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.