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?
233
226
239
279
229
293
216
182
the range of the sample data is 111.0 dollars
(round to one decimal place as needed.)
the standard deviation of the sample data is 35.0 dollars.
(round to one decimal place as needed.)
the variance of the sample data is ▼
(round to one decimal place as needed.)

Explanation:

Step1: Calculate the mean

First, find the sum of the data values: \(233 + 226+239 + 279+229+293+216+182=\sum x = 1997\).
The number of data points \(n = 8\).
The mean \(\bar{x}=\frac{\sum x}{n}=\frac{1997}{8}=249.625\).

Step2: Calculate the variance

The formula for sample variance \(s^{2}=\frac{\sum(x - \bar{x})^{2}}{n - 1}\).
\((233 - 249.625)^{2}=(- 16.625)^{2}=276.390625\)
\((226 - 249.625)^{2}=(-23.625)^{2}=558.140625\)
\((239 - 249.625)^{2}=(-10.625)^{2}=112.890625\)
\((279 - 249.625)^{2}=(29.375)^{2}=863.984375\)
\((229 - 249.625)^{2}=(-20.625)^{2}=425.390625\)
\((293 - 249.625)^{2}=(43.375)^{2}=1881.390625\)
\((216 - 249.625)^{2}=(-33.625)^{2}=1130.640625\)
\((182 - 249.625)^{2}=(-67.625)^{2}=4572.140625\)

\(\sum(x - \bar{x})^{2}=276.390625+558.140625 + 112.890625+863.984375+425.390625+1881.390625+1130.640625+4572.140625 = 9821.969375\)

\(s^{2}=\frac{9821.969375}{8 - 1}=\frac{9821.969375}{7}\approx1403.1\)

Answer:

The variance of the sample data is \(1403.1\) dollars.