Probability = 0.2682 or 26.82% Subtract area to the left of the larger z-score minus the area to the left of the smaller z-score. the area under the normal “Find curve between and .” Answer: Area = 0.9494 Probability = 0.9494 or 94.94% What z-scores bound the middle 2ND DISTR 3:invNorm(area) =NORMSINV(0.25) 50% of the area? z-score has that much area “What to the left of it?” Think and draw: 50% = 0.5000 shaded in middle 1 - 0.5000 = 0.5000 not shaded And by symmetry, the right side Half of that is 0.2500 boundary is at z = positive 0.6745. 0.2500 in the left tail Check can be done with 0.2500 in the right tail Check: NORMSDIST: Answer: to What z-scores bound the middle Solution: =NORMSINV(0.1600) 68% of the area? This is why the Empirical Rule says That’s the z-value to use in the Think and draw: a Normal Distribution, about answer. We rounded to ±0.9945. “In 68% = 0.6800 shaded in middle 68% of the data lies within 1 1 0.6800 = 0.3200 not shaded standard deviation of the mean.” Check: – Half of that is 0.1600 0.1600 in the left tail Yes, that rounds to 68%. 0.1600 in the right tail Answer: to
Probability = 0.2682 or 26.82% Subtract area to the left of the larger z-score minus the area to the left of the smaller z-score. “Find the area under the normal curve between and .” Answer: Area = 0.9494 Probability = 0.9494 or 94.94% What z-scores bound the middle 50% of the area? Think and draw: 50% = 0.5000 shaded in middle 1 - 0.5000 = 0.5000 not shaded Half of that is 0.2500 0.2500 in the left tail 0.2500 in the right tail Answer: to 2ND DISTR 3:invNorm(area) “What z-score has that much area to the left of it?” And by symmetry, the right side boundary is at z = positive 0.6745. Check: =NORMSINV(0.25) Check can be done with NORMSDIST: What z-scores bound the middle 68% of the area? Think and draw: 68% = 0.6800 shaded in middle 1 – 0.6800 = 0.3200 not shaded Half of that is 0.1600 0.1600 in the left tail 0.1600 in the right tail Answer: to This is why the Empirical Rule says “In a Normal Distribution, about 68% of the data lies within 1 standard deviation of the mean.” Solution: That’s the z-value to use in the answer. We rounded to ±0.9945. Check: Yes, that rounds to 68%. =NORMSINV(0.1600)
TI84_and_Excel.docx