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Chapter 7

THE. Normal. PROBABILITY DISTRIBUTION. Chapter 7. …the set of all the values in any interval is uncountable or infinite!. Recall…. Continuous Random Variable. ….we will now study the class of continuous probability distributions. Quantitative. Numerical Observations. Also Recall that….

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Chapter 7

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  1. THE Normal PROBABILITYDISTRIBUTION Chapter 7

  2. …the set ofall the valuesin any intervalisuncountable or infinite! Recall… Continuous Random Variable ….we will now study the class of continuous probability distributions.

  3. Quantitative Numerical Observations Also Recall that… Variables … can be classified as either Discrete or Continuous Characteristics Continuous … can assumeanyvaluewithin a specified range! e.g. - Pressure in a tire - Weight of a pork chop - Height of students in a class

  4. Continuous Random Variable …the Total sum of probabilities should always be 1! When dealing with aContinuous Random Variablewe assume thatthe probabilitythat the variablewill take on any particular valueis 0! Instead, Probabilitiesare assigned to intervals of values!

  5. Continuous Probability Distributions Range of Values f(x) P(a<X<b) x b a

  6. See Histograms Relative Frequency Histogram  Probability Function • Data: • Heights of adult Canadian males… • n = 50,class interval= 2 • n = 500,class interval= 1 • n = 5000,class interval= 0.4 • Probability function

  7. (A) n = 50 (B) n = 500 0.07 - 0.06 - 0.05 - 0.04 - 0.03 - 0.02 - 0.01 - 0.00 - 0.08 - 0.07 - 0.06 - 0.05 - 0.04 - 0.03 - 0.02 - 0.01 - 0.00 - Chart 7-1(C) Chart 7-1(D) (C) n = 5000 Probability Function 140 150 160 170 180 190 200 145 155 165 175 185 195 0.08 - 0.07 - 0.06 - 0.05 - 0.04 - 0.03 - 0.02 - 0.01 - 0.00 - 0.08 - 0.07 - 0.06 - 0.05 - 0.04 - 0.03 - 0.02 - 0.01 - 0.00 - Chart 7-1(A) Chart 7-1(B) 150 160 170 180 190 155 165 175 185 195 Histograms

  8. Reminder …that all probability functionsare not normal! Aprobability functioncan 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)

  9. Characteristics Normal Probability Distribution The arithmetic mean, median, and mode of the distribution are equal and located at the peak. The normal curve is bell-shaped and has a single peak at the exact centre of the distribution. …thus half the area under the curve is above themean and half isbelow it.

  10. Characteristics Normal Probability Distribution The normal probability distribution is asymptotic, The normal probability distribution is symmetrical about its mean. i.e. the curve gets closer and closer to the x-axis but never actually touches it.

  11. Characteristics Normal Probability Distribution Theoretically, curve extends to infinity Summary Normal curve is symmetrical + Mean, median, and mode are equal

  12. Q uestion s = 2, 3, 4 Normal Probability Distribution part of a“family”of curves How does the Standard Deviation values affect the shape of f(x)? s = 2 s =3 s =4

  13. Q Normal Probability Distribution uestion part of a “family” of curves m = 10 m = 11 m = 12 How does the following Expected values affect the location of f(x)? m = 10, 11, 12

  14. - µ) z (X = s 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 populationStandard Deviation, . A z-value is, therefore, a location on the standard normal curve. Formula

  15. Area = +0.5 Area = + 0.5 The Standard Normal Probability Distribution Total Area underthe curve is 100%or 1 …since the mean is located at the peak of the curve,half the areaunder the curve isabove themean and half isbelow the mean

  16. 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? Q uestion Formula - µ) z (X = s Show curve The Standard Normal Probability Distribution z - $3,300 $3,000 = $300 z = 1.00 A z-value of 1 indicates that the value of $3,300 is one standarddeviationabovethe mean of $3,000.

  17. A z-value of 1 indicates that the value of $3,300 is one standarddeviationabovethe mean of $3,000. $3300 z= 1.0 $3000 z= 0

  18. Q uestion Formula - µ) z (X = s Show curve 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. - $2,550 $3,000 z = $300 Z = -1.50 A z-value of –1.50 indicates that the value of $2,550 is one and a half (1.5) standarddeviationsbelowthe mean of $3,000. What is the z-value for a salary of $2,550?

