An Astronomer Makes N Measurements Of The Distance Between J
An Astronomer Makes N Measurements Of The Distance Between Jupiter And
An astronomer makes n measurements of the distance between Jupiter and a particular one of its moons. Experience with the instruments used leads her to believe that for the proper units, the measurements will be normally distributed with mean d, the true distance, and variance 16. She performs a series of n measurements. Let An = (X1 + X2 + ... + Xn)/n be the average of these measurements. (a) Show that P{An - 8/n^0.5 ≤ d ≤ An + 8/n^0.5} approximately equals .95.
Paper For Above instruction
In this problem, we analyze the statistical properties of measurements taken by an astronomer to estimate the distance between Jupiter and one of its moons. Understanding the distribution of the sample mean, especially in the context of known variance and normality, allows us to construct confidence intervals for the true distance d. The goal here is to show that with a certain probability, the true distance d lies within a specific interval centered at the sample mean.
Given, the measurements \(X_1, X_2, ..., X_n\) are independent and identically distributed (i.i.d.) normal random variables with mean \(d\) and variance \(\sigma^2 = 16\). The sample mean, \(A_n = \frac{X_1 + X_2 + \cdots + X_n}{n}\), also follows a normal distribution with mean \(d\) and variance \(\frac{\sigma^2}{n} = \frac{16}{n}\>. According to the properties of normal distributions, the standardized form of \(A_n\) is a standard normal variable:
\[
Z = \frac{A_n - d}{\sqrt{\frac{\sigma^2}{n}}} = \frac{A_n - d}{4 / \sqrt{n}}
\]
which follows a standard normal distribution \(N(0,1)\). To compute the probability that \(d\) lies within the interval \([A_n - 8 / \sqrt{n}, A_n + 8 / \sqrt{n}]\), we note that:
\[
P \left( A_n - \frac{8}{\sqrt{n}} \leq d \leq A_n + \frac{8}{\sqrt{n}} \right)
= P \left( - \frac{8}{\sqrt{n}} \leq d - A_n \leq \frac{8}{\sqrt{n}} \right)
\]
Rearranging, this becomes:
\[
P \left( - \frac{8}{\sqrt{n}} \leq d - A_n \leq \frac{8}{\sqrt{n}} \right) = P \left( - \frac{8}{\sqrt{n}} \leq -(A_n - d) \leq \frac{8}{\sqrt{n}} \right)
\]
and recognizing that \(A_n - d\) is normally distributed with mean 0 and variance \(16/n\), the standardized variable \(Z = \frac{A_n - d}{4/\sqrt{n}}\) follows a standard normal distribution. Then:
\[
P \left( -2 \leq Z \leq 2 \right)
\]
since \(\frac{8 / \sqrt{n}}{4 / \sqrt{n}} = 2\). From standard normal distribution tables or properties, the probability that \(Z\) falls within \(-2\) and \(2\) is approximately 0.95, corresponding to a 95% confidence interval. Thus, the probability that the true distance \(d\) lies within this interval centered at the sample mean \(A_n\) is approximately 95%, confirming the specified coverage probability.
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