Transcript Document

Chapter Six Normal Curves and Sampling Probability Distributions

Chapter 6 Section 2 Standard Units and Areas Under the Standard Normal Distribution

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.

z

Formula for z score 

x

  

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.

z z z z

x

    21  2   4 2   2.00

25

Calculating z-scores Mean delivery time = 25 minutes Standard deviation = 2 minutes Convert 29.7 minutes to a z score.

z

x

   29.7

 25

z

z z

 4.7

2  2.35

2

Interpreting z-scores Mean delivery time = 25 minutes Standard deviation = 2 minutes Interpret a z score of 1.6.

The delivery time is 28.2 minutes.

z

 1.6

3.2

x

  

x

 

x

25  2 25 28.2

x

Standard Normal Distribution: μ

= 0

σ

= 1

 2  1 1 2 3 4  4  3

Values are converted to z scores where

z

x

  

Importance of the Standard Normal Distribution: Standard Normal Distribution: Any Normal Distribution: 0 1 The areas are equal.

 1 

Use of the Normal Probability Table

(Table 4) - Appendix I Entries give the probability that a standard normally distributed random variable will assume a value between the mean (zero) and a given z-score.

z To find the area between z = 0 and z = 1.34

0.00

0.01

Z-Scores

0.02

0.03

0.04

0.05

0.06

1.1

1.2

1.3

1.4

0.3643

0.3849

0.4032

0.4192

0.3665

0.3869

0.4049

0.4207

0.3686

0.3888

0.4066

0.4222

0.3708

0.3907

0.4082

0.4236

0.3729

0.2925

0.4099

0.4251

0.3749

0.3944

0.4115

0.4365

0.3770

0.3962

0.4131

0.4279

Patterns for Finding Areas Under the Standard Normal Curve To find the area between a given z value and zero: Use Table 4 (Appendix I) directly.

0 z

Patterns for Finding Areas Under the Standard Normal Curve To find the area between z values on either side of zero: Add area from z 1 to zero to area from zero to z 2 .

z 1 0 z 2

Patterns for Finding Areas Under the Standard Normal Curve To find the area between z values on the same side of zero: Subtract area from zero to z 1 the area from zero to z 2 .

from 0 z 1 z 2

Patterns for Finding Areas Under the Standard Normal Curve To find the area to the right of a positive z value or to the left of a negative z value: Subtract the area from zero to z from 0.5000 .

0.5000

0 z

Patterns for Finding Areas Under the Standard Normal Curve To find the area to the left of a positive z value or to the right of a negative z value: Add 0.5000 to the area from zero to z .

0.5000

z table 0

Use of the Normal Probability Table

Normal Probability

Normal Probability

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: b. less than 30 minutes.

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.

P

  25 2 25 

z

 27  2 25  

P P

  0  2 0  

z z

  2 2 1    0.3413

0 1

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: b.

less than 30 minutes.

P z

 30  2 25   

P z

  5 2   2.5

 0.5000

 0.4938

0.9938

2.5

c.

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: less than 22.7 minutes.

P z

 22.7

 2 25   

P z

  2.3

2     1.15

 0.5000

 0.3749

0.1251

-1.15

Homework Assignments

Chapter 6 Section 2 Pages 274 - 276 Exercises: 1 - 49, odd Exercises: 2 - 50, even