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Chapter 7 Section 2

Chapter 7 Section 2. The Standard Normal Distribution. 1. 2. 3. Chapter 7 – Section 2. Learning objectives Find the area under the standard normal curve Find Z -scores for a given area Interpret the area under the standard normal curve as a probability. 1. 2. 3.

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Chapter 7 Section 2

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  1. Chapter 7Section 2 The Standard Normal Distribution

  2. 1 2 3 Chapter 7 – Section 2 • Learning objectives • Find the area under the standard normal curve • Find Z-scores for a given area • Interpret the area under the standard normal curve as a probability

  3. 1 2 3 Chapter 7 – Section 2 • Learning objectives • Find the area under the standard normal curve • Find Z-scores for a given area • Interpret the area under the standard normal curve as a probability

  4. Chapter 7 – Section 2 • The standard normal curve is the one with mean μ = 0 and standard deviation σ = 1 • We have related the general normal random variable to the standard normal random variable through the Z-score • In this section, we discuss how to compute with the standard normal random variable

  5. Chapter 7 – Section 2 • There are several ways to calculate the area under the standard normal curve • What does not work – some kind of a simple formula • We can use a table (such as Table IV on the inside back cover) • We can use technology (a calculator or software) • Using technology is preferred

  6. Chapter 7 – Section 2 • Three different area calculations • Find the area to the left of • Find the area to the right of • Find the area between • Three different area calculations • Find the area to the left of • Find the area to the right of • Find the area between • Three different methods shown here • From a table • Using Excel • Using StatCrunch

  7. Chapter 7 – Section 2 • "To the left of" – using a table • Calculate the area to the left of Z = 1.68 • "To the left of" – using a table • Calculate the area to the left of Z = 1.68 • Break up 1.68 as 1.6 + .08 • "To the left of" – using a table • Calculate the area to the left of Z = 1.68 • Break up 1.68 as 1.6 + .08 • Find the row 1.6 • "To the left of" – using a table • Calculate the area to the left of Z = 1.68 • Break up 1.68 as 1.6 + .08 • Find the row 1.6 • Find the column .08 • "To the left of" – using a table • Calculate the area to the left of Z = 1.68 • Break up 1.68 as 1.6 + .08 • Find the row 1.6 • Find the column .08 • The probability is 0.9535 Enter Enter Enter Enter Read

  8. Chapter 7 – Section 2 • "To the left of" – using Excel • The function in Excel for the standard normal is =NORMSDIST(Z-score) • NORM (normal) S (standard) DIST (distribution) Read Enter Enter Read

  9. Chapter 7 – Section 2 • To the left of 1.68 – using StatCrunch • The function is Stat – Calculators – Normal Read Enter Enter

  10. Chapter 7 – Section 2 • "To the right of" – using a table • The area to the left of Z = 1.68 is 0.9535 • "To the right of" – using a table • The area to the left of Z = 1.68 is 0.9535 • The right of … that’s the remaining amount • The two add up to 1, so the right of is 1 – 0.9535 = 0.0465 Enter Enter Read

  11. Chapter 7 – Section 2 • "To the right of" – using Excel • "To the right of" – using Excel • The right of … that’s the remaining amount of to the left of, subtract from 1 1 – 0.9535 = 0.0465 Read Enter Read

  12. "To the right of" – using StatCrunch • Change the <= to >= (the picture and number change) Read Enter Chapter 7 – Section 2 • "To the right of" – using StatCrunch • "To the right of" – using StatCrunch • Change the <= to >= Read Enter

  13. Chapter 7 – Section 2 • “Between” • Between Z = – 0.51 and Z = 1.87 • This is not a one step calculation

  14. Included too much Chapter 7 – Section 2 • The left hand picture … to the left of 1.87 … includes too much • It is too much by the right hand picture … to the left of -0.51

  15. We want We start out with, but it’s too much We correct by Chapter 7 – Section 2 • Between Z = – 0.51 and Z = 1.87

  16. Chapter 7 – Section 2 • We can use any of the three methods to compute the normal probabilities to get • The area between -0.51 and 1.87 • We can use any of the three methods to compute the normal probabilities to get • The area between -0.51 and 1.87 • The area to the left of 1.87, or 0.9693 … minus • We can use any of the three methods to compute the normal probabilities to get • The area between -0.51 and 1.87 • The area to the left of 1.87, or 0.9693 … minus • The area to the left of -0.51, or 0.3050 … which equals • We can use any of the three methods to compute the normal probabilities to get • The area between -0.51 and 1.87 • The area to the left of 1.87, or 0.9693 … minus • The area to the left of -0.51, or 0.3050 … which equals • The difference of 0.6643 • We can use any of the three methods to compute the normal probabilities to get • The area between -0.51 and 1.87 • The area to the left of 1.87, or 0.9693 … minus • The area to the left of -0.51, or 0.3050 … which equals • The difference of 0.6643 • Thus the area under the standard normal curve between -0.51 and 1.87 is 0.6643

  17. We want We delete the extra on the left We delete the extra on the right Chapter 7 – Section 2 • A different way for “between”