  19. The Standard Normal Probability Distribution A z-value of –1.50 indicates that the value of $2,550 is one and a half standarddeviationsbelow the mean of $3,000. $2550 z= -1.50 $3000 z= 0

  20. 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

  21. Understanding Standard Deviations and z-scores

  22. Understanding Standard Deviations and z-scores Q uestion How many  is 40 away from  ? 0 20 40 50 60 80 100 If…  = 50 = 8 +  40 is 10 units(-10)away from 50 40 is –1.25 away from 50 If 1 standard deviation ()is 8 units,then 10 units must be 1.25 SD Using z

  23. … divide bysto getz m - z X z – 10 8 Formula = = = Note … distance from the  s The Standard Normal Probability Distribution the Z Distribution Recall A z-value is the distance between a selected value (designated X) and the populationmeanµ, divided by the populationStandard Deviation, . 40 – 50 8 = -1.25

  24. The Standard Normal Probability Distribution Q uestion the Z Distribution - z-value +z-value Right of mean Left of mean 0 20 40 50 60 80 100 Note Recall 0 Z-scale The meanµ, of a standard normal distribution is 0,andStandard Deviation  is1.

  25. Q uestion 0 20 40 50 60 80 100 - µ) z (X = s z-table The Standard Normal Probability Distribution Calculate P(50 < X < 60) Transferring each value into z-scores, If…  = 50 = 8 0 1.25 Z1= (60-50)/8 = 1.25 = P(0 < z < 1.25) P(50 < X < 60)

  26. Provides the second decimal place Provides the z-score to 1 decimal place Normal Distribution Table To find the area between a zscore of0 and a score of1.25 0.05 1.20 0.3944 …thus

  27. The Standard Normal Probability Distribution the Z Distribution New 0.3944 0 20 40 50 60 80 100 … 0.3944 is the area between the mean and a positivez score 1.25 0 1.25 Therefore, the Probability of 50<X<60 is 39.44%

  28. Q uestion 0 20 40 50 60 80 100 - µ) z (X = s The Standard Normal Probability Distribution Calculate P(40 < X < 60) Transferring each value into z-scores, If…  = 50 = 8 0 -1.25 1.25 Z1= (40-50)/8 = -1.25 Z2= (60-50)/8 = 1.25 …and

  29. 0.3944 0 20 40 50 60 80 100 The Standard Normal Probability Distribution … because both sides are symmetrical, the left side (the area between the mean and a negativez score -1.25) must have the same area………….. 0.3944. Z2= 1.25 (determined earlier) 0 -1.25 1.25 = for a Total area of: 0.3944 + 0.3944 or 0.7888

  30. The Standard Normal Probability Distribution Q uestion the Z Distribution 0 20 40 50 60 80 100 Note Calculate P(40 < X < 60) If…  = 50 = 8 P(40 < X < 60) = P(40 X 60) …since P(X= 40 or 60) is zero

  31. Using the Normal Distribution Table PractiseFinding Areas

  32. Find …the area between: Z1 = 0 and Z2 = 1.00 1 2 0 Note z-table Practise using the Normal Distribution Table LocateArea on the normal curve Look up 1.00 in Table 1.00 Be sure to always sketch the curve, insert the given values and shade the required area.

  33. Provides the second decimal place Provides the z-score to 1 decimal place Normal Distribution Table To find the area between a zscore of 0 and a score of1.00 0.00 The Area between 0<Z<1.00 is 0.3413 1.00 Therefore, the Probability of 0<Z<1.00 is 34.13% 0.3413

  34. Find …the area between: Z1 = 0 and Z2 = 1.64 1 2 Practise using the Normal Distribution Table LocateArea on the normal curve Look up 1.64 in Table 0 1.64 The Area between 0<Z<1.64 is 0.4495 Therefore, the Probability of 0<Z<1.64 is 44.95%

  35. Find …the area between: Z1 = 0 and Z2 = 0.47 1 2 Practise using the Normal Distribution Table LocateArea on the normal curve Look up 0.47 in Table 0 0.47 The Area between 0<Z<0.47 is 0.1808 Therefore, the Probability of 0<Z<0.47 is 18.08%

  36. Find …the area between: Z1 = 0 and Z2 = -1.64 1 2 Note Practise using the Normal Distribution Table LocateArea on the normal curve Look up 1.64 in Table 0 The Area between -1.64<Z<0 is 0.4495 -1.64 Therefore, the Probability of -1.64<Z<0 is 44.95% Because of symmetry, this area is the same as betweenz of 0 and positive 1.64

  37. Find …the area between: Z1 = 0 and Z2 = -2.22 LocateArea on the normal curve 1 2 Look up 2.22 in Table Practise using the Normal Distribution Table 0 -2.22 2.22 The Area between –2.22<Z<0 is 0.4868 Therefore, the Probability of –2.22<Z<0 is 48.68%

  38. Practise using the Normal Distribution Table 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,areas and probabilitiesare always positive values!

