Weibull exercise

Lifetimes in years of an electric motor have a Weibull distribution with b = 3 and a = 5.

Find a guarantee time, g, such that 95% of the motors last more than g years. That is, find g so that P(X > g) = 0.95

What is called a linear combination of random variables?

aX+b, aX+bY+c, etc. 2X+3 5X+9Y+10

5.5.3 Means and Variances for Linear Combinations

If a and b are constants, then E(aX+b)=aE(X)+b.__________________________

• Corollary 1. E(b)=b. • Corollary 2. E(aX)=aE(X).

Let 𝑈= 𝑎0 + 𝑎1𝑋+ 𝑎2𝑌+ 𝑎3𝑍+ ⋯

Means of Linear Combinations

𝐸ሺ𝑈ሻ= 𝑎0 + 𝑎1𝐸𝑋+ 𝑎2𝐸𝑌+ 𝑎3𝐸𝑍+ ⋯

Example X, Y, Z = values of three dice You win 𝑈= 5+ 2𝑋+ 3𝑌+ 4𝑍 dollars 𝐸ሺ𝑋ሻ= 𝐸ሺ𝑌ሻ= 𝐸ሺ𝑍ሻ= 3.5 Your total expected winnings are 𝐸ሺ𝑈ሻ= 5+ 2ሺ3.5ሻ+ 3ሺ3.5ሻ+ 4ሺ3.5ሻ= 36.50

Expected value (constant) = constant 𝐸ሺ5ሻ= 5

𝐸ሺ∑ሻ= ∑ሺ𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑣𝑎𝑙𝑢𝑒𝑠ሻ 𝐸ሺ5+ 3𝑋+ 4𝑌+ 5𝑍ሻ= 𝐸ሺ5ሻ+ 𝐸ሺ3𝑋ሻ+ 𝐸ሺ4𝑌ሻ+ 𝐸ሺ5𝑍ሻ 𝐸ሺ𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡∙𝑋ሻ= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡∙𝐸𝑋 𝐸ሺ3𝑋ሻ= 3𝐸ሺ𝑋ሻ 𝐸ሺ4𝑌ሻ= 4𝐸ሺ𝑌ሻ Eሺ5𝑍ሻ= 5𝐸ሺ𝑍ሻ 𝐸ሺ𝑎0 + 𝑎1𝑋+ 𝑎2𝑌+ 𝑎3𝑍ሻ = 𝐸ሺ𝑎0ሻ+ 𝐸ሺ𝑎1𝑋ሻ+ 𝐸ሺ𝑎2𝑌ሻ+ 𝐸ሺ𝑎3𝑍ሻ = 𝑎0 + 𝑎1𝐸𝑋+ 𝑎2𝐸𝑌+ 𝑎3𝐸𝑍

Variances of Linear Combinations

𝑉𝑎𝑟ሺ𝑎0 + 𝑎1𝑋+ 𝑎2𝑌+ 𝑎3𝑍ሻ = 𝑎12𝑉𝑎𝑟ሺ𝑋ሻ+ 𝑎22𝑉𝑎𝑟ሺ𝑌ሻ+ 𝑎32𝑉𝑎𝑟ሺ𝑍ሻ when X, Y and Z are independent.

𝑉𝑎𝑟ሺ𝑎0ሻ= 0 𝑉𝑎𝑟ሺ𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡ሻ= 0

𝑉𝑎𝑟ሺ𝑎0 + 𝑋ሻ= 𝑉𝑎𝑟ሺ𝑋ሻ Adding a constant doesn’t change variance

𝑉𝑎𝑟ሺ𝑎1𝑋ሻ= 𝑎12𝑉𝑎𝑟𝑋 Variances are in units2.

𝑆𝑡.𝐷𝑒𝑣ሺ𝑎1𝑋ሻ= ȁ&𝑎1ȁ& 𝑆𝑡.𝐷𝑒𝑣ሺ𝑋ሻ

If X, Y, Z are independent or uncorrelated. 𝑉𝑎𝑟ሺ𝑋+ 𝑌+ 𝑍ሻ= 𝑉𝑎𝑟ሺ𝑋ሻ+ 𝑉𝑎𝑟ሺ𝑌ሻ+ 𝑉𝑎𝑟ሺ𝑍ሻ Putting these facts together 𝑉𝑎𝑟ሺ𝑎0 + 𝑎1𝑋+ 𝑎2𝑌+ 𝑎3𝑍ሻ = ሺif independentሻ 𝑉𝑎𝑟ሺ𝑎0ሻ+ 𝑉𝑎𝑟ሺ𝑎1𝑋ሻ+ 𝑉𝑎𝑟ሺ𝑎2𝑌ሻ+ 𝑉𝑎𝑟ሺ𝑎3𝑍ሻ = 𝑎12𝑉𝑎𝑟ሺ𝑋ሻ+ 𝑎22𝑉𝑎𝑟ሺ𝑌ሻ+ 𝑎32𝑉𝑎𝑟ሺ𝑍ሻ For independent X, Y

