Chapter 13, Problem 22CCSR

Solution

To find: The standard deviation of the given data to the nearest hundredth.

The standard deviation is 7.67.

Given:

The daily water usage in gallons for 2 weeks is as shown in the table below:

92.3	81.3
85.3	81.6
89.7	76.9
101.2	94.0
80.3	89.6
91.4	96.3
88.8	102.1

Concept used:

The standard deviation of an ungrouped data is,

  $s=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n-1}}$ , where $\overline{x}$ is the mean of the given data and $n$ is the number of observations.

Calculation:

The mean of the given data is,

  $\begin{array}{c}\overline{x}=\frac{92.3+85.3+89.7+101.2+80.3+91.4+88.8+81.3+81.6+76.9+94.0+89.6+96.3+102.1}{14}\\ =\frac{1250.8}{14}\\ =\text{89}\text{.342}8\end{array}$

Now the square of the deviations from the mean are,

  $\begin{array}{c}{\left(92.3-89.3428\right)}^2=8.7450\\ {\left(85.3-89.3428\right)}^2=16.3442\\ {\left(89.7-89.3428\right)}^2=0.1275\\ {\left(101.2-89.3428\right)}^2=140.5931\\ {\left(80.3-89.3428\right)}^2=81.7722\\ {\left(91.4-89.3428\right)}^2=4.2320\\ {\left(88.8-89.3428\right)}^2=0.2946\\ {\left(81.3-89.3428\right)}^2=64.6866\\ {\left(81.6-89.3428\right)}^2=59.9509\\ {\left(76.9-89.3428\right)}^2=154.8232\\ {\left(94.0-89.3428\right)}^2=21.6895\\ {\left(89.6-89.3428\right)}^2=0.0661\\ {\left(96.3-89.3428\right)}^2=48.4026\\ {\left(102.1-89.3428\right)}^2=162.7461\end{array}$

Now the sum of these squared deviations is,

  ${\displaystyle \sum {\left(x-\overline{x}\right)}^2}=764.4736$

Now the standard deviation is,

  $\begin{array}{c}s=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n-1}}\\ =\sqrt{\frac{764.4736}{14-1}}\\ \approx 7.67\end{array}$