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
[tex]\text{Range} = 234-119 = 115\$[/tex]
Standard Deviation:
[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]
where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]Mean =\displaystyle\frac{1356}{8} = 169.5[/tex]
Sum of squares of differences = 4160.25 + 90.25 + 2550.25 + 1482.25 + 2352.25 + 1406.25 + 552.25 + 812.25 = 13406
[tex]\sigma = \sqrt{\dfrac{13406}{7}} = 43.76\$[/tex]
Variance =
[tex]\sigma^2 = 1915.142\$[/tex]
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.