# Homework Help: Help with use of Chebyshev's inequality and sample size

1. Mar 14, 2014

### penguinnnnnx5

1. The problem statement, all variables and given/known data

2. Relevant equations

P (|Y - μ| < kσ) ≥ 1 - Var(Y)/(k2σ2) = 1 - 1/k2

??

3. The attempt at a solution

using the equation above

1 - 1/k2 = .9

.1 = 1/k2

k2 = 10

k = √10 = 3.162

k = number of standard deviations. After this I don't know where to go.

I don't have a solid understanding of Chebyshev's inequality either.

2. Mar 14, 2014

### Ray Vickson

First figure out what probability you need to find, then worry about how to find it using Chebyshev or some other method. So, if your measurements are $M_1, M_2, \ldots, M_n$, what is their average, $\bar{M}$? In terms of $c$ and $U_1, U_2, \ldots, U_n$, what would be your formula for $\bar{M}$? What would be the mean and variance of $\bar{M}$ (expressed in terms of $c, \:n$ and other given quantities)? Now how would you express the event "the average is within half a degree of $c$"? At this point you are ready to apply some probability!

3. Mar 14, 2014

### penguinnnnnx5

I would say that the average $\bar{M}$ $=$ ($n c$ + $U_1, U_2, \ldots, U_n$) / $n$ since you are finding $c$ $n$ times and adding that to all the $U_i$ from each sample. Then to average it, you'd need to divide it by $n$ of course.

If it is half a degree of $c$, would that mean that $|\bar{M}$ - $c| >= .5?$

4. Mar 14, 2014

### penguinnnnnx5

This would make c the mean, yes? Because it is the expected value/ the value you expect to be correct. Would that make $U_n$ the variance then? But if so, how will we find the variance if $U_n$ is not given? We only know that $Var(U_n) = 3$

But given what I know now, would this mean then that $P (|\bar{M} - c| >= .5) = 1 -P (|\bar{M} - c| < .5) = 1 - σ^2 / nε^2$ where ε = .5?