section 6.4 finding z-values using the normal curve ( with enhancements by d.r.s. ) hawkes learning...
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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 leftTRANSCRIPT
Section 6.4Finding 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 0.7357 to its left?
Continuous Random Variables
6.4 Finding z-Values Using the Normal
Curve
Standard Normal Distribution Table from – to positive zz 0.00 0.01 0.02 0.03 0.04
0.0 0.5000 0.5040 0.5080 0.5120 0.5160
0.1 0.5398 0.5438 0.5478 0.5517 0.5557
0.2 0.5793 0.5832 0.5871 0.5910 0.5948
0.3 0.6179 0.6217 0.6255 0.6293 0.6331
0.4 0.6554 0.6591 0.6628 0.6664 0.6700
0.5 0.6915 0.6950 0.6985 0.7019 0.7054
0.6 0.7257 0.7291 0.7324 0.7357 0.7389
0.7 0.7580 0.7611 0.7642 0.7673 0.7704
0.8 0.7881 0.7910 0.7939 0.7967 0.7995
z = 0.63
Standard Normal Distribution Table from – to positive zz 0.00 0.01 0.02 0.03 0.04
0.0 0.5000 0.5040 0.5080 0.5120 0.5160
0.1 0.5398 0.5438 0.5478 0.5517 0.5557
0.2 0.5793 0.5832 0.5871 0.5910 0.5948
0.3 0.6179 0.6217 0.6255 0.6293 0.6331
0.4 0.6554 0.6591 0.6628 0.6664 0.6700
0.5 0.6915 0.6950 0.6985 0.7019 0.7054
0.6 0.7257 0.7291 0.7324 0.7357 0.7389
0.7 0.7580 0.7611 0.7642 0.7673 0.7704
0.8 0.7881 0.7910 0.7939 0.7967 0.7995
HAWKES LEARNING SYSTEMS
math courseware specialists
TI-84 Plus Instructions:
1. Press 2nd, then VARS2. 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
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 zz 0.08 0.07 0.06 0.05 0.04
-1.1 0.1190 0.1210 0.1230 0.1251 0.1271
-1.0 0.1401 0.1423 0.1446 0.1469 0.1492
-0.9 0.1635 0.1660 0.1685 0.1711 0.1736
-0.8 0.1894 0.1922 0.1949 0.1977 0.2005
-0.7 0.2177 0.2206 0.2236 0.2266 0.2296
Standard Normal Distribution Table from – to positive zz 0.08 0.07 0.06 0.05 0.04
-1.1 0.1190 0.1210 0.1230 0.1251 0.1271
-1.0 0.1401 0.1423 0.1446 0.1469 0.1492
-0.9 0.1635 0.1660 0.1685 0.1711 0.1736
-0.8 0.1894 0.1922 0.1949 0.1977 0.2005
-0.7 0.2177 0.2206 0.2236 0.2266 0.2296
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]
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.
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 zz 0.08 0.07 0.06 0.05 0.04
-2.7 0.0027 0.0028 0.0029 0.0030 0.0031
-2.6 0.0037 0.0038 0.0039 0.0040 0.0041
-2.5 0.0049 0.0051 0.0052 0.0054 0.0055
-2.4 0.0066 0.0068 0.0069 0.0071 0.0073
-2.3 0.0087 0.0089 0.0091 0.0094 0.0096
-2.2 0.0113 0.0116 0.0119 0.0122 0.0125
Standard Normal Distribution Table from – to positive zz 0.08 0.07 0.06 0.05 0.04
-2.7 0.0027 0.0028 0.0029 0.0030 0.0031
-2.6 0.0037 0.0038 0.0039 0.0040 0.0041
-2.5 0.0049 0.0051 0.0052 0.0054 0.0055
-2.4 0.0066 0.0068 0.0069 0.0071 0.0073
-2.3 0.0087 0.0089 0.0091 0.0094 0.0096
-2.2 0.0113 0.0116 0.0119 0.0122 0.0125
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.
HAWKES LEARNING SYSTEMS
math courseware specialists
TI-84 Plus Instructions:
1. Press 2nd, then VARS2. 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
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 PICTUREto 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 0.90.
Continuous Random Variables
6.4 Finding z-Values Using the Normal
Curve
z = 1.645.
Standard Normal Distribution Table from – to positive zz 0.08 0.07 0.06 0.05 0.04
-1.8 0.0301 0.0307 0.0314 0.0322 0.0329
-1.7 0.0375 0.0384 0.0392 0.0401 0.0409
-1.6 0.0465 0.0475 0.0485 0.0495 0.0505
-1.5 0.0571 0.0582 0.0594 0.0606 0.0618
-1.4 0.0694 0.0708 0.0721 0.0735 0.0749
Standard Normal Distribution Table from – to positive zz 0.08 0.07 0.06 0.05 0.04
-1.8 0.0301 0.0307 0.0314 0.0322 0.0329
-1.7 0.0375 0.0384 0.0392 0.0401 0.0409
-1.6 0.0465 0.0475 0.0485 0.0495 0.0505
-1.5 0.0571 0.0582 0.0594 0.0606 0.0618
-1.4 0.0694 0.0708 0.0721 0.0735 0.0749
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 !!!!!
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
HAWKES LEARNING SYSTEMS
math courseware specialists
TI-84 Plus Instructions:
1. First find 1 – area between –z and z2. Divide answer in step 1 by two3. Press 2nd, then VARS4. 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 0.1616.
Continuous Random Variables
6.4 Finding z-Values Using the Normal
Curve
Standard Normal Distribution Table from – to positive zz 0.04 0.03 0.02 0.01 0.00
-1.8 0.0329 0.0336 0.0344 0.0351 0.0359
-1.7 0.0409 0.0418 0.0427 0.0436 0.0446
-1.6 0.0505 0.0516 0.0526 0.0537 0.0548
-1.5 0.0618 0.0630 0.0643 0.0655 0.0668
-1.4 0.0749 0.0764 0.0778 0.0793 0.0808
Standard Normal Distribution Table from – to positive zz 0.04 0.03 0.02 0.01 0.00
-1.8 0.0329 0.0336 0.0344 0.0351 0.0359
-1.7 0.0409 0.0418 0.0427 0.0436 0.0446
-1.6 0.0505 0.0516 0.0526 0.0537 0.0548
-1.5 0.0618 0.0630 0.0643 0.0655 0.0668
-1.4 0.0749 0.0764 0.0778 0.0793 0.0808
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.
HAWKES LEARNING SYSTEMS
math courseware specialists
TI-84 Plus Instructions:
1. First divide the area in the tails by two2. Press 2nd, then VARS3. 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
CurveAgain, 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
HAWKES LEARNING SYSTEMS
math courseware specialists
What z-value represents the 90th percentile?
Continuous Random Variables
6.4 Finding z-Values Using the Normal
Curve
The 90th 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 90th percentile. TI-84: invNorm(0.9000)= 1.281551567
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 90th percentile? Solution:
To determine the temperature that represents the 90th percentile, we first need to find the z-value that represents the 90th 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 !!!
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 90th 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 90th 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