1. Section 6.4 Finding z-Values Using the Normal Curve ( with enhancements by D.R.S. ) HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2008 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved.

2. HAWKES LEARNING SYSTEMS math courseware specialists What z-value has an area of 0.7357 to its left? Continuous Random Variables 6.4 Finding z-Values Using the Normal Curve Standard Normal Distribution Table from –  to positive z z0.000.010.020.030.04 0.00.50000.50400.50800.51200.5160 0.10.53980.54380.54780.55170.5557 0.20.57930.58320.58710.59100.5948 0.30.61790.62170.62550.62930.6331 0.40.65540.65910.66280.66640.6700 0.50.69150.69500.69850.70190.7054 0.60.72570.72910.73240.73570.7389 0.70.75800.76110.76420.76730.7704 0.80.78810.79100.79390.79670.7995 z  0.63 Standard Normal Distribution Table from –  to positive z z0.000.010.020.030.04 0.00.50000.50400.50800.51200.5160 0.10.53980.54380.54780.55170.5557 0.20.57930.58320.58710.59100.5948 0.30.61790.62170.62550.62930.6331 0.40.65540.65910.66280.66640.6700 0.50.69150.69500.69850.70190.7054 0.60.72570.72910.73240.73570.7389 0.70.75800.76110.76420.76730.7704 0.80.78810.79100.79390.79670.7995

3. HAWKES LEARNING SYSTEMS math courseware specialists TI-84 Plus Instructions: 1.Press 2 nd, then VARS 2.Choose 3: invNorm( 3.The format for entering the statistics is invNorm(area) In the previous example we could have entered invNorm(0.7357). Calculator result:.6301445679 Continuous Random Variables 6.4 Finding z-Values Using the Normal Curve You tell invNorm an area invNorm tells you the z value that has that amount of area to its left

4. HAWKES LEARNING SYSTEMS math courseware specialists What z-value has an area of 0.2000 to its left? Continuous Random Variables 6.4 Finding z-Values Using the Normal Curve z   0.84 Standard Normal Distribution Table from –  to positive z z0.080.070.060.050.04  1.1 0.11900.12100.12300.12510.1271  1.0 0.14010.14230.14460.14690.1492  0.9 0.16350.16600.16850.17110.1736  0.8 0.18940.19220.19490.19770.2005  0.7 0.21770.22060.22360.22660.2296 Standard Normal Distribution Table from –  to positive z z0.080.070.060.050.04  1.1 0.11900.12100.12300.12510.1271  1.0 0.14010.14230.14460.14690.1492  0.9 0.16350.16600.16850.17110.1736  0.8 0.18940.19220.19490.19770.2005  0.7 0.21770.22060.22360.22660.2296 Using the tables, the area we are looking for falls between 0.1977 and 0.2005. [.2005 is closer to.2000 than is.1977] We will use the area closest to the one we want. [so use.2005]

5. HAWKES LEARNING SYSTEMS math courseware specialists What z-value has an area of 0.2000 to its left? Continuous Random Variables 6.4 Finding z-Values Using the Normal Curve z   0.84 TI-84: invNorm(0.2000)= -.8416212335 Round to -0.84 This time it agreed with the table result. But again, TI-84 will often be more precise.

6. HAWKES LEARNING SYSTEMS math courseware specialists What z-value has an area of 0.0096 to its right? Continuous Random Variables 6.4 Finding z-Values Using the Normal Curve z  2.34 Standard Normal Distribution Table from –  to positive z z0.080.070.060.050.04  2.7 0.00270.00280.00290.00300.0031  2.6 0.00370.00380.00390.00400.0041  2.5 0.00490.00510.00520.00540.0055  2.4 0.00660.00680.00690.00710.0073  2.3 0.00870.00890.00910.00940.0096  2.2 0.01130.01160.01190.01220.0125 Standard Normal Distribution Table from –  to positive z z0.080.070.060.050.04  2.7 0.00270.00280.00290.00300.0031  2.6 0.00370.00380.00390.00400.0041  2.5 0.00490.00510.00520.00540.0055  2.4 0.00660.00680.00690.00710.0073  2.3 0.00870.00890.00910.00940.0096  2.2 0.01130.01160.01190.01220.0125 z   2.34, however the table assumes that the area is to the left of z. Since the standard normal curve is symmetric, we can simply change the sign of the z-value to obtain the correct answer.

7. HAWKES LEARNING SYSTEMS math courseware specialists TI-84 Plus Instructions: 1.Press 2 nd, then VARS 2.Choose 3: invNorm( 3.The format for entering the statistics is invNorm(1  area) In the previous example we could have entered invNorm(1  0.0096). Continuous Random Variables 6.4 Finding z-Values Using the Normal Curve

8. HAWKES LEARNING SYSTEMS math courseware specialists What z-value has an area of 0.0096 to its right? Continuous Random Variables 6.4 Finding z-Values Using the Normal Curve TI-84: invNorm(0.0096)= - 2.341624909 alternatively: if area to RIGHT is 0.0096, Then area to LEFT is 1 – 0.0096 = 0.9904 TI-84: invNorm(0.9904)= 2.341624909 Strongly suggested: DRAW A PICTURE to help you get the signs right !!!

