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? 207 226 289 233 102 180 131 275 the range of the sample data is 187.0 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

First, find the sum of the data values: $207 + 226+289+233+102+180+131+275 = 1643$. There are $n = 8$ data - points. The mean $\bar{x}=\frac{1643}{8}=205.375$.

Step2: Calculate the squared differences

For each data - point $x_i$, calculate $(x_i-\bar{x})^2$.
For $x_1 = 207$: $(207 - 205.375)^2=(1.625)^2 = 2.640625$.
For $x_2 = 226$: $(226-205.375)^2=(20.625)^2 = 425.44140625$.
For $x_3 = 289$: $(289 - 205.375)^2=(83.625)^2 = 6993.140625$.
For $x_4 = 233$: $(233-205.375)^2=(27.625)^2 = 763.140625$.
For $x_5 = 102$: $(102-205.375)^2=(-103.375)^2 = 10686.421875$.
For $x_6 = 180$: $(180 - 205.375)^2=(-25.375)^2 = 643.94140625$.
For $x_7 = 131$: $(131-205.375)^2=(-74.375)^2 = 5532.640625$.
For $x_8 = 275$: $(275-205.375)^2=(69.625)^2 = 4847.640625$.

Step3: Calculate the sample variance

The sample variance $s^2=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}$.
$\sum_{i = 1}^{8}(x_i-\bar{x})^2=2.640625+425.44140625+6993.140625+763.140625+10686.421875+643.94140625+5532.640625+4847.640625 = 29894.9075$.
$s^2=\frac{29894.9075}{8 - 1}=\frac{29894.9075}{7}=4270.701071\approx4270.7$.

Step4: Calculate the sample standard deviation

The sample standard deviation $s=\sqrt{s^2}=\sqrt{4270.701071}\approx65.3$.

Answer:

The standard deviation of the sample data is $65.3$ dollars.