Questions: Question 14, 3.2.11 HW Score: 47.41%, 16.12 of 34 points Points: 0.71 of 1 Listed below are prices in dollars per night of stay at hotels. 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? 142 234 184 252 223 197 262 The range of the sample data is = (Round to one decimal place as needed.) Clear all Check answer

Solution Steps

The range of a dataset is the difference between the maximum and minimum values. For the given hotel prices:

\[ \text{Range} = \max(142, 234, 184, 252, 223, 197, 262) - \min(142, 234, 184, 252, 223, 197, 262) = 262 - 142 = 120 \]

Thus, the range of the sample data is \(\boxed{120.0}\).

The mean \(\mu\) of a dataset is calculated as the sum of all data points divided by the number of data points:

\[ \mu = \frac{\sum x_i}{n} = \frac{142 + 234 + 184 + 252 + 223 + 197 + 262}{7} = \frac{1494}{7} = 213.4 \]

The sample variance \(\sigma^2\) is calculated using the formula:

\[ \sigma^2 = \frac{\sum (x_i - \mu)^2}{n-1} \]

Substituting the values:

\[ \sigma^2 = \frac{(142 - 213.4)^2 + (234 - 213.4)^2 + (184 - 213.4)^2 + (252 - 213.4)^2 + (223 - 213.4)^2 + (197 - 213.4)^2 + (262 - 213.4)^2}{6} = 1766.6 \]

Thus, the variance of the sample data is \(\boxed{1766.6}\).

The standard deviation is the square root of the variance:

\[ \text{Standard Deviation} = \sqrt{\sigma^2} = \sqrt{1766.6} = 42.0 \]

Thus, the standard deviation of the sample data is \(\boxed{42.0}\).

Final Answer