  39. Find …the area between: Z1 = 0 and Z2 = -2.96 1 2 Practise using the Normal Distribution Table LocateArea on the normal curve Look up 2.96 in Table 0 -2.96 2.96 The Area between –2.96<Z<0 is 0.4985 Therefore, the Probability of –2.96<Z<0 is 49.85%

  40. Find …the area between: Z1 = -1.65 and Z2 = 1.65 1 2 Add together Practise using the Normal Distribution Table LocateArea on the normal curve Look up 1.65 in Table 0 -1.65 1.65 The Area (a1) between –1.65<Z<0 is 0.4505 The Area (a2) between 0<Z<1.65 is 0.4505 Therefore, the required Total Area is 0.9010

  41. Find …the area between: LocateArea on the normal curve 1 2 Look up – 2.00 then 1.00 in Table Add together Practise using the Normal Distribution Table Z1 = -2.00 and Z2 = 1.00 0 -2.00 1.00 The Area (a1) between –2.00<Z<0 is 0.4772 The Area (a2) between 0<Z<1.00 is 0.3413 Therefore, the required Total Area is 0.8185

  42. Find …the area between: LocateArea on the normal curve 1 2 Look up – 0.44 then 1.96 in Table Add together Practise using the Normal Distribution Table Z1 = -0.44 and Z2 = 1.96 0 -0.44 1.96 The Area (a1) between –0.44<Z<0 is 0.1700 The Area (a2) between 0<Z<1.96 is 0.4750 Therefore, the required Total Area is 0.6450

  43. Total Area underthe curve is 100%or 1 Area = 0.5 Area = 0.5 …since the mean is located at the peak of the curve,half the areaunder the curve isabove themean and half isbelow the mean The Standard Normal Probability Distribution

  44. LocateArea on the normal curve 1 2 3 a1 Look up 2.00 in Table Area = 0.5 Adjust as needed Practise using the Normal Distribution Table Find …the area to the Left of Z1 = 2.00 0 2 The Area (a1) between 0<Z<2.0 is 0.4772 The Area to the left of Z1 isa1+.5 = 0.4772 + 0.5 Therefore, the required Total Area is 0.9772

  45. LocateArea on the normal curve 1 2 3 a1 Area = 0.5 Look up 1.96 in Table Adjust as needed Practise using the Normal Distribution Table Find …the area to the Left of Z1 = 1.96 0 1.96 The Area (a1) between 0<Z<1.96 is 0.4750 The Area to the left of Z1 isa1+.5= 0.4750 + 0.5 Therefore, the required Total Area is 0.9750

  46. LocateArea on the normal curve 1 2 3 Area = 0.5 Look up 1.64 in Table Adjust as needed Practise using the Normal Distribution Table Find …the area to the Left of Z1 = -1.64 a1 0 -1.64 The Area (a1) between –1.64<Z<0 is 0.4495 The Area to the left of Z1 is 0.5 -a1 = 0.5 - 0.4495 Therefore, the required Total Area is 0.0505

  47. LocateArea on the normal curve 1 2 Area = 0.5 a1 Look up 0.95 in Table 3 Adjust as needed Practise using the Normal Distribution Table Find …the area to the Left of Z1 = -0.95 0 -0.95 The Area (a1) between –0.95<Z<0 is 0.3289 The Area to the left of Z1 is 0.5 -a1 = 0.5 - 0.3289 Therefore, the required Total Area is 0.1711

  48. LocateArea on the normal curve 1 2 Area = 0.5 a1 Look up 0.95 in Table 3 Adjust as needed Practise using the Normal Distribution Table Find …the area to the Right of Z1 = -0.95 0 -0.95 The Area (a1) between –0.95<Z<0 is 0.3289 The Area to the left of Z1 is 0.5 +a1 = 0.5 + 0.3289 Therefore, the required Total Area is 0.8289

  49. LocateArea on the normal curve 1 2 Look up 1.00 in Table 3 Adjust as needed Practise using the Normal Distribution Table Find …the area to the Right of Z1 = 1.00 A Area = 0.5 a1 0 0 1.00 The Area (a1) between 0<Z<1.00 is 0.3413 The Area to the left of Z1 is 0.5 -a1 = 0.5 - 0.3413 Therefore, the required Total Area is 0.1587

  50. Find …the area between: LocateArea on the normal curve a1 Find 1 2 3 Look up 1.96 then 2.58 in Table a2 Adjust as needed Subtract Practise using the Normal Distribution Table Z1 = 1.96 and Z2 = 2.58 0 0 1.96 2.58 The Area (a1) between 0<Z<1.96 is 0.4750 The Area (a2) between 0<Z<2.58 is 0.4951 Therefore, the required Total Area is 0.0201

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