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?
208 162 236 209 222 77 268 185
the range of the sample data is
(round to one decimal place as needed.)
the standard deviation of the sample data is
(round to one decimal place as needed.)
the variance of the sample data is
(round to one decimal place as needed.)
how useful are the measures of variation for someone searching for a room?
a. the measures of variation are very useful because a person does not want to buy a room where the variation is too high.
b. the measures of variation are not very useful because when searching for a room, low prices, location, and good accommodations are more important than the amount of variation in the area.
c. the measures of variation are very useful because a person does not want to buy a room where the variation is too low.

Explanation:

Step1: Find the range

Range = Max - Min. The data set is \(208,162,236,209,222,77,268,185\). Max = 268, Min = 77. So Range = \(268 - 77=191.0\) dollars.

Step2: Calculate the mean

\(\bar{x}=\frac{208 + 162+236+209+222+77+268+185}{8}=\frac{1567}{8}=195.875\) dollars.

Step3: Calculate the squared - differences

\((208 - 195.875)^2=(12.125)^2 = 147.015625\), \((162-195.875)^2=(- 33.875)^2 = 1147.515625\), \((236 - 195.875)^2=(40.125)^2 = 1610.015625\), \((209 - 195.875)^2=(13.125)^2 = 172.265625\), \((222 - 195.875)^2=(26.125)^2 = 682.515625\), \((77 - 195.875)^2=(-118.875)^2 = 14130.640625\), \((268 - 195.875)^2=(72.125)^2 = 5201.015625\), \((185 - 195.875)^2=(-10.875)^2 = 118.265625\).

Step4: Calculate the variance

The sample variance \(s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}\), where \(n = 8\). \(\sum_{i=1}^{8}(x_{i}-\bar{x})^{2}=147.015625+1147.515625+1610.015625+172.265625+682.515625+14130.640625+5201.015625+118.265625 = 23209.25\). \(s^{2}=\frac{23209.25}{7}\approx3315.6\) (rounded to one decimal place) dollars².

Step5: Calculate the standard deviation

The sample standard deviation \(s=\sqrt{s^{2}}=\sqrt{3315.6}\approx57.6\) (rounded to one decimal place) dollars.

Step6: Evaluate the usefulness of measures of variation

The measures of variation are very useful because a person does not want to buy a room where the variation is too high. High variation means there is a large spread in prices, and a person may be able to find a better - priced room if they know the spread.

Answer:

The range of the sample data is 191.0 dollars.
The standard deviation of the sample data is 57.6 dollars.
The variance of the sample data is 3315.6 dollars².
A. The measures of variation are very useful because a person does not want to buy a room where the variation is too high.