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Area under the Normal Curve with TI-84 and Excel Example problem The picture TI-84 solution π(π§ < 1.37) βFind the area under the normal curve to the left of π§ = 1.37.β Answer: Area = 0.9147 Probability = 0.9147 or 91.47% 2ND DISTR (on the VARS key) (-) 1 2ND EE (on comma key) 9 9 (calculatorβs way to say ββ) , comma 1.37 ) right paren ENTER π(π§ > 1.37) βFind the area under the normal curve to the right of π§ = 1.37.β Answer: Area = 0.0853 Probability = 0.0853 or 8.53% π(0.55 < π§ < 1.85) βFind the area under the normal curve between π§ = 0.55 and π§ = 1.85.β Answer: Area = 0.2590 Probability = 0.2590 or 25.90% Excel solution 0.914657 =NORMSDIST(1.37) =NORMSDIST(z-score) gives you the area to the left of that z-score. Use =NORM.S.DIST(z-score) in Excel 2010+ 0.085343451 =1-NORMSDIST(1.37) 2ND DISTR (on the VARS key) 1.37 , comma 1 2ND EE (on comma key) 9 9 (calculatorβs way to say β) ) right paren ENTER To get area to the RIGHT of a certain z-score, tell Excel to subtract 1 minus the area to the left of that z-score. 0.259002912 =NORMSDIST(1.85)NORMSDIST(0.55) To get the area between two zscores, tell Excel to subtract The area to the left of the higher z-score, minus the area to the left of the lower z-score. Document1 2/9/2016 10:44 PM D.R.S. Example problem The picture TI-84 solution βFind the total area under the normal curve.β You should know this very basic special fact, that the total area under a probability distribution is always exactly precisely = 1. βWhat is the area under the right half of the normal curve?β and βWhat is the area under the left half of the normal curve?β You should know these very special facts about the Normal Distribution: ο· it is symmetric, ο· π§ = 0 is in the middle, ο· half of the area (0.5) is to the left ο· and half of the area (0.5) is to the right. Excel solution =NORMSDIST(1E99)-NORMSDIST(1E99) Use the same 1E99 for β and the same -1E99 for ββ in Excel as you use in the TI-84. =NORMSDIST(1E99)NORMSDIST(0) =NORMSDIST(0)-NORMSDIST(1E99) The calculator has rounding errors in the 10th decimal position. The correct answers are both exactly precisely 0.5000000000000000000000000. Document1 2/9/2016 10:44 PM D.R.S. Example problem The picture TI-84 solution π(π§ < β2.02) βFind the area under the normal curve to the left of π§ = β2.02. Excel solution =NORMSDIST(-2.02) Answer: Area = 0.0217 Probability = 0.0217 or 2.17% Area to the left of β2.02 plus area to the right of 2.02 totals the entire area under the curve, exactly precisely 1.000000000000. π(π§ > β2.02) βFind the area under the normal curve to the right of π§ = β2.02.β Answer: Area = 0.9783 Probability = 0.9783 or 97.83% π(β1.38 < π§ < β0.38) βFind the area under the normal curve between π§ = β1.38 and π§ = β0.38.β Answer: Area = 0.2682 Probability = 0.2682 or 26.82% π(β1.79 < π§ < 2.20) βFind the area under the normal curve between π§ = β1.79 and π§ = 2.20.β = 1 β NORMSDIST(-2.02) Area to the right of a z-score: subtract 1 minus the area to its left. =NORMSDIST(-0.38)NORMSDIST(-1.38) Area between two z-scores: Subtract area to the left of the larger z-score minus the area to the left of the smaller z-score. =NORMSDIST(2.20)-NORMSDIST(1.79) Answer: Area = 0.9494 Probability = 0.9494 or 94.94% Document1 2/9/2016 10:44 PM D.R.S. Example problem The picture TI-84 solution 2ND DISTR 3:invNorm(area) βWhat z-score has that much area to the left of it?