033 normal distribution

7/26/2019 033 Normal Distribution

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Normal Distribution“Awareness Program”

Dr. Att ia Hussi en Gom aa

Maintenance Engin eer ing Consultant

2008

Engin eer ing service - American Univers i ty in Cairo (AUC)



Data Measurement

Lower SpecificationLimit

Upper SpecificationLimit

Histogram Plot

0$ 500$

N o .

o f M

i n o r M a i n t . W O s



The Normal Distribution

By far the most important and widely used density curveis that describing the normal distribution. Thenormal distribution is the “bell curve” that is found in

many settings.

A normal distribution is symmetric, bell-shaped, and iscompletely defined by its mean, , and its standarddeviation, .

f( x) 1

2 e

1

2

x

2



Mean & Standard Deviation

From:http://www.gifted.uconn.edu/siegle/research/Normal/instructornotes.html



N(10,2)

N(12,2)

N(10,4)



Process A B C

Mean 10 10 10

Range 6-14 8-12 9-11Variance 4 1 0.25

Sigma 2 1 0.50



99.7%

95%

68%

2 3 2 3

The 68-95-99.7 Rule

N( , )



Suppose that the prices of new homes in Wakecounty are described well by a normaldistribution with mean $170,000 and standarddeviation $20,000. What percentage of housescost between $150,000 and $190,000?

150 190170

68%



Suppose that the prices of new machines aredescribed well by a normal distribution withmean $170,000 and standard deviation$20,000. What percentage of machine cost lessthan $210,000?

170 210

50% 47.5%

50%+47.5%=97.5%



Standardizing Observations: z-scores

The universality of the 68-95-99.7 Rule points out thefact that all normal distributions share certainproperties. It turns out that if we measure in units ofstandard deviations, , all normal distributions are the

same. We standardize a value by calculating a z-score telling us how many standard deviation unitsaway from the mean, , it is:

z x



Suppose that the prices of new machines aredescribed well by a normal distribution withmean $170,000 and standard deviation$20,000. What percentage of machine costbetween $150,000 and $190,000?

For $150,000: z

150,000 170,000

20,000 1

For $190,000: z

190,000 170,000

20,0001



The Standard Normal Distribution

The N(0,1) distribution is called the standard normaldistribution. If a variable X has a N(, ) distribution,then the standardized variable

has a N(0,1) distribution. This property allows us tocarry out virtually all calculations involving normaldistributions by using the N(0,1) distribution.

Z X



Find the area under the standard normal density curveleft of -1.54.

-1.54

Answer: 0.0630



Weighting times on a customer service phone linefollow a normal distribution with mean 8 minutes andstandard deviation 2 minutes. What percentage of

callers wait more than 11 minutes?

z 1110

2

0.5

The area right of 11 is equal to 1minus the area left of 11, which isequal to 1 minus the area left of

0.5 under the N(0,1) curve. UsingTable A we find:

1 - 0.6915 = 0.3085



Calculations using the standard normal distribution fallinto one of three categories:

1. L(z) = Area left of z. Directly from Table A.

2. R(z) = Area right of z. R(z) = 1 - L(z)

3. B(zLo, zHi) = Area between zLo and zHi.

B(zLo, zHi) = L(zHi) - L(zLo)



Weighting times on a customer service phone linefollow a normal distribution with mean 8 minutes andstandard deviation 2 minutes. Above what length are

the longest 25% of the calls?

0.25

?

Using Table A, we find that the z-scorecorresponding to R(z) = 0.25 is z=0.675.We can now invert the standardizationFormula to find the call length:

0.675 x 10

2

x 2 0.675 10 11.35



Areas and Probabilities

• Cumulative probability:)()( a X pa F

320-1-3

Z

p r o b a b i l i t y

d e n s i t y

Normal CurveCumulativ e Probabili ty

a=X

)()(1 X a pa F



Areas and Probabilities (2)

• Probability of an Interval

43210-1-2-3-4Z

p r o b a b i l i t y

d e n s i t y

Normal CurveInterval Probability

)21()1()2( X p F F

033 normal distribution

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