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}