# Problem

Take this (easy) problem as an example:

An astronomer is interested in measuring the distance, in light-years, from his observatory to a distant star. Although the astronomer has a measuring technique, he knows that, because of changing atmospheric conditions and normal error, each time a measurement is made it will not yield the exact distance, but merely an estimate. As a result, the astronomer plans to make a series of measurements and then use the average value of these measurements as his estimated value of the actual distance. If the astronomer believes that the values of the measurements are independent and identically distributed random variables having a common mean d (the actual distance) and a common variance of 4 (light-years), how many measurements need he make to be reasonably sure that his estimated distance is accurate to within 0.5 light-year?

# Approaches

At first sight, what seemed to me more reasonable, was to use the Chebyshev's inequality as follows:

$X_{i}$ = distance in l.y. observed in experiment $i$

$E(X_{i})=\mu =d$

$Var(X_{i})=\sigma^{2} = 4$

Sample Mean:

$E(\bar{X}_{n})=\mu =d$

$Var(\bar{X}_{n})=\frac{\sigma^{2}}{n} = \frac{4}{n}$

So by Chebyshev's inequality we have:

$P(\left | \bar{X}_{n}-d \right | <0.5)>1-\frac{4}{n \cdot 0.5^{2}} \\ = 1- \frac{1}{n}\cdot \frac{4}{0.5^{2}} \\ = 1- \frac{16}{n}$

So if we consider the sentence "reasonably sure" as $0.95$, then $1- \frac{16}{n} = 0.95$ when $n= 320$.

So I would answer with: $n=320$ is enough.

But, Using the Central Limit Theorem we have that:

$P(\left | \bar{X}_{n}-d \right |\leq 0.5) \\ = P\left \{ \sqrt{n}\frac{\left | \bar{X}_{n}-d \right |}{2} \leq \frac{\sqrt{n}}{4}\right \}\\ =P\left ( \left | Z \right | \leq \frac{\sqrt{n}}{4} \right ) \\ \approx \Phi \left ( \frac{\sqrt{n}}{4} \right )-\Phi \left (- \frac{\sqrt{n}}{4} \right ) \\ =2\Phi \left ( \frac{\sqrt{n}}{4} \right )-1$

where symbol $\Phi(z)$ denotes the cumulative distribution function of a standard normal variable. Hence, $n$ should be, approximatively, the value such that:

$2\Phi \left ( \frac{\sqrt{n}}{4} \right )=0.975$

In other terms $\frac{\sqrt{n}}{4}$ is, approximatively, equal to the $97.5$% quantile of standard normal distribution. Using the normal table, we find that $\frac{\sqrt{n}}{4} = 1.96$ and thus $n = (1.96 \cdot 4)^{2} = 61.466$, thus $n$ must be equal to $62$, which is very different from the result I ended up with, which is $320$.

# Questions

What's the reasoning behind this? This was an exercice I had to solve, but I used the Chebychev ineq. approach instead of the central limit theorem (used by my prof) and the results are very different. Is it correct? Am I missing something important?

Any clarification is appreciated.

• Your inequality is reversed. As $n$ increases, the probability that $X$ ls close to $d$ is decreasing as per your calculations. – Dilip Sarwate Dec 21 '15 at 12:41
• yes I forgot 1-16/n, I edit – francescop Dec 21 '15 at 13:20
• You wrote "variance of 4 (light-years)", which s dimensionally incorrect. Is the variance 4 light years squared, or is 4 light years the standard deviation? – Mark L. Stone Dec 21 '15 at 13:36
• well it's an exercice taken from Ross, S. (2010). A First Course in Probability, 8th edition. Pearson Prentice Hall. I assume it's $\sigma^{2}$ 4 l.y. the variance of the random variables as I wrote. – francescop Dec 21 '15 at 13:53