Question

Let \(Y_{1}

Short answer

The probability \(P\left(Y_{1}<\mu<Y_{2}\right)=\frac{1}{2}\). The expected length of the interval \(Y_{2}-Y_{1}\) is approximately \(1.13* \sigma\). The constant that solves \(P(\bar{X}-c \sigma<\mu<\bar{X}+c \sigma)=\frac{1}{2}\) is \(c=0.674\), and the expected length of this interval is approximately \(1.35* \sigma\), which is larger than the expected length from part (a).

Step-by-step solution

Compute Probability

Since the two sample points are randomly picked from a normally distributed population, they're exchangeable, meaning each point has the same chance of being the first order statistic \(Y_{1}\) or the second order statistic \(Y_{2}\). Consequently, \(P\left(Y_{1}<\mu<Y_{2}\right)= P\left(\mu<Y_{1}<Y_{2}\right)\). These are mutually exclusive and exhaustive events; hence their total probability equals 1, so \(P\left(Y_{1}<\mu<Y_{2}\right)= \frac{1}{2}\).

Calculate Expected Length

To calculate the expected value of \(Y_{2}-Y_{1}\), we can use the formula for the difference of 2nd and 1st order statistics for a sample size of 2 from a standard normal distribution, which is \(E[Y_{2}-Y_{1}] = 2 \cdot \frac{\sqrt{2}}{\sqrt{\pi}} \approx 1.13\). The scale is multiplied by \(\sigma\) for \(N\left(\mu, \sigma^{2}\right)\) which leads to \(1.13*\sigma\).

Solve for the Constant c

To find the constant \(c\), we use the definition of the probability \(P(\bar{X}-c \sigma<\mu<\bar{X}+c \sigma)=\frac{1}{2}\). This is equivalent to seeing that the mean is within \(c\) standard deviations of the sample mean, which by symmetry is half of all possible samples. By definition, this probability corresponds to a Z-score of \(c\). Computing the inverse of the standard normal distribution function for 0.75 (which reflects the sum of the lower tail probability of 0.5 and half of the probability in the middle), we find \(c=0.674\).

Compare Lengths

We can compare the lengths by calculating the expected length of the random interval for part (b) using the constant \(c\) which is \(2*c*\sigma = 2*0.674*\sigma \approx 1.35*\sigma\), which is larger than the expected length of the random interval in part (a) \(1.13*\sigma\). The interval around the sample mean is wider, suggesting that using the sample mean and standard deviation gives us an interval estimate for the population mean that has a larger expected length than that obtained by using the order statistics.