1. 3.5 Applying the Normal Distribution – Z Scores Example 1 Determine the number of standard deviations above or below the mean each piece of data is. (This is known as a z-score) a) mean = 100, st dev = 20, data = 110 110 is 0.5 standard deviations above the mean.

2. b) mean = 25, st dev = 5, data = 18 25 is 1.4 standard deviations below the mean. c) X ~ N(40, 4), data = 50 50 is 5 standard deviations above the mean.

3. Example 2 Determine the percent and percentile corresponding to each z-score from question 1 and sketch. You will need page 398/399 of your textbook. a) mean = 100, st dev = 20, data = 110, z = 0.5 z = 0.5 => 0.6915 = 69.15% the 69 th percentile 100120140160806040 110 69.15% 30.85%

4. b) mean = 25, st dev = 5, data = 18, z = -1.4 z = -1.4 => 0.0808 = 8.08% the 8 th percentile 25 30 3540201510 18 8.08% 91.92%

5. c) mean = 40, st dev = 2, data = 50, z = 5 z = 5 => too high = 100% 100 th percentile (technically slightly less) 40 42 4446383634 50 100% 0%

6. Example 3 The heights of Doyle students are normally distributed. X ~ N (150, 144). How tall does someone need to be to be in the top 20% ? mean = 150, st dev = 12 data = ?, z = ? To be in the top 20% you must be in at least the ____ percentile. 80 th z = ____ corresponds most closely to 80%. 0.84 solve for x To be in the top 20% you must be at least 160.08 cm 80% 20%