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.

HAWKES LEARNING SYSTEMS math courseware specialists What z-value has an area of to its left? Continuous Random Variables 6.4 Finding z-Values Using the Normal Curve Standard Normal Distribution Table from –  to positive z z z  0.63 Standard Normal Distribution Table from –  to positive z z

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: 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

HAWKES LEARNING SYSTEMS math courseware specialists What z-value has an area of 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 z      Standard Normal Distribution Table from –  to positive z z      Using the tables, the area we are looking for falls between and [.2005 is closer to.2000 than is.1977] We will use the area closest to the one we want. [so use.2005]

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

HAWKES LEARNING SYSTEMS math courseware specialists What z-value has an area of 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 z       Standard Normal Distribution Table from –  to positive z z       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.

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  ). Continuous Random Variables 6.4 Finding z-Values Using the Normal Curve

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

HAWKES LEARNING SYSTEMS math courseware specialists Find the value of z such that the area between –z and z is Continuous Random Variables 6.4 Finding z-Values Using the Normal Curve z  Standard Normal Distribution Table from –  to positive z z      Standard Normal Distribution Table from –  to positive z z      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 = Because of symmetry each tail will only have half of 0.10 in its area, DRAW A PICTURE !!!!!

HAWKES LEARNING SYSTEMS math courseware specialists Find the value of z such that the area between –z and z is Continuous Random Variables 6.4 Finding z-Values Using the Normal Curve z = 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 = Because of symmetry each tail will only have half of 0.10 in its area, THAT REASONING MUST BE DONE WITH TI-84 SOLUTION, TOO !!! DRAWING A PICTURE IS STRONGLY RECOMMENDED !!! TI-84: invNorm(0.0500)= invNorm(0.9500)=

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.

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 Continuous Random Variables 6.4 Finding z-Values Using the Normal Curve Standard Normal Distribution Table from –  to positive z z      Standard Normal Distribution Table from –  to positive z z      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 is z  If the area in both tails is , then the area in one tail would be

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)= Take positive and round: z=1.40

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 in the tables, or , which is extremeley close to Doing so we find z  Thus z  1.28 represents the 90 th percentile. TI-84: invNorm(0.9000)= rounded: z=1.28

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,  THIS IS AN x PROBLEM !!! Convert it to a z problem and find area !!!

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

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) =