Question

listed below are prices in dollars for one night at different hotels in a certain region. find the range, variance, and standard deviation for the given sample data. include appropriate units in the results. how useful are the measures of variation for someone searching for a room?
233 226 239 279 229 293 216 182
the range of the sample data is 111.0 dollars.
(round to one decimal place as needed.)
the standard deviation of the sample data is
(round to one decimal place as needed.)

Explanation:

Step1: Calculate the mean

The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$.
Here, $n = 9$, and $x=\{233,226,239,279,229,293,216,182\}$.
$\sum_{i=1}^{9}x_{i}=233 + 226+239+279+229+293+216+182$
$=233+226 = 459$; $459+239=698$; $698+279 = 977$; $977+229=1206$; $1206+293 = 1499$; $1499+216=1715$; $1715+182=1897$
$\bar{x}=\frac{1897}{9}\approx210.8$

Step2: Calculate the variance

The variance $s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}$
$(233 - 210.8)^{2}=(22.2)^{2}=492.84$
$(226-210.8)^{2}=(15.2)^{2}=231.04$
$(239 - 210.8)^{2}=(28.2)^{2}=795.24$
$(279-210.8)^{2}=(68.2)^{2}=4651.24$
$(229 - 210.8)^{2}=(18.2)^{2}=331.24$
$(293-210.8)^{2}=(82.2)^{2}=6756.84$
$(216-210.8)^{2}=(5.2)^{2}=27.04$
$(182-210.8)^{2}=(- 28.8)^{2}=829.44$
$\sum_{i = 1}^{9}(x_{i}-\bar{x})^{2}=492.84+231.04+795.24+4651.24+331.24+6756.84+27.04+829.44$
$=492.84+231.04 = 723.88$; $723.88+795.24=1519.12$; $1519.12+4651.24=6170.36$; $6170.36+331.24=6501.6$; $6501.6+6756.84=13258.44$; $13258.44+27.04=13285.48$; $13285.48+829.44=14114.92$
$s^{2}=\frac{14114.92}{9 - 1}=\frac{14114.92}{8}=1764.4$

Step3: Calculate the standard deviation

The standard deviation $s=\sqrt{s^{2}}$
$s=\sqrt{1764.4}\approx42.0$

Answer:

The standard deviation of the sample data is $42.0$ dollars.