Question

In: Statistics and Probability

The lifespan of an electrical component has an exponential distribution with parameter lambda = 0.013. Suppose...

The lifespan of an electrical component has an exponential distribution with parameter lambda = 0.013. Suppose we have an iid sample of size 100 of these components

Some hints: P(X < c) for an exponential(lambda) can be found via pexp(c,lambda) E[X] = 1/lambda and Var[X] = 1/lambda^2

Round all answers to 4 decimals

Using the exact probability distribution, what is the probability that a single component will be within 15.38 units of the population mean?

Using Chebyshev's inequality, what is a lower bound on the probability that the sample mean will be within 15.38 units of the population mean?

Using the CLT, what is an approximation to the probability that the sample mean will be within 15.38 units of the population mean?

Solutions

Expert Solution

SOLUTION:


Related Solutions

(Exponential Distribution) The life, in years, of a certain type of electrical switch has an exponential...
(Exponential Distribution) The life, in years, of a certain type of electrical switch has an exponential distribution with an average life of ?? = 2 years. i) What is the probability that a given switch is still functioning after 5 years? ii) If 100 of these switches are installed in different systems, what is the probability that at most 30 fail during the first year?(also Binomial Distribution
The life, in years, of a certain type of electrical switch has an exponential distribution with...
The life, in years, of a certain type of electrical switch has an exponential distribution with an average life 2 years. If 100 of these switches are installed in different systems, what is the probability that at most 2 fail during the first year?
Suppose the time to failure follows an exponential distribution with parameter λ. Based on type II...
Suppose the time to failure follows an exponential distribution with parameter λ. Based on type II censoring determine the likelihood and the maximum likelihood estimator for λ. Assume that n = 3 and the experiment ends after 2 failures.
let X~GAM(3,2), when X = x, conditional distribution of Y has parameter as lambda = x,...
let X~GAM(3,2), when X = x, conditional distribution of Y has parameter as lambda = x, ~ POI(lambda) (a) what is E(Y) and Var(Y)? (b) what is marginal distribution of Y? (c) what is E(X|Y=15) and Var(X|Y=15)?
For a random sample of sizenfrom an Exponential distribution with rate parameter λ (so that the...
For a random sample of sizenfrom an Exponential distribution with rate parameter λ (so that the density is fY(y) =λe−λy), derive the maximum likelihood estimator, the methods of moments estimator, and the Bayes estimator (that is, the posterior mean) using a prior proportional to λe−λ, for λ >0. (Hint: the posterior distribution will be a Gamma.)
The random variable X has an Exponential distribution with parameter beta= 5. The P(X > 18|X...
The random variable X has an Exponential distribution with parameter beta= 5. The P(X > 18|X > 12) is equal to
The usable lifetime of a particular electronic component is known to follow an exponential distribution with...
The usable lifetime of a particular electronic component is known to follow an exponential distribution with a mean of 6.1 years. Let X = the usable lifetime of a randomly selected component. (a) The proportion of these components that have a usable lifetime between 5.9 and 8.1 years is __. (b) The probability that a randomly selected component will have a usable life more than 7.5 years is __. (c) The variance of X is __.
The lifespan of a system component follows a Weibull distribution with α (unknown) and β=1. (hint:...
The lifespan of a system component follows a Weibull distribution with α (unknown) and β=1. (hint: start with a confidence interval for μ) f(x)=αβXβ-1 exp(-αXβ) a.      Derive 92% large sample confidence interval for α. b.      Find the maximum likelihood estimator of α.
The lifespan of a system component follows a Weibull distribution with α (unknown) and β=1. (hint:...
The lifespan of a system component follows a Weibull distribution with α (unknown) and β=1. (hint: start with confidence interval for μ) f(x)=αβXβ-1 exp(-αXβ) a.      Derive 98% large sample confidence interval for α. b.      Find maximum likelihood estimator of α.
The lifespan of a system component follows a Gamma distribution with β (unknown) and α=1. (hint:...
The lifespan of a system component follows a Gamma distribution with β (unknown) and α=1. (hint: start with confidence interval for μ) f(x)=(1/βα Γ(α)) Xα-1 e-X/β a.      Derive 98% large sample confidence interval for β. b.      Find maximum likelihood estimator of β. a) 98% large sample confidence interval fo β is b) MLE for β is
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT