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alex edited untitled.tex
about 8 years ago
Commit id: 0bf27e4fa374e6a78c43fd0df36448da2965cd19
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}
\item{There are a few steps to this.
\begin{itemize}
\item{}
\item{}
\item{} \item{Find mean radius by dividing mean diameter by 2:
\begin{align*}
\mu_D = \frac{\sum{D_i}}{N} = \frac{24.75+24.82+24.87+24.79+24.95+24.71+24.81+24.65+24.69}{9}\\
&= \frac{223.04}{9} = 24.78\text{mm}\\
\text{Therefore, } \mu_r = \frac{\mu_D}{2} &= \frac{24.78}{2}\\
&= 12.39\text{mm}\\
\end{align*}
}
\item{Here, we use
\begin{align*}
\sigma &= \sqrt{\frac{\sum{(r_i - \mu_r)^2}}{N-1}}\\
&= 0.0469\text{mm}
\end{align*}
Make sure that they use the radii here, and not the raw diameter values.
}
\item{
\begin{equation*}
\sigma_{error} = \frac{\sigma}{\sqrt{N}} = 0.0156\text{mm}
\end{equation*}
}
\item{Note that the relative error for the volume is three times larger than the error of $\sigma_{error}$ as a fraction of the mean.
\begin{align*}
V &= \frac{4\pi \mu_r^3}{3} = \frac{4\pi(12.39)^3}{3} = 7967\text{mm}\\
\delta V &= V(\frac{3\sigma_{error}}{\mu_r}) = 7967(\frac{3\times0.0156}{12.39})=30.1\\
\text{Therefore, } V &= 7970\pm30\text{mm}^3
\end{align*}
}
\end{itemize}
}
\end{enumerate}