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Chapter 4: Problem 28

Let \(Y_{1}

Short Answer

Expert verified
The expected value of the random length \(Y_{2}-Y_{1}\) is equal to \(\sqrt{ 2\pi } \sigma\). The constant \(c\) that solves the equation is approximately 0.6745. The length of the random interval for \(\bar{X}\) is smaller than the expected length for \(Y_{2}-Y_{1}\), indicating that the interval centered at the sample mean is more efficient for estimating \(\mu\).

Step by step solution

01

Calculate the Joint Distribution of \(Y_{1}\) and \(Y_{2}\)

As \(Y_{1}y_{1},Y_{2}<y_{2})\). Thus, the joint density function of \(Y_{1}\) and \(Y_{2}\) can be obtained by taking the partial derivative of \(F_{Y_{1},Y_{2}}(y_{1},y_{2})\) with respect to \(y_{1}\) and \(y_{2}\).
02

Calculate the Probability and Expected Value in Part (a)

Utilize the joint distribution of \(Y_{1}\) and \(Y_{2}\) to calculate \(P\left(Y_{1}<\mu<Y_{2}\right)\) by integrating the density function over \(Y_{1}<\mu<Y_{2}\). Next, calculate the expected value of the random length \(Y_{2}-Y_{1}\) by integrating \(y_{2} - y_{1}\) over all possible values of \(Y_{1}\) and \(Y_{2}\).
03

Utilize the properties of the Mean to Find \(c\) in Part (b)

Recall that \(\bar{X}\) is the sample mean of \(Y_{1}\) and \(Y_{2}\), and likewise follows a Normal distribution. Manipulate the property \(P(\bar{X}-c \sigma<\mu<\bar{X}+c \sigma)=\frac{1}{2}\) to find \(c\) by using the cumulative distribution function of \(\bar{X}\). Lastly, compare the length of this random interval with the expected value of \(Y_{2}-Y_{1}\) calculated in part (a).

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