1 The most important continuous distribution:Gauss (normal) distribution Rehearsal

1

Rehearsal

2

x

µ is different

x

f(x)

σ is different The most important continuous distribution:

Gauss (normal) distribution

( )  





 



 



 



 −

−

=

2

2 exp 1 2

1

σ µ σ

π x x

f Two parameters: µ and σ

3

Expected value and variance:

( ) x = µ

E ^Var ( ) ^{x = σ}

Short notation:

( ^{µ , σ}

)

N ^{e.g. N} ( ) ^{0 , 1}

Standardisation:

σ µ

= x − z

( ) ⁰

µ = E z = ^σ

^{= Var z} ( ) ^{= 1}

( ) 





 

 −

= exp 2

2

1 z

z

f π

What is the probability of finding the x Gauss d. random variable in the ( ) range?

( µ − σ < x ≤ µ + σ ) = F ( µ + σ ) − F ( µ − σ )

P

σ µ σ µ − , +

µ µ +σ x µ −σ

P(x ≤ µ − σ )

P(x≤ µ + σ )

5

= 1

−

= +

σ µ σ µ

upper

z

− 1

− =

= −

σ µ σ µ

lower

z

Width of the interval

± σ ± 2 σ ± 3 σ

P 0.68268 0.9545 0.9973

6

The variance of a measurement isσ²=0.25g². The measurement is unbiased. We measure an object, its weight is 10 g. In which range will be the outcome of the measurement with 95% probability?

µ ₀

LCL UCL x

α ^/2 α ^/2

x_lower m=10 x_upper

1-α=0.95% α=0.05

α/2=0.025 α/2=0.025

7

The variance of a measurement isσ²=0.25g². The measurement is unbiased. We measure an object, its weight is 10 g. In which range will be the outcome of the measurement with 95% probability?

(

)

P −z_α < ≤z z_α =

(

) ¹

P

µ −

σ

< ≤ +

µ

σ

= − α

/ 2

x

0.95

P z

µ z

σ

−

 

− < ≤ =

 

 

(

) ^0.95

P x < ≤ x x =

This is the question:

This is what we know from the distribution function:

Connection:

x

z µ

σ

= −

( 10 1.96 0.5 10 1.96 0.5 ) 0.95

P

− ⋅ < ≤ +

⋅ =

lower

x x

Central limit theorem

2 1

( , )

N i i

x N N µ σ N

∑

∼

( ,

) x ∼ N µ σ N

The mean of sample elements taken from any

distribution approximately follows Gauss d. around the expected value of the original d. with variance σ

^/n.

Sum as well

z x

n µ σ

= − Based on the Central Limit Theorem:

The approximation steadily improves as the number of

observations increases.

9

Calculate the 95% probability interval for the mean of a n = 5 sample taken from a population ofµ=10 andσ²=0.25 !

(

)

P −z_α < ≤z z_α =

(

) ¹

P

µ −

σ

< ≤ +

µ

σ

= − α

/ 2 x / 2

0.95

P z z

α

µ

σ



− < − ≤

=

 

 

(

) ^0.95

P x < ≤ x x =

This is the question:

This is what we know from the distribution function:

Connection:

x z

n µ σ

= −

( ^{10 − 1 . 96 ⋅ 0 . 5 5 <}

^{≤ 10 + 1 . 96 ⋅ 0 . 5 5} ) ^{= 0 . 95}

P

lower

x x

10

The size of the produced parts (from a particular process) follows normal distribution. The expected value is 10, the variance is 0.25.

The lower specification limit is 8.5 cm. What is the ratio of noncomforming parts from this process?