section 6.4 finding z-values using the normal curve ( with enhancements by d.r.s. ) hawkes learning...

Section 6.4Finding z-Values Using the

Normal Curve(with enhancements by D.R.S.)

HAWKES LEARNING SYSTEMS

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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


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



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



Curve

You tell invNorm an area

invNorm tells you the z value that has that amount of area to its left






Curve

z -0.84


-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


-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]






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.



What z-value has an area of 0.0096 to its right?



Curve

z = 2.34


-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


-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.




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).



Curve



What z-value has an area of 0.0096 to its right?



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 !!!



Find the value of z such that the area between –z and z is 0.90.



Curve

z = 1.645.


-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


-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 !!!!!



Find the value of z such that the area between –z and z is 0.90.



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




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



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.



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.



Curve


-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


-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.




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



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



What z-value represents the 90th percentile?



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



Determine the following:



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 !!!






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






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

section 6.4 finding z-values using the normal curve ( with enhancements by d.r.s. ) hawkes learning...

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