The life of automobile voltage regulators has an exponential distribution with a mean life
of six years. You purchase an automobile that is four years old, with a working voltage
regulator, and plan to own it for four years. [1 mark]
(a) What is the probability that the voltage regulator fails during your ownership?
(b) If your regulator fails after you own the automobile three years and it is replaced,
what is the mean time until the next failure?.
(a) The probability of the voltage regulator failing during your ownership can be calculated using the cumulative distribution function of the exponential distribution:
P(X < 8) = 1 - e^(-8/6) ≈ 0.488
where X is the time until failure in years. Therefore, the probability of the voltage regulator failing during your ownership is approximately 0.488 or 48.8%.
(b) After the replacement of the regulator, the life expectancy of the new regulator is still six years. Therefore, the mean time until the next failure is also six years.
Suppose that ̂ and ̂ are estimators of the parameter . We know that
( ̂ ) ( ̂ )
( ̂ ) ( ̂ ) Which estimator is best? In what sense
is it best?
The second estimator, ̂ , is the best in the sense of being more efficient than ̂ . Efficiency in this context means that ̂ produces smaller variance or mean squared error than the other estimator, for a given sample size.
To see this, we can calculate the variances of the two estimators:
Var( ̂ ) = Var( ̂ /2) = (1/4)Var( ̂ ) = (1/4)σ^2/n,
Var( ̂ ) = Var(( ̂ + ̂ )/2) = (1/4)(Var( ̂ )+Var( ̂ )+2Cov( ̂ , ̂ ))
= (1/4)[(σ^2/n)+(σ^2/n)+2ρσ^2/n]
= (1/4)(2+2ρ)σ^2/n
where Cov( ̂ , ̂ ) = ρσ^2/n is the covariance between the two estimators, assuming they are unbiased.
Since ρ ≤ 1, it follows that Var( ̂ ) ≥ Var( ̂ ), with equality only when ρ = 1 (i.e., when ̂ and ̂ are perfectly correlated). Thus, ̂ is more efficient than ̂ in general, and is the best estimator in the sense of having smaller variance or mean squared error.
. Find the equation of regression line. Estimate
for the following data set.
Without the data set provided, we cannot find the equation of the regression line or estimate the value of . Please provide the necessary data to answer this question.
To answer these questions, we can make use of the properties of exponential distributions.
In an exponential distribution, the mean (μ) is equal to the reciprocal of the failure rate (λ). In this case, we are given that the mean life of automobile voltage regulators is six years, so the failure rate λ can be calculated as 1/6.
(a) To find the probability that the voltage regulator fails during your ownership (4 years), we need to calculate the cumulative distribution function (CDF) of the exponential distribution up to time 4.
P(regulator fails during ownership) = 1 - P(regulator survives for 4 years)
= 1 - e^(-λ * 4)
= 1 - e^(-(1/6) * 4)
Therefore, the probability that the voltage regulator fails during your ownership is 1 - e^(-(1/6) * 4).
(b) If the regulator fails after you own the automobile for three years and is replaced, we want to find the mean time until the next failure. Since the exponential distribution is memoryless, the time until the next failure will have the same distribution as the original one.
So, the mean time until the next failure would still be equal to the mean life of the voltage regulator, which is six years.
In summary:
(a) The probability of the voltage regulator failing during your ownership is 1 - e^(-(1/6) * 4).
(b) The mean time until the next failure, after the regulator is replaced, is six years.