Normal Distribution

TUM School of Management Production and Supply Chain Management Prof Martin Grunow Technische Universitt Mnchen

Areas Under any Normal Distribution

Holly Ott

( )2If ( , ), then 0 1X X N ~ N ,

:

21

Theorem: ~ According to this theorem, if a random variable is normally distributed, then a function of it, (X )/, has the standard normal distribution.

Quality Engineering & Management Module 3


-

0.0020

0.0040

0.0060

0.0080

0.0100

0.0120

0.0140

0.0160

0.0180

0 25 50 75 100 125 150 175 200

0

0.002

0.004

0.006

0.008

0.01

0.012

0.014

0.016

0.018

-100 -75 -50 -25 0 25 50 75 100

0 .00

0 .05

0 .10

0 .15

0 .20

0 .25

0 .30

0 .35

0 .40

0 .45

-4 -3 -2 -1 0 1 2 3 4

Converting between Normal distributions Start with = 100 = 25 X = 125

Center the distribution over 0 by subtracting the mean

Rescale the x and y axes by dividing by the standard deviation

z = X !!

=125!10025

=1

Holly Ott 22 Quality Engineering & Management Module 3


A random variable X ~ N(2.0, 0.0025). (a) P(X 1.87) (b) P(X > 2.2)


= P X ! 2.00.05 "1.87! 2.00.05

#

$%

&

'( ( ) 0047.06.2P05.0

13.0P ==

= ZZ

= P Z > 2.2! 2.00.05

"

#$

%

&'= P(Z > 4.0) = 0.0



(c) Find P(1.9 X 2.1) (d) Find t such that P(X t) = 0.05


P Z ! t " 20.05#

$%

&

'(= 0.05 ! t " 20.05 = "1.645! t =1.918

( ) 1.9 2.0 2.1 2.0P 1.9 2.1 P0.05 0.05

0.972 0.0228 0.9544

X Z =

= =



D is the diameter of bolts (inches), and, D ~ N(0.25, 0.012) Bolt specs call for 0.24 0.02 inches. What proportion of the bolts are outside specification? We need:

i.e., 16% of the bolts are outside specification.

Application of Normal Distribution

( ) ( )P 0 22 P 0 26D < . D .+ >

0.22 0.25 0.26 0.25P P0.01 0.01

( 3) ( 1)0.00135 0.1587 0.16

Z Z

P Z P Z

= < + >

= < + >

= + =



b) If the process mean is moved to coincide with the center of spec, what proportion will be defective?

When process mean coincides with spec center: P(D < 0.22) + P(D > 0.26) i.e., 4.56% will be outside specification

Centering a process will improve process conditions; further

improvement has to come from reducing variability.

Centering Improves a Process

0.22 0.24 0.26 0.24P P0.01 0.01

2 0.0228 0.0456

Z Z = < + >

= =



X is the random variable that denotes the life of batteries in years. X ~ N(5, 0.25)

a) If any battery failing before 4 years is replaced under warranty what proportion of batteries need replacement?

i.e., 2.28% will have to be replaced during warranty.

Reducing Variability

( ) ( )4 5P X 4 P P 2.0 0.02280.5

Z Z < = < = < =



b) What should be the standard deviation if no more than 0.5% should require replacement?

Let be the new standard deviation. Find such that P(X


Coming Up

Lecture 3.2: The Central Limit Theorem



Practice

Now let's do a practice problem to apply the normal distribution.

Please complete the next "Practice" module before continuing with Lecture 3.2.


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Normal Distribution