𝜎𝑋+𝑌 = ඥ𝑉𝑎𝑟ሺ𝑋ሻ+ 𝑉𝑎𝑟ሺ𝑌ሻ

𝜎𝑋+𝑌2 = 𝜎𝑋2 + 𝜎𝑌2

𝜎𝑋+𝑌 = ට𝜎𝑋2 + 𝜎𝑌2

Like Pythagorean Theorem 𝑐2 = ȁ2 + b2, 𝑐= ξ𝑎2 + 𝑏2

Suppose that “20 ohm” resistors have a mean of 20 ohms and a standard deviation of 0.2 ohms. Four independent resistors are to be chosen a connected in series.

R1 = resistance of resistor 1R2 = resistance of resistor 2R3 = resistance of resistor 3R4 = resistance of resistor 4

 The system’s resistance, S, is

 S = R1 + R2 + R3 + R4

 Find the mean and standard deviation for the system resistance, S.

Most useful facts:

If X1, X2, X3, … , Xn are independent measurements each with 𝐸(𝑋1) = 𝜇 and 𝑉𝑎𝑟ሺ𝑥𝑖ሻ= 𝜎2 then

𝐸ሺ𝑋തሻ= 𝜇 𝑉𝑎𝑟ሺ𝑋തሻ= 𝜎2𝑛 𝑆𝑡.𝐷𝑒𝑣ሺ𝑋തሻ= 𝜎ξ𝑛

To see this:

𝐸൬1𝑛 ∑ 𝑋𝑖൰= 1𝑛 𝐸ሺ∑ 𝑋𝑖ሻ= 1𝑛 ∑ 𝐸ሺ𝑋𝑖ሻ= 1𝑛 𝜇𝑛

𝑖=1 = 1𝑛ሺ𝑛𝜇ሻ= 𝜇

The sample mean 𝑋ത is an unbiased estimator of the population mean 𝜇.

𝑉𝑎𝑟൬1𝑛 ∑ 𝑋𝑖൰= ൬1𝑛൰2 𝑉𝑎𝑟ሺ∑ 𝑋𝑖ሻ= 1𝑛2 𝑉𝑎𝑟ሺ𝑋𝑖ሻ= 1𝑛2 𝜎2𝑛

𝑖=1 = 1𝑛2ሺ𝑛𝜎2ሻ= 𝜎2𝑛

5.5.4 Propagation of Error

• Often done in chemistry or physics labs.• In last section we talked about the case when

U=g(x) is linear in x

When U=g(X,Y,…Z) is not linear

• We can get useful approximations to the mean and variance of U.

• Using Taylor expansion of a nonlinear g,

Where the partial derivatives are evaluated at (x,y,…,z)=(EX, EY,…,EZ).

( , ,..., ) ( , ,..., ) ( ) ( ) ... ( )g g g

g x y z g EX EY EZ x EX y EY z EZx y z

The right hand side of the equation is linear in x,y,…z.

( , ,..., ) ( , ,..., ) ( ) ( ) ... ( )g g g

g x y z g EX EY EZ x EX y EY z EZx y z

For the mean E(U)

since

( , ,..., ) ( , ,..., ) ( ) ( ) ... ( )

( ) ( ( , ,..., )) ( ( , ,..., )) ( ) ( ) ... ( )

( , ,..., ) 0 0 ... 0 ( , ,..., )

g g gg x y z g EX EY EZ x EX y EY z EZ

x y z

g g gE U E g x y z E g EX EY EZ E x EX E y EY E z EZ

x y z

g EX EY EZ g EX EY EZ

( ) ( ) *0 0g g g

E x EX EX EXx x x

For Var(U)

2 2 2( ) ( ( , ,..., )) ( ) ( ) ... ( )g g g

Var U Var g x y z VarX VarY VarZx y z

Example

Example 24 on page 311The assembly resistance R is related to R1, R2 and R3 by

A large lot of resistors is manufactured and has a mean resistance of 100 Ω with a standard deviation of 2 Ω . If 3 resistors are taken at random , can you approximate mean and variance for the resulting assembly resistance?

