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.

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Presentation transcript:

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? 5.5.3 Means and Variances for Linear Combinations What is called a linear combination of random variables? aX+b, aX+bY+c, etc. 2X+3 5X+9Y+10

If a and b are constants, then E(aX+b)=aE(X)+b. __________________________ Corollary 1. E(b)=b. Corollary 2. E(aX)=aE(X).

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

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).

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

For the mean E(U) since

For Var(U)

Example

Example 24 on page 311 The 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?

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.

Central Limit Theorem If the population is normal or sample size is large, mean follows a normal distribution and follows a standard normal distribution.

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.

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.

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?