β 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 And by symmetry, the right side boundary is at z = positive 0.6745. Check: Answer: π§ = β0.6745 to π§ = 0.6745 Check can be done with NORMSDIST: =NORMSDIST(0.6745)NORMSDIST(-0.6745) What z-scores bound the middle 68% of the area? Solution: Think and draw: 68% = 0.6800 shaded in middle 1 β 0.6800 = 0.3200 not shaded Half of that is 0.1600 0.1600 in the left tail 0.1600 in the right tail Thatβs the z-value to use in the answer. We rounded to ±0.9945. Answer: π§ = β0.9945 to π§ = 0.9945 Excel solution =NORMSINV(0.25) =NORMSINV(0.1600) Check: This is why the Empirical Rule says βIn a Normal Distribution, about 68% of the data lies within 1 standard deviation of the mean.β Yes, that rounds to 68%. Document1 2/9/2016 10:44 PM D.R.S. Example problem The picture TI-84 solution What z-scores bound the middle 95% of the area? Think and draw: 95% = 0.9500 shaded in middle 1 β 0.9500 = 0.0500 not shaded Half of that is 0.0250 0.0250 in the left tail 0.0250 in the right tail Answer: π§ = β1.9600 to π§ = 1.9600 Or π§ = β1.96 to π§ = 1.96 What z-scores bound the middle 99.7% of the area? Think and draw: 99.7% = 0.9970 1 β 0.9970 = 0.0030 Half of that is 0.0015 0.0015 in the left tail 0.0015 in the right tail Answer: π§ = β2.967 to π§ = 2.967 Or π§ = β2.97 to π§ = 2.97 Excel solution =NORMSINV(0.0250) Rounded to π§ = ±1.9600 Check: This is why the Empirical Rule says βIn a Normal Distribution, about 95% of the data lies within 2 standard deviations of the mean.β =NORMSINV(0.0015) Check: This is why the Empirical Rule says βIn a Normal Distribution, about 95% of the data lies within 2 standard deviations of the mean.β Check Empircal Ruleβs rounded value of π§ = ±3: Document1 2/9/2016 10:44 PM D.R.S. Example problem The picture TI-84 solution What z-score separates the top 10% of the data from the bottom 90%? Excel solution =NORMSINV(0.9000) Check: Think and draw: 90% = 0.9000 area is to the left. 10% = 0.1000 area is to the right Answer: π§ = 1.2816 or π§ = 1.28 What score separates the bottom third of the data from the top two-thirds? =NORMSINV(1/3) Think and draw: 1/3 to the left. You can use the fraction in TI-84 or in Excel. Answer: π§ = β0.4307 or π§ = β0.43 Document1 2/9/2016 10:44 PM D.R.S. TI-84 Production notes The TI-84 graphs were done as described below. Perhaps the ShadeNorm command should have been used instead. Thatβs something that can be investigated another day. ShadeNorm documentation is shown on the next page. The π₯-axis has the π§ values from β4 to 4 with tickmarks each π§ = 1 unit. The π¦-axis goes from β0.1 to 0.4 with tickmarks every 0.1 unit. The β0.1 was chosen to leave space at the bottom of the screen. For shading, 2ND CALC Lower Limit and Upper Limit were given and a value was displayed. As discussed somewhere else, this value is not always accurate enough because the infinitely long tails are missed with our choice of finite Xmin and Xmax values. So we clear the screen and press GRAPH again to get the shaded graph without the inadequate β« π(π₯) ππ₯ result. Then 2ND DRAW 1:ClrDraw clears the shading in preparation for another problem. Document1 2/9/2016 10:44 PM D.R.S. TI-84 ShadeNorm command Example: 2ND DISTR, DRAW submenu, 1:ShadeNorm ShadeNorm(-1,1) Document1 2/9/2016 10:44 PM D.R.S.