Chapter L, Problem 67IL

Interpretation Introduction

Interpretation:

The average value and the standard deviation has to be determined and the number of values that fall within one standard deviation of the average value has to be given.

Concept Introduction:

Standard Deviation: A series of measurements is equal to the square root of the sum of the squares of the deviations for each measurement from the average, divided by one less than the number of measurements.

  $\begin{array}{l}SD=\sqrt{\frac{\sum |x-\mu {|}^2}{N}}\\ \text{where},x\text{is values in the data set,}\\ \mu is\text{average of the data set (mean value),}\\ \text{ N is no}\text{.of data points}\text{.}\end{array}$

Answer to Problem 67IL

The average value is $\text{5}\text{.24 \%}$

The standard deviation obtained is $0.0506\%$

Seven of the ten values fall within the region of $\text{ 5}\text{.19 }\le \text{ x }\le \text{ 5}\text{.29}\text{.}$

Explanation of Solution

The average value and the standard deviation is calculated as,

  $\begin{array}{l}SD=\sqrt{\frac{\sum |x-\mu {|}^2}{N}}\\ \text{where},x\text{is values in the data set,}\\ \mu is\text{average of the data set (mean value),}\\ \text{ N is no}\text{.of data points}\text{.}\\ \\ \text{Given : x = 5}\text{.22\%,5}\text{.28\%,5}\text{.22\%,5}\text{.30\%,5}\text{.19\%,}\\ \text{ 5}\text{.23\%,5}\text{.33\%,5}\text{.26\%,5}\text{.15\%,5}\text{.22\%}\text{.}\\ \text{Find mean value (}\mu ):\\ \mu =\frac{\text{5}\text{.22\%+5}\text{.28\%+5}\text{.22\%+5}\text{.30\%+5}\text{.19\%+5}\text{.23\%+5}\text{.33\%+5}\text{.26\%+5}\text{.15\%+5}\text{.22\%}}{10}\\ \text{ = 5}\text{.24 \%}\\ \begin{array}{cccc}\text{Determination}& \text{Measured values}& \begin{array}{l}\text{Difference}\text{between}\\ \text{Measurement}\text{and}\text{Average}\\ (x-\mu )\end{array}& \begin{array}{l}\text{Square of Difference}\\ |x-\mu {|}^2\end{array}\\ \text{1}& \text{5}\text{.22\%}& -0.02& 4\times {10}^{-4}\\ 2& \text{5}\text{.28\%}& 0.04& 1.6\times {10}^{-3}\\ 3& \text{5}\text{.22\%}& -0.02& 4\times {10}^{-4}\\ 4& \text{5}\text{.30\%}& 0.06& 3.6\times {10}^{-3}\\ 5& \text{5}\text{.19\%}& -0.05& 2.5\times {10}^{-3}\\ 6& \text{5}\text{.23\%}& -0.01& 1\times {10}^{-4}\\ 7& \text{5}\text{.33\%}& 0.09& 8.1\times {10}^{-3}\\ 8& \text{5}\text{.26\%}& 0.02& 4\times {10}^{-4}\\ 9& \text{5}\text{.15\%}& -0.09& 8.1\times {10}^{-3}\\ 10& \text{5}\text{.22\%}& -0.02& 4\times {10}^{-4}\end{array}\\ \sum |x-\mu {|}^2=4\times {10}^{-4}+1.6\times {10}^{-3}+4\times {10}^{-4}+3.6\times {10}^{-3}+2.5\times {10}^{-3}+1\times {10}^{-4}+8.1\times {10}^{-3}+4\times {10}^{-4}+8.1\times {10}^{-3}+4\times {10}^{-4}\\ \text{ = 0}\text{.0256 \%}\\ SD=\sqrt{\frac{\text{0}\text{.0256}}{10}}=0.0506\%\\ \text{Seven of the ten values fall within the region of 5}\text{.19 }\le \text{ x }\le \text{ 5}\text{.29}\text{.}\end{array}$

As shown above, the mean value of the given data and also find the standard deviation value by substituting the obtained values in the equation.

Conclusion

The average value was $\text{5}\text{.24 \%}$

The standard deviation obtained was $0.0506\%$

Seven of the ten values fall within the region of $\text{ 5}\text{.19 }\le \text{ x }\le \text{ 5}\text{.29}\text{.}$