copyright (c) 2002 houghton mifflin company. all rights reserved. 1 z score the z value or z score...

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 1

Z Score

• The z value or z score tells the number of standard deviations the original measurement is from the mean.

• The z value is in standard units.


Formula for z score

x

z


Calculating z-scores

The amount of time it takes for a pizza delivery is approximately normally distributed with a mean of 25 minutes and a standard deviation of 2 minutes. Convert 21 minutes to a z score.

00.22

2521xz


Calculating z-scores

Mean delivery time = 25 minutes Standard deviation = 2 minutes

Convert 29.7 minutes to a z score.

35.22

257.29xz


Interpreting z-scores

Mean delivery time = 25 minutes Standard deviation = 2 minutes

Interpret a z score of 1.6.

2.2825)2(6.1zx The delivery time is 28.2 minutes.


Standard Normal Distribution:

= 0

= 1

x

-1 1

Values are converted to

z scores where

0


Importance of the Standard Normal Distribution:

10

1

Areas will be equal.

Any Normal Distribution:

Standard Normal Distribution:


Use of the Normal Probability Table

(Table 5) - Appendix II

Entries give the probability that a standard normally distributed random variable will assume a

value to the left of a given negative z-score.



(Table 5a) - Appendix II

Entries give the probability that a standard normally distributed random variable will assume a

value to the left of a given positive z value.


To find the area to the left of z = 1.34

_____________________________________z… 0.03 0.04 0.05 ..…

_____________________________________ . .

1.2 … .8907 .8925 .8944 ….

1.3 … .9082 .9099 .9115 ….

1.4 … .9236 .9251 .9265 ….

.


Patterns for Finding Areas Under the Standard Normal

CurveTo find the area to the left of a given

negative z :

Use Table 5 (Appendix II) directly.

z 0



CurveTo find the area to the left of a given

positive z :

Use Table 5 a (Appendix II) directly.

z0



CurveTo find the area between z values

Subtract area to left of z1 from area to left of z2 .

z20z1



CurveTo find the area between z values, subtract

area to left of z1 from area to left of z2 .

z20 z1



CurveTo find the area to the right of a positive z

value or to the right of a negative z value:

Subtract from 1.0000 the area to the left of the given z.

z0

Area under entire curve = 1.000.



a. P(z < 1.24) = ______

b. P(0 < z < 1.60) = _______

c. P( - 2.37 < z < 0) = ______

.8925

.4452

.4911


Normal Probability

d. P( - 3 < z < 3 ) = ________

e. P( - 2.34 < z < 1.57 ) = _____

f. P( 1.24 < z < 1.88 ) = _______

.9974

.9322

.0774


Normal Probability

g. P( - 2.44 < z < - 0.73 ) = _______

h. P( z < 1.64 ) = __________

i . P( z > 2.39 ) = _________

.9495

.0084

.2254


Normal Probability

j. P ( z > - 1.43 ) = __________

k. P( z < - 2.71 ) = __________

.9236

.0034


Application of the Normal Curve

The amount of time it takes for a pizza delivery is approximately normally distributed with a mean of 25 minutes and a standard deviation of 2 minutes. If you order a pizza, find the probability that the delivery time will be:

a. between 25 and 27 minutes. a. ___________

b. less than 30 minutes. b. __________

c. less than 22.7 minutes. c. __________

.3413

.9938

.1251


Inverse Normal Distribution

Finding z scores when probabilities (areas) are given


Find the indicated z score:


.8907

0 z = 1.23



.6331

z = – 0.34

.3669



.3560

0 z = 1.06

.8560



.4792

z = 0– 2.04

.0208



0 z =

.4900

2.33



z = 0

.005

– 2.575



If area A + area B = .01, z = __________

A B

– z 0 z

2.575 or 2.58

= .005


Application of Determining z Scores

The Verbal SAT test has a mean score of 500 and a standard deviation of 100. Scores are normally distributed. A

major university determines that it will accept only students whose Verbal SAT scores are in the top 4%. What is the

minimum score that a student must earn to be accepted?


...students whose Verbal SAT scores are in the top 4%.

Mean = 500, standard deviation = 100

= .04.9600

z = 1.75

The cut-off score is 1.75 standard deviations above the mean.


Application of Determining z Scores

Mean = 500, standard deviation = 100

= .04.9600

z = 1.75

The cut-off score is 500 + 1.75(100) = 675.

copyright (c) 2002 houghton mifflin company. all rights reserved. 1 z score the z value or z score...

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