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CH 06 Sec 2

The z score tells the number of standard deviations a measurement is from the mean. It is calculated by taking the measurement minus the mean and dividing by the standard deviation. The document provides examples of calculating z scores and using a standard normal distribution table to find probabilities associated with various z score ranges.

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0% found this document useful (0 votes)
60 views20 pages

CH 06 Sec 2

The z score tells the number of standard deviations a measurement is from the mean. It is calculated by taking the measurement minus the mean and dividing by the standard deviation. The document provides examples of calculating z scores and using a standard normal distribution table to find probabilities associated with various z score ranges.

Uploaded by

ahsan ali
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PPT, PDF, TXT or read online on Scribd
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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.

x   21  25
z    2.00
 2
Calculating z-scores
Mean delivery time = 25 minutes
Standard deviation = 2 minutes
Convert 29.7 minutes to a z score.

x   29 .7  25
z   2.35
 2
Interpreting z-scores
Mean delivery time = 25 minutes
Standard deviation = 2 minutes
Interpret a z score of 1.6.

x  z     1 .6( 2 )  25  28 .2
The delivery time is 28.2 minutes.
Standard Normal Distribution:

 =0

 =1
-1 0 1

Values are converted to


x z =
z scores where

Importance of the Standard
Normal Distribution:
Standard
Normal
Distribution:

Any Normal 0 1
Distribution:
Areas will be equal.

 1
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.
Use of the Normal Probability
Table

(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
Curve
To find the area to the left of a given
negative z :
Use Table 5 (Appendix II) directly.

z 0
Patterns for Finding Areas
Under the Standard Normal
Curve
To find the area to the left of a given positive
z:
Use Table 5 a (Appendix II) directly.

0 z
Patterns for Finding Areas
Under the Standard Normal
Curve
To find the area between z values on either
side of zero:
Subtract area to left of z1 from area to left
of z2 .

z1 0 z2
Patterns for Finding Areas
Under the Standard Normal
Curve
To find the area between z values on the
same side of zero:
Subtract area to left of z1 from area to left
of z2 .

0 z1 z2
Patterns for Finding Areas
Under the Standard Normal
Curve
To 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.
Area under
entire curve
= 1.000.

0 z
Use of the Normal Probability
Table

a. .8925
P(z < 1.24) = ______

b. .4452
P(0 < z < 1.60) = _______

c. .4911
P( - 2.37 < z < 0) = ______
Normal Probability

d. .9974
P( - 3 < z < 3 ) = ________

e. .9322
P( - 2.34 < z < 1.57 ) = _____

f. .0774
P( 1.24 < z < 1.88 ) = _______
Normal Probability

g. .2254
P( - 2.44 < z < - 0.73 ) = _______

h. .9495
P( z < 1.64 ) = __________

i. .0084
P( z > 2.39 ) = _________
Normal Probability

j. .9236
P ( z > - 1.43 ) = __________

k. .0034
P( z < - 2.71 ) = __________
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. .3413


a. ___________

b. less than 30 minutes. .9938


b. __________

c. less than 22.7 minutes. .1251


c. __________

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