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?
213, 209, 164, 78, 97, 181, 291, 153, 9
the range of the sample data is 183 dollars. (round to one decimal place as needed.)
the standard deviation of the sample data is 60.8 dollars. (round to one decimal place as needed.)
the variance of the sample data is □ (round to one decimal place as needed.)
options for units: nights, dollars, dollars², nights²

Explanation:

Step1: Recall the formula for sample variance

The formula for sample variance \( s^2 \) is \( s^2=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1} \), where \( x_i \) are the data points, \( \bar{x} \) is the sample mean, and \( n \) is the number of data points. We know the standard deviation \( s=60.8 \) dollars. And the relationship between standard deviation and variance is \( s^2=\text{variance} \) (since variance is the square of standard deviation).

Step2: Calculate the variance

We know that variance \( =s^2 \). Given \( s = 60.8 \) dollars, then \( s^2=(60.8)^2 \).
\[
(60.8)^2=60.8\times60.8 = 3696.64
\]
But wait, maybe we should calculate it from the data. Let's first find the data points: 213, 209, 164, 78, 97, 181, 291, 153, 9 (Wait, maybe I missed some? Wait the original data: Let's list all the data points from the image. Let's assume the data is: 213, 209, 164, 78, 97, 181, 291, 153, 9? Wait no, maybe the data is 213, 209, 164, 78, 97, 181, 291, 153, and maybe another? Wait the range is 183, which is max - min. Let's check: max is 291, min is 9, 291 - 9=282? No, the given range is 183. Wait maybe the data is 213, 209, 164, 78, 97, 181, 153, and maybe two more? Wait the standard deviation is given as 60.8. Let's use the formula for variance from standard deviation. Since standard deviation \( s = 60.8 \), then variance \( s^2=(60.8)^2=3696.64 \), and rounding to one decimal place, it is 3696.6 (wait no, 60.8 squared: 6060=3600, 600.8=48, 0.860=48, 0.80.8=0.64, so (60 + 0.8)^2=60^2+2600.8 + 0.8^2=3600+96 + 0.64=3696.64. So variance is 3696.6 (rounded to one decimal place) dollars².

Wait, maybe the data points are: 213, 209, 164, 78, 97, 181, 291, 153, and let's count the number of data points. Let's see: the numbers are 213, 209, 164, 78, 97, 181, 291, 153, and maybe 9? Wait no, the range is 183, so max - min=183. If max is 213, min is 30? No, the given range is 183. Wait the problem says "the range of the sample data is 183 dollars". So range = max - min=183. Let's find max and min from the data. Let's list the data: 213, 209, 164, 78, 97, 181, 291, 153, 9? No, 291 - 9=282≠183. Maybe the data is 213, 209, 164, 78, 97, 181, 153, and two more? Wait maybe I made a mistake. Alternatively, since standard deviation is given as 60.8, variance is standard deviation squared. So \( s^2=(60.8)^2 = 3696.64\approx3696.6 \) (rounded to one decimal place) dollars².

Answer:

\( 3696.6 \) dollars²