  18. Chapter 7 – Section 2 • Again, we can use any of the three methods to compute the normal probabilities to get • Again, we can use any of the three methods to compute the normal probabilities to get • The area between -0.51 and 1.87 • The area to the left of -0.51, or 0.3050 … plus • The area to the right of 1.87, or .0307 … which equals • The total area to get rid of which equals 0.3357 • Again, we can use any of the three methods to compute the normal probabilities to get • The area between -0.51 and 1.87 • The area to the left of -0.51, or 0.3050 … plus • The area to the right of 1.87, or .0307 … which equals • The total area to get rid of which equals 0.3357 • Thus the area under the standard normal curve between -0.51 and 1.87 is 1 – 0.3357 = 0.6643

  19. 1 2 3 Chapter 7 – Section 2 • Learning objectives • Find the area under the standard normal curve • Find Z-scores for a given area • Interpret the area under the standard normal curve as a probability

  20. Chapter 7 – Section 2 • We did the problem: Z-Score  Area • Now we will do the reverse of that Area  Z-Score • We did the problem: Z-Score  Area • Now we will do the reverse of that Area  Z-Score • This is finding the Z-score (value) that corresponds to a specified area (percentile) • And … no surprise … we can do this with a table, with Excel, with StatCrunch, with …

  21. “To the left of” – using a table • Find the Z-score for which the area to the left of it is 0.32 • Look in the middle of the table … find 0.32 • “To the left of” – using a table • Find the Z-score for which the area to the left of it is 0.32 • Look in the middle of the table … find 0.32 • The nearest to 0.32 is 0.3192 … a Z-Score of -.47 Read Find Read Chapter 7 – Section 2 • “To the left of” – using a table • Find the Z-score for which the area to the left of it is 0.32

  22. Read Enter Read Chapter 7 – Section 2 • "To the left of" – using Excel • The function in Excel for the standard normal is =NORMSINV(probability) • NORM (normal) S (standard) INV (inverse)

  23. Enter Read Chapter 7 – Section 2 • "To the left of" – using StatCrunch • The function is Stat – Calculators – Normal

  24. Read Read Enter Chapter 7 – Section 2 • "To the right of" – using a table • Find the Z-score for which the area to the right of it is 0.4332 • Right of it is .4332 … left of it would be .5668 • A value of .17

  25. Read Enter Read Chapter 7 – Section 2 • "To the right of" – using Excel • To the right is .4332 … to the left would be .5668 (the same as for the table)

  26. Chapter 7 – Section 2 • "To the right of" – using StatCrunch • Change the <= to >= (watch the picture change) Enter Read

  27. Chapter 7 – Section 2 • We will often want to find a middle range, to find the middle 90% or the middle 95% or the middle 99%, of the standard normal • The middle 90% would be

  28. Chapter 7 – Section 2 • 90% in the middle is 10% outside the middle, i.e. 5% off each end • These problems can be solved in either of two equivalent ways • We could find • The number for which 5% is to the left, or • The number for which 5% is to the right

  29. Chapter 7 – Section 2 • The two possible ways • The number for which 5% is to the left, or • The number for which 5% is to the right 5% is to the right 5% is to the left

  30. Chapter 7 – Section 2 • The number zα is the Z-score such that the area to the right of zα is α • The number zα is the Z-score such that the area to the right of zα is α • Some useful values are • z.10 = 1.28, the area between -1.28 and 1.28 is 0.80 • z.05 = 1.64, the area between -1.64 and 1.64 is 0.90 • z.025 = 1.96, the area between -1.96 and 1.96 is 0.95 • z.01 = 2.33, the area between -2.33 and 2.33 is 0.98 • z.005 = 2.58, the area between -2.58 and 2.58 is 0.99

  31. 1 2 3 Chapter 7 – Section 2 • Learning objectives • Find the area under the standard normal curve • Find Z-scores for a given area • Interpret the area under the standard normal curve as a probability

  32. Chapter 7 – Section 2 • The area under a normal curve can be interpreted as a probability • The standard normal curve can be interpreted as a probability density function • The area under a normal curve can be interpreted as a probability • The standard normal curve can be interpreted as a probability density function • We will use Z to represent a standard normal random variable, so it has probabilities such as • P(a < Z < b) • P(Z < a) • P(Z > a)

  33. Summary: Chapter 7 – Section 2 • Calculations for the standard normal curve can be done using tables or using technology • One can calculate the area under the standard normal curve, to the left of or to the right of each Z-score • One can calculate the Z-score so that the area to the left of it or to the right of it is a certain value • Areas and probabilities are two different representations of the same concept

  34. Example: Chapter 7 – Section 2 • Determine the area under the standard normal curve that lies • a. to the left of Z = –2.31. (0.0104) • b. to the right of Z = –1.47. (0.9292) • c. between Z = –2.31 and Z = 0. (0.4896) • d. between Z = –2.31 and Z = –1.47. (0.0603) • e. between Z = 1.47 and Z = 2.31. (0.0603) • f. between Z = –2.31 and Z = 1.47. (0.9188) • g. to the left of Z = –2.31 or to the right of Z = 1.47. (0.0812)

  35. Example: Chapter 7 – Section 2 • The Graduate Record Examination (GRE) is a test required for admission to many U.S. graduate schools. The Department of Molecular Genetics at Ohio State University requires a GRE score no less than the 60th percentile. (Source: www.biosci.ohio-state.edu/~molgen/html/admission_criteria.html.) • a. Find the Z-score corresponding to the 60th percentile. In other words, find the Z-score such that the area under the standard normal curve to the left is 0.60. (0.25) • b. How many standard deviations above the mean is the 60th percentile? (0.25)

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