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