  76 a1a1 a2a2 Z -scale -.4 1.20    (X(X z1z1    (X(X z1z1

60. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 60 robability that… a person from Newmarket, selected at random, will use between 76 and 92 litres per day, is…? The area under a normal curve between a z-value of 76 and a z-value of 92 is 0.5403, therefore, we conclude that 54.03% of the residents will use between 76 to 92 litres per day! 3 Look up 0.40 then 1.20 in Table a2a2 a1a1 =.1554 =.3849 A =.5403 Add together New

61. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 61 Professor Mann has determined that the scores in his statistics course are approximately normally distributed with a mean of 72 and a standard deviation of 5. Let X be the score that separates an A from a B. Let X be the score that separates an A from a B. He announces to the class that the top 15 percent of the scores will earn an A. What is the lowest score a student can earn and still receive an A?  72 X A F D C B Sketch given information onto the normal curve 1

62. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 62 To begin, let X be the score that separates an A from a B. If 15 percent of the students score more than X, then 35 percent must score between the mean of 72 and X  a 1 =.35 72 X A A =.15 Now, use the table “backwards” to find the z-score corresponding to a 1 =.35 2 Determine the Z score from the given areas

63. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 63 Normal Distribution Table Search in the centre of the table for the area of 0.35 1.00 0.3508 0.04 Substitution

64. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 64  a 1 =.35 72 X A A =.15 …mean of 72 and a standard deviation of 5. He announces to the class that the top 15 percent of the scores will earn an A. What is the lowest score a student can earn and still receive an A? …mean of 72 and a standard deviation of 5. He announces to the class that the top 15 percent of the scores will earn an A. What is the lowest score a student can earn and still receive an A? Z -scale 0 1.04 5 - 72 X 1.04  X    A student needs a score of 77.2 to receive an A 3 Substitute values into equation    (X(X z1z1

65. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 65 Using the Normal Curve to Approximate the Binomial Distribution

66. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. The Normal Approximation to the Binomial The normal distribution (a continuous distribution ) yields a good approximation of the binomial distribution (a discrete distribution) for large values of n. Use when np and n(1- p ) are both greater than 5!

67. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 67 The probability of success stays the same for each trial The trials are independent Interested in the number of successes The experiment consists of n Bernoulli trials The outcomes are classified into one of two mutually exclusive categories, …such as success or failure Binomial Experiment

68. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 68 Mean and Variance of a Binomial Probability Distribution Mean and Variance of a Binomial Probability Distribution 2   npnp   Formula )1( p npnp   2  From Chapter 6 If.60 of students do their homework at night, find the probability that at least 50 students in a class of 80 did their homework last night.         = 19.2 Sketch given information onto the normal curve 1 1

69. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 69 49.5 Locate 50 on the normal curve 1 48 If we want at least 50, we start at 49.5 on the normal curve! 5050 50 students is represented by the narrow strip, not a single line… This is adjustment of.5 is called the continuity correction factor …find the probability that at least 50 students in a class of 80 did their homework last night.

70. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 70 …find the probability that at most 50 students in a class of 80 did their homework last night. Locate 50 on the normal curve 1 If we want at most 50, we want to include all of 50, so we use 50.5 on the normal curve! 50.5 48 5050 49.5 48 5050 Therefore if we want less than 50, we do not want to include all of 50, so we use 49.5 on the normal curve More… continuity correction factor

71. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. The value.5 is subtracted or added, depending on the problem, to a selected value when a binomial probability distribution (a discrete probability distribution) is being approximated by a continuous probability distribution (the normal distribution) Examples Continuity Correction Factor

72. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. This is the mean of a binomial distribution A recent study by a marketing research firm showed that 15% of Canadian households owned a video camera. For a sample of 200 homes, how many of the homes would you expect to have video cameras? What is the variance? What is the standard deviation? 5.25  0498.5   = np(1 – p) = (30)(1 -.15) = 25.5  = np = (.15)(200) = 30

73. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 73 a1a1 What is the probability that less than 40 homes in the sample have video cameras? “less than” does not include all of 40 – so use 39.5 The likelihood that less than 40 of the 200 homes have a video camera is about 97%. 30 39.5 Locate A rea on the normal curve 12 Calculate the Z scores 3 Look up 1.88 in Table =.4699 a1a1 A =.9699 5.0498 - 30 39.5  1.88  +.5 =    (X(X z1z1

74. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 74 Test your learning … www.mcgrawhill.ca/college/lind Click on… Online Learning Centre for quizzes extra content data sets searchable glossary access to Statistics Canada’s E-Stat data …and much more!

75. Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved. 7 - 75 This completes Chapter 7