## Elementary Technical Mathematics

The sample standard deviation of the data measured by the technician is $0.0001$ mm.
$7.0036\text{mm},$ $7.0035$mm, $7.0038$mm, $7.0035$mm, $7.0036$mm The sum of the all measured values is, $7.0036+7.0035+7.0038+7.0035+7.0036=35.018$ The number of the measured value is $5$. Thus, the mean of the data is, \begin{align} & \text{Mean}=\frac{\text{Sum of the all measured value}}{\text{Number of the measured value}} \\ & \text{Mean}=\frac{35.018}{5} \\ & \text{Mean}=7.0036 \\ \end{align} The difference of the data about the mean is, $7.0036-7.0036=0$ And, $7.0035-7.0036=-0.0001$ And, $7.0038-7.0036=-0.0002$ And, $7.0035-7.0036=-0.0001$ And, $7.0036-7.0036=0$ Squaring the differenced data gives: ${{\left( 0 \right)}^{2}}=0$ And, ${{\left( -0.0001 \right)}^{2}}=0.00000001$ And, ${{\left( -0.0002 \right)}^{2}}=0.00000004$ And, ${{\left( -0.0001 \right)}^{2}}=0.00000001$ And, ${{\left( 0 \right)}^{2}}=0$ Sum of all squared data above is $0+0.00000001+0.00000004+0.00000001+0=0.00000006$ Dividing the sum of the square data by the number of observations less one gives $\frac{0.00000006}{5-1}=0.000000015$ Therefore, the sample standard deviation of the data is $\sqrt{0.000000015}=0.0001$ Thus, the sample standard deviation of the data measured by the technician is $0.0001$ mm.