9. HAWKES LEARNING SYSTEMS math courseware specialists Find the value of z such that the area between –z and z is 0.90. Continuous Random Variables 6.4 Finding z-Values Using the Normal Curve z  1.645. Standard Normal Distribution Table from –  to positive z z0.080.070.060.050.04  1.8 0.03010.03070.03140.03220.0329  1.7 0.03750.03840.03920.04010.0409  1.6 0.04650.04750.04850.04950.0505  1.5 0.05710.05820.05940.06060.0618  1.4 0.06940.07080.07210.07350.0749 Standard Normal Distribution Table from –  to positive z z0.080.070.060.050.04  1.8 0.03010.03070.03140.03220.0329  1.7 0.03750.03840.03920.04010.0409  1.6 0.04650.04750.04850.04950.0505  1.5 0.05710.05820.05940.06060.0618  1.4 0.06940.07080.07210.07350.0749 Since 0.05 value exactly between –1.64 and –1.65 we have If the area between –z and z is 0.90, then the area in the tails would be 1 – 0.90 = 0.10. Because of symmetry each tail will only have half of 0.10 in its area, 0.05. DRAW A PICTURE !!!!!

10. HAWKES LEARNING SYSTEMS math courseware specialists Find the value of z such that the area between –z and z is 0.90. Continuous Random Variables 6.4 Finding z-Values Using the Normal Curve z = 1.645 turns out to be same as the table answer this time. If the area between –z and z is 0.90, then the area in the tails would be 1 – 0.90 = 0.10. Because of symmetry each tail will only have half of 0.10 in its area, 0.05. THAT REASONING MUST BE DONE WITH TI-84 SOLUTION, TOO !!! DRAWING A PICTURE IS STRONGLY RECOMMENDED !!! TI-84: invNorm(0.0500)= -1.644853626 invNorm(0.9500)= 1.644853626

11. HAWKES LEARNING SYSTEMS math courseware specialists TI-84 Plus Instructions: 1.First find 1 – area between –z and z 2.Divide answer in step 1 by two 3.Press 2 nd, then VARS 4.Choose 3: invNorm( 5.The format for entering the statistics is invNorm(step 2 answer) 6.Take the absolute value of the answer from step 5 Continuous Random Variables 6.4 Finding z-Values Using the Normal Curve * DRAW A PICTURE AND DO IT AS ON THE PREVIOUS SLIDE !!!!!!!! These steps will work but don’t memorize steps like this! Instead, be able to draw a picture and reason it out.

12. HAWKES LEARNING SYSTEMS math courseware specialists Find the value of z such that the area to the left of –z plus the area to the right of z is 0.1616. Continuous Random Variables 6.4 Finding z-Values Using the Normal Curve Standard Normal Distribution Table from –  to positive z z0.040.030.020.010.00  1.8 0.03290.03360.03440.03510.0359  1.7 0.04090.04180.04270.04360.0446  1.6 0.05050.05160.05260.05370.0548  1.5 0.06180.06300.06430.06550.0668  1.4 0.07490.07640.07780.07930.0808 Standard Normal Distribution Table from –  to positive z z0.040.030.020.010.00  1.8 0.03290.03360.03440.03510.0359  1.7 0.04090.04180.04270.04360.0446  1.6 0.05050.05160.05260.05370.0548  1.5 0.06180.06300.06430.06550.0668  1.4 0.07490.07640.07780.07930.0808 The corresponding z-value is –1.40. Then the z-value such that the area is to the left of –z plus the area to the right of z is 0.1616 is z  1.40. If the area in both tails is 0.1616, then the area in one tail would be 0.0808.

13. HAWKES LEARNING SYSTEMS math courseware specialists TI-84 Plus Instructions: 1.First divide the area in the tails by two 2.Press 2 nd, then VARS 3.Choose 3: invNorm( 4.The format for entering the statistics is invNorm(step 1 answer) 5.Take the absolute value of the answer from step 4 Continuous Random Variables 6.4 Finding z-Values Using the Normal Curve Again, don’t memorize steps! Draw a picture and reason it out! TI-84: invNorm(0.0808)= -1.399710595 Take positive and round: z=1.40

14. HAWKES LEARNING SYSTEMS math courseware specialists What z-value represents the 90 th percentile? Continuous Random Variables 6.4 Finding z-Values Using the Normal Curve The 90 th percentile is the z-value for which 90% of the area under the standard normal curve is to the left of z. We will look for 0. 0900 in the tables, or 0.8997, which is extremeley close to 0.9000. Doing so we find z  1.28. Thus z  1.28 represents the 90 th percentile. TI-84: invNorm(0.9000)= 1.281551567 rounded: z=1.28

15. HAWKES LEARNING SYSTEMS math courseware specialists Determine the following: Continuous Random Variables 6.4 Finding z-Values Using the Normal Curve The body temperatures of adults are normally distributed with a mean of 98.6° F and a standard deviation of 0.73° F. What temperature represents the 90 th percentile? Solution: To determine the temperature that represents the 90 th percentile, we first need to find the z-value that represents the 90 th percentile. Once we have the z-value we can substitute z, , and  into the standard score formula and solve for x. From the previous example, we found z  1.28,  98.6,  0.73. THIS IS AN x PROBLEM !!! Convert it to a z problem and find area !!!

16. HAWKES LEARNING SYSTEMS math courseware specialists Determine the following: Continuous Random Variables 6.4 Finding z-Values Using the Normal Curve The body temperatures of adults are normally distributed with a mean of 98.6° F and a standard deviation of 0.73° F. What temperature represents the 90 th percentile? TI-84 SHORTCUT Solution – You need this background information: TI-84 can do it as an x problem, so you don’t have to convert it to a z problem. For z problems (you already know this): invNorm(area to left)= z answer For x problems (this is new): invNorm(area to left, mean, stdev) = x answer

17. HAWKES LEARNING SYSTEMS math courseware specialists Determine the following: Continuous Random Variables 6.4 Finding z-Values Using the Normal Curve The body temperatures of adults are normally distributed with a mean of 98.6° F and a standard deviation of 0.73° F. What temperature represents the 90 th percentile? TI-84 SHORTCUT Solution: You still need to realize that it’s 90% of area to the left. invNorm(area to left, mean, stdev) = x answer invNorm(0.9000, 98.6, 0.73) = 99.53553264