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? 209 164 107 255 277 291 148 250 The range of the sample data is 1 dollars. dollars. nights squared . nights. dollars squared . (Round to one decimal place as needed.) The standard deviation of the sample data is 2 dollars squared . dollars squared . nights squared . dollars. nights. (Round to one decimal place as needed.) The variance of the sample data is 0 ▼ dollars. dollars squared . nights. nights squared . (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 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. B. The measures of variation are very useful because a person does not want to buy a room where the variation is too high. C. The measures of variation are very useful because a person does not want to buy a room where the variation is too low. D. The measures of variation are not very useful because the values are nominal data that do not measure or count anything, so the resulting statistics are meaningless.

Answer #1

Range of the data = Max - Min = 291 - 107 = 184

Variance of the data = 1/N-1 * Sum(Xi - )^{2}
= 4455.7

Std deviation of the data = SQRT(Variance) = 66.8

The measures of variation here tell us that the amount of
variability in the room prices in the region are quite high (as
high as 66.8 $). This suggests that the quality of living in the
area cannot be judged precisely, making it difficult to shortlist a
certain room. The customer might as well ignore this area given the
high amount of variability and hence unreliability. Hence,
**B** is the correct option.

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