Answer:
Range = 115$
Standard Deviation = 43.76$
Variance = 1915.142$
Option 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.
Step-by-step explanation:
We are given the data for prices in dollars for one night at different hotels in a certain region.
234, 160, 119, 131, 218, 207, 146, 141
Range:
Sorted data: 119, 131, 141, 146, 160, 207, 218, 234
![\text{Range} = 234-119 = 115\$](https://tex.z-dn.net/?f=%5Ctext%7BRange%7D%20%3D%20234-119%20%3D%20115%5C%24)
Standard Deviation:
where
are data points,
is the mean and n is the number of observations.
Sum of squares of differences = 4160.25 + 90.25 + 2550.25 + 1482.25 + 2352.25 + 1406.25 + 552.25 + 812.25 = 13406
![\sigma = \sqrt{\dfrac{13406}{7}} = 43.76\$](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%7B%5Cdfrac%7B13406%7D%7B7%7D%7D%20%3D%2043.76%5C%24)
Variance =
![\sigma^2 = 1915.142\$](https://tex.z-dn.net/?f=%5Csigma%5E2%20%3D%201915.142%5C%24)
Measure of variance for someone searching for room:
Option 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.