Chapter 3.2, Problem 16E

To determine

To find:

The sample standard deviation for the given data set.

Answer to Problem 16E

Solution:

The sample standard deviation for the given data set is 4.74$° \text{F}$.

Explanation of Solution

Given information:

The following data represent high temperature for the cities in the southeast (in degrees Fahrenheit).

High Temperatures for Cities in the Southeast

$\begin{array}{l}\begin{array}{cc}Stem& Leaves\\ \begin{array}{l}7\\ 7\\ 8\\ 8\\ 9\\ 9\end{array}& \begin{array}{l}\\ 79\\ 234\\ 5578899\\ 0011223\\ 5\end{array}\end{array}\\ Key:\begin{array}{cc}7& 7\end{array}=77°F\end{array}$

Approach:

A stem leaf plot is combination of two columns. Last significant digit of each data should be placed in the column of leaves and others digit in the column of stem. Each stem should be listed in numerical order and each leaf should be placed next to its stem.

Formula used:

The formula to calculate the sample standard deviation for given set of data is,

$s=\sqrt{\frac{{\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}}{n-1}}$

Where, $x_i$ is the $i^{\text{th}}$ data value, $\overline{x}$ is the sample mean, and $N$ is the total number of collection.

Calculation:

Following is the set of data corresponding to the given stem and-leaf plot.

$\begin{array}{l}77°,79°,82°,83°,84°,85°,85°,87°,88°,88°,\\ 89°,89°,90°,90°,91°,91°,92°,92°,93°,95°\end{array}$.

Here, $n=20$, Compute $\overline{x}$ for the given set of data,

$\begin{array}{rll}\overline{x}& =& \frac{\begin{array}{l}77+79+82+83+84+85+85+87+88+88+\\ 89+89+90+90+91+91+92+92+93+95\end{array}}{20}\\ \hspace{0.17em}& =& \frac{1750}{20}\\ \hspace{0.17em}& =& 87.5\end{array}$

Construct the table for sample standard deviation of the given set of data.

$x_i$	$x_i-\overline{x}$	${\left(x_i-\overline{x}\right)}^2$
$77$	$77-87.5=-10.5$	110.25
$79$	$79-87.5=-8.5$	72.25
$82$	$82-87.5=-5.5$	30.25
$83$	$83-87.5=-4.5$	20.25
$84$	$84-87.5=-3.5$	12.25
$85$	$85-87.5=-2.5$	6.25
$85$	$85-87.5=-2.5$	6.25
$87$	$87-87.5=-0.5$	0.25
$88$	$88-87.5=0.5$	0.25
$88$	$88-87.5=0.5$	0.25
$89$	$89-87.5=1.5$	2.25
$89$	$89-87.5=1.5$	2.25
$90$	$90-87.5=2.5$	6.25
$90$	$90-87.5=2.5$	6.25
$91$	$91-87.5=3.5$	12.25
$91$	$91-87.5=3.5$	12.25
92	$92-87.5=4.5$	20.25
92	$92-87.5=4.5$	20.25
93	$93-87.5=5.5$	30.25
95	$95-87.5=7.5$	56.25

Calculate the sum of ${\left(x_i-\overline{x}\right)}^2$.

$\begin{array}{l}{\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}\\ =110.25+72.25+30.25+20.25+12.25+6.25+6.25+0.25+0.25+0.25\\ +2.25+2.25+6.25+6.25+12.25+12.25+20.25+20.25+30.25+56.25\\ =427\end{array}$

Substitute 427 for ${\left(x_i-\overline{x}\right)}^2$ and 20 for $n$ in the formula of the standard deviation.

$\begin{array}{rll}s& =& \sqrt{\frac{{\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}}{n-1}}\\ & =& \sqrt{\frac{427}{19}}\\ & =& \sqrt{22.47368}\\ \hspace{0.17em}& \approx & 4.74\end{array}$

The sample standard deviation for the given data set is 4.74$° \text{F}$.