𝐸ሺ𝑅𝑖ሻ= 100Ω 𝑆𝑡.𝐷𝑒𝑣ሺ𝑅𝑖ሻ= 2Ω 𝑉𝑎𝑟ሺ𝑅𝑖ሻ= 4Ω2

𝐸ሺ𝑅ሻ≈ 100+ቀ1100 + 1100ቁ−1

= 100+ቀ2100ቁ−1

= 100+ 50 = 150

𝑉𝑎𝑟ቆ𝑅1 +൬1𝑅2 + 1𝑅3൰

−1ቇ= ሺsince ind′tሻ 𝑉𝑎𝑟ሺ𝑅1ሻ+ 𝑉𝑎𝑟ቆ൬1𝑅2 + 1𝑅3൰

−1ቇ

𝜕𝜕𝑅2൬𝑅2𝑅3𝑅2 + 𝑅3൰= 𝑅32

ሺ𝑅2 + 𝑅3ሻ2 𝜕𝜕𝑅2൬𝑅2𝑅3𝑅2 + 𝑅3൰= 𝑅22

ሺ𝑅2 + 𝑅3ሻ2

𝑉𝑎𝑟ቀ𝑅2𝑅3𝑅2+𝑅3ቁ≈ ቀ1002

ሺ100+100ሻ2ቁ2 𝑉𝑎𝑟ሺ𝑅2ሻ+ቀ1002

ሺ100+100ሻ2ቁ2 𝑉𝑎𝑟ሺ𝑅3ሻ

= ቀ100416×1004ቁ∙4× 2

= 12

𝑉𝑎𝑟ሺ𝑅ሻ= 𝑉𝑎𝑟ሺ𝑅1ሻ+ 𝑉𝑎𝑟ቀ𝑅2𝑅3𝑅2+𝑅3ቁ

≈ 4+ 0.5 = 4.5 𝑆𝑡.𝐷𝑒𝑣ሺ𝑅ሻ≈ ξ4.5 = 2.12Ω

5.5.5 The Central Limit Effect

• If X1, X2,…, Xn are independent random variables with mean m and variance s2, then for large n, the variable is approximately normally distributed.

If a population has the N(m,s) distribution, then the sample mean x of n independent observations has the distribution.

For ANY population with mean m and standard deviation s the sample mean of a random sample of size n is approximately

when n is LARGE.( , )Nn

( , )Nn

x

Central Limit Theorem

If the population is normal or sample size is large, mean follows a normal distribution

and

follows a standard normal distribution.

x( , )N

n

n

xxz

x

x

• The closer x’s distribution is to a normal distribution, the smaller n can be and have the sample mean nearly normal.

• If x is normal, then the sample mean is normal for any n, even n=1.

• Usually n=30 is big enough so that the sample mean is approximately normal unless the distribution of x is very asymmetrical.

• If x is not normal, there are often better procedures than using a normal approximation, but we won’t study those options.

• Even if the underlying distribution is not normal, probabilities involving means can be approximated with the normal table.

• HOWEVER, if transformation, e.g. , makes the distribution more normal or if another distribution, e.g. Weibull, fits better, then relying on the Central Limit Theorem, a normal approximation is less precise. We will do a better job if we use the more precise method.

If X1, X2, … , Xn are normal, then 𝑋ത is exactly normal.

Example: Bags of potatoes weigh

𝜇= 5.0 pounds

𝜎= 0.1 pounds

If we buy 4 bags, what is Pሺ𝑋ത< 4.9ሻ?

𝐸ሺ𝑋തሻ= 𝜇= 5 ඥ𝑉𝑎𝑟ሺ𝑋തሻ= 𝜎ξ𝑛 = 0.1ξ4 = 0.05

Pሺ𝑋ത< 4.9ሻ= Pቆ𝑋ത− 𝐸𝑋തඥVȁrሺXഥሻ< 4.9− 5.00.05 ቇ= Pሺ𝑧< −2ሻ= 0.0228

Example• X=ball bearing diameter• X is normal distributed with m=1.0cm and s=0.02cm• =mean diameter of n=25• Find out what is the probability that will be off by

less than 0.01 from the true population mean.

%76.98)5.25.2(

)004.0

0.101.1

004.0

0.199.0()01.199.0(

004.025

02.0

0.1

zP

zPxP

nx

x

xx

Exercise

The mean of a random sample of size n=100 is going to be used to estimate the mean daily milk production of a very large herd of dairy cows. Given that the standard deviation of the population to be sampled is s=3.6 quarts, what can we assert about the probabilities that the error of this estimate will be more then 0.72 quart?