X45 69ㄱ8P(X) 0.12 0.18/032/0.25find:P(x=8)mecinVarianceStandard deviation

From the provided information,X45678P (X)0.120.180.320.25

Answered: 4 1 68 5 8 P(X) 0.12/0.18/032/0.25…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A random sample of 10 observations from population A has sample mean of 152.3 and a sample standard deviation of 1.83. Another random sample of 8 observations from population B has a sample standard deviation of 1.94. Assuming equal variances in those two populations, a 99% confidence interval for μA − μB is (-0.19, 4.99), where μA is the mean in population A and μB is the mean in population B. (a) What is the sample mean of the observations from population B? (b) If we test H0 : μA ≤ μB against Ha : μA > μB, using α = 0.02, what is your conclusion?
Q1: Lifetimes of a certain component are lognormally distributed with parameters u = 1 day and o = 0.5 days. Find the mean lifetime of these components. 1-Find the standard deviation of the lifetimes. 2-Find CDF 3- Find reliability at t= 4 day Q2: The lifetime (in days) of a certain electronic component that operates in a high-temperature environment is lognormally distributed with u = 1.2 and o = 2. a. Find the mean lifetime. b. Find the probability that a component lasts between three and six days. c. Find the 95th percentile of the lifetimes. d- find CDF Q3: The authors suggest using a Weibull distribution to model the duration of a bake step in the manufacture of a semiconductor. Let T represent the duration in hours of the bake step for a randomly chosen lot. If T Weibull(0.3, 0.1), what is the probability that the bake step takes longer than four hours? What is the probability that it takes between two and seven hours? What reliability at t= 0.8 hours
Heteroskedasticity arises because of non-constant variance of the error terms. We said proportional heteroskedasticity exists when the error variance takes the following structure: Var(et)=σt^2=σ^2 xt. But as we know, that is only one of many forms of heteroskedasticity. To get rid of that specific form of heteroskedasticity using Generalized Least Squares, we employed a specific correction – we divided by the square root of our independent variable x. And the reason why that specific correction worked, and yielded a variance of our GLS estimates that was sigma-squared, was because of the following math: (Picture 1) Where var(et)=σ^2 according to our LS assumptions. In other words, dividing everything by the square root of x made this correction work to give us sigma squared at the end of the expression. But if we have a different form of heteroskedasticity (i.e. a difference variance structure), we have to do a different correction to get rid of it. (a) what correction would you use…

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4 1 68 5 8 P(X) 0.12/0.18/032/0.25 find: P(x=8) mecin variance Standard Clevicition