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C.M. Pascual

E LEMENTARY. S TATISTICS. Chapter 7 Normal Distributions: Finding Probabilities. C.M. Pascual. Properties of the Normal Distribution. The normal distribution curve is bell-shaped; The mean, median, and mode are equal and located at the center of the distribution;

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C.M. Pascual

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  1. ELEMENTARY STATISTICS Chapter 7 Normal Distributions: Finding Probabilities C.M. Pascual

  2. Properties of the Normal Distribution • The normal distribution curve is bell-shaped; • The mean, median, and mode are equal and located at the center of the distribution; • The normal distribution curve is unimodal; • The curve is symmetrical about the mean; • The curve is continuous; • The curve never touches the x-axis; • The total area under the normal distribution curve is equal to 1 or 100%.

  3. Other Normal Distributions If  0or 1 (or both), we will convert values to standard scores using Formula 5-2, then procedures for working with all normal distributions are the same as those for the standard normal distribution.

  4. x - µ z =  Other Normal Distributions If  0or  1 (or both), we will convert values to standard scores using Formula 5-2, then procedures for working with all normal distributions are the same as those for the standard normal distribution. Formula 7-2

  5. Converting to Standard Normal Distribution P  x (a) Figure 7-13

  6. x -  z =  Converting to Standard Normal Distribution P P  x z 0 (a) (b) Figure 5-13

  7. Probability of Weight between 143 pounds and 201 pounds 201 - 143 z = = 2.00 29 x = 143 s=29 Weight 143201 z 0 2.00 Figure 7-14

  8. Probability of Weight between 143 pounds and 201 pounds Value found in Appendix B x = 143 s=29 Weight 143201 z 0 2.00 Figure 7-14

  9. Probability of Weight between 143 pounds and 201 pounds 0.4772 x = 143 s=29 Weight 143201 z 0 2.00 Figure 7-14

  10. Probability of Weight between 143 pounds and 201 pounds There is a 0.4772 probability of randomly selecting a woman with a weight between 143 and 201 lbs. x = 143 s=29 Weight 143201 z 0 2.00 Figure 7-14

  11. Probability of Weight between 143 pounds and 201 pounds OR - 47.72% of women have weights between 143 lb and 201 lb. x = 143 s=29 Weight 143201 z 0 2.00 Figure 7-14

  12. Example 1 • Determine the area under the standard normal curve between 0 to 1.96 • Solution: • Using Appendix B, the area = 0.4750 0 z=1.96

  13. Example 2 • Determine the area under the standard normal curve between z= - 1.53 • Solution: By symmetry • Using Appendix B, the area = 0.4370 z= -1.53 0 z=1.53

  14. Example 3 • Determine the area under the standard normal curve right of z= 0.71 • Solution: Area to the Right of 0 is 0.5 • Using Appendix B, at z=0.71, the area = 0.2611; • At right of z = 0.5 – 0,2611 = 0.2389 Area = 0.2389 0 0.71

  15. Example 4 • Find the area under the standard normal curve left of z= -2.12 • Solution: By symmetry Area to the left of 0 is 0.5 • Using Appendix B, at z=0.71, the area = 0.4830; • At right of z = 0.5 – 0.4830 = 0.0170 Area = 0.4830 Area = 0.2389 z = - 2.12 0

  16. Example 5 • Find the area under the standard normal curve between z= 1.08 and z = 2.96 • Solution: • Area between 0 and 1.08 = 0.3599; Area between 0 and 2.96 = 0.4985 • Area between 1.08 and 2.96 = 0.4985-0.3599 = 0.1386 Area = 0.1386 0 z-=1.08 z=2.96

  17. Example 6 • Find the area under the standard normal curve between z=-1.3 and z = 0.99 • Solution: • Area between -1.3 and 0 = 0.4032; Area between 0 and 0.99 = 0.3389 • Area between -1.3 and 0.99 = 0.4032+0.3389 = 0.7421 Area = 0.7421 z=-1.3 0 z=0.99

  18. Example 7 • Find the value of z for the normal curve with area of 0.029 • Solution: • Area between 0 and z = 0.5 – 0.029 = 0.471; From Appendix B search for area 0f 0.471 with z value of 1.90 • Therefore the value of z = 1.90 Area = 0.029 0 z =?

  19. Example 8 • Find the probability P(0 < z < 1.65) using the standard normal distribution. • Solution: • Area between 0 and z = 1.65 = 0.4505 or 45.05%; • Therefore the value of z = 1.90 Area = 0.4505 0 z =1.65

  20. Example 9 • Determine the value of z when x=12, µ=16 and σ = 3 • Solution: • Using the formula z = (x - µ)/ σ • = (12 – 16)/3 • = - 1.33

  21. Example 10 • For a continuous random variable that has a normal distribution with mean of 20 and a standard deviation of 4, find the area under the normal curve from x=20 and x=27. • Solution: • Find standardize the normal distribution by converting the x values to z values: • Using the formula z = (x - µ)/ σ • z= (20 – 20)/4 = 0 and z = (27 – 20)/4 = 1.75 • Then using Appendix B; at z= 1.75; Area = 0.4608 • The area between 0 and 1.75 is 0.4608

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