251Distrex2 1/14/08Use of Standard Normal TablesBefore we start, there are a few things that you have to know about the Standardized Normal Distribution.In the following z represents a variable with the Standardized Normal Distribution. Any other random variable is x.1) The Distribution is symmetrical about zero. This means that you never have to look up a probability between, say -2 and zero on a conventional Normal table. You instead look up the probability between zero and 2. 2002 zPzP2) The area under the Normal curve, like the area under the curve for all continuous distributions, is 1.Since the zero is the halfway point under the curve, this implies the following: 5.0 zP and 5.0 zP3) The standard notation to say that a given distribution is Normal with a certain mean and standard deviation is ,~ N.This means that, if z represents a variable with the Standardized Normal Distribution, we write 1,0~ Nz and if x is Normally distributed with a mean of 50 and a standard deviation of 100 we write 100,50~ Nx.4) Any probability for a Normally distributed variable can be found using the Standardized Normal Distribution by using the transformation xz.This means that if 100,50~ Nx and we want 250200 xP, we replace values of x with 10050xz. We thus say: 00.250.11005025010050200250200 zPzPxPThe following pages include 8 problems that represent all the Standardized Normal Distribution problemsthat you will ever see. The diagrams shown were generated by NormArea, a Global Macro routine written by Jon Cryer at the University of Iowa. Because it only shows areas under the curve for two points or areas below one point, some of the diagrams are reversed. The process of creating the diagrams in Minitabis shown at the end. If you do diagrams like this by hand, it is extremely important that you add a vertical line at zero.251Distrex2 9/7/06Problem (i) 05.10 zP This is exactly what is given on the Normal table. If you go to the 1.5 line of the table and then read across to the .05 column you will find .3531. 3531.05.10 zPNote that students are often confused when a calculation yields a zero and they remember that in most problems they should add or subtract. If the interval for the standardized distribution begins or ends with zero, you only have to look up one number. Note also that 005.1 zP 3531.05.10 zP. normData AxisDensity43210-1-2-3-4-50.40.30.20.10.0Normal Curve with Mean 0 and Standard Deviation 1The Area Between 0 and 1.05 is 0.35312251Distrex2 9/7/06Problem (ii) 05.107.2 zP. This time we shade areas on both sides of zero and will add the area from -2.07 to zero to the area from zero to 1.05. Remember that all probabilities are positive. 8339.3531.4808.05.1007.2005.10007.205.107.2 zPzPzPzPzPIn general, if the area is on both sides of zero, we add probabilities. normData AxisDensity43210-1-2-3-4-50.40.30.20.10.0Normal Curve with Mean 0 and Standard Deviation 1The Area Between -2.07 and 1.05 is 0.83393251Distrex2 9/7/06Problem (iii) 05.172.0 zPRemember to add a vertical line at zero. We have shaded an area between two numbers on one side of zero, so we look up the larger of the two areas (zero to 1.05 here) and subtract the smaller area. Rememberthat all probabilities are positive. Be careful that you don’t try to find 7.2 instead of 0.72. 0889.2642.3531.72.0005.1005.172.0 zPzPzP normIn general, if the area is on one side of zero and does not start or end with zero, we subtract probabilities.Data AxisDensity43210-1-2-3-4-50.40.30.20.10.0Normal Curve with Mean 0 and Standard Deviation 1The Area Between 0.72 and 1.05 is 0.08894251Distrex2 9/7/06Problem (iv) 72.005.1 zPIf you remembered to draw a vertical line at zero, you would immediately see that this is the mirror image of the last problem and thus has the same solution. Of course, if you did it from scratch, you would do the following. 072.0005.172.005.1 zPzPzP 0889.2642.3531.72.0005.10 zPzPnormData AxisDensity43210-1-2-3-4-50.40.30.20.10.0Normal Curve with Mean 0 and Standard Deviation 1The Area Between -1.05 and -0.72 is 0.08895251Distrex2 9/7/06Problem (v) zP 2This is weird-looking, but could happen if some calculations end up on the wrong side. If you make a diagram with a vertical line at zero and think about what this says, you should realize that the values ofz that you want are above 2. Since this is all on one side of zero, we take the entire area above zero and subtract the area between zero and 2. 0228.4772..5.00.20022 zPzPzPzPnorm6251Distrex2 9/7/06Problem (vi) 00.705.1 zP This is pretty clearly one where we need to subtract. But what is the area (probability) between zero and 7?At the bottom of your table you will find the pair of columns copied here. normFor values above 3.09, see belowIf z0 is between 00 zzP is 3.08 and 3.10 .4990 3.11 and 3.13 .4991 3.14 and 3.17 .4992 3.18 and 3.21 .4993 3.22 and 3.26 .4994 3.27 and 3.32 .4995 3.33 and 3.38 .4996 3.39 and 3.48 .4997 3.49 and 3.61 .4998 3.62 and 3.89 .4999 3.90 and up .5000 So, for example, 4993.20.30 zP. But 7 is above 3.90, so 5000.00.70 zP and 1469.3531.5000.05.1000.7000.705.1 zPzPzP Data AxisDensity7.55.02.50.0-2.5-5.00.40.30.20.10.0Normal Curve with Mean 0 and Standard Deviation 1The Area Between 1.05 and 7 is 0.14697251Distrex2 9/7/06Problem (vii) 2zP Please shade the entire area above -2. If you have remembered to add a vertical line for zero, you will see that you have shaded an area on both sides of zero and are adding. 9772.5.4772.0022 zPzPzPnorm8251Distrex2 9/7/06Problem (viii α) 1F Remember that the capital F is a common notation for a
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