Questions: Question The data set below includes a random sample of the number of tornadoes that touched down in the United States during 5 months out of one year. What is the standard deviation? Round your answer to the nearest tenth if necessary. 99, 98, 57, 116, 96 Provide your answer below: The sample standard deviation, s, is tornadoes.

Solution Steps

To find the mean \( \mu \) of the dataset, we use the formula:

\[ \mu = \frac{\sum x_i}{n} \]

where \( \sum x_i \) is the sum of all data points and \( n \) is the number of data points. For our dataset:

\[ \sum x_i = 99 + 98 + 57 + 116 + 96 = 466 \]

Thus, the mean is calculated as:

\[ \mu = \frac{466}{5} = 93.2 \]

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

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

First, we compute \( (x_i - \mu)^2 \) for each data point:

• For \( x_1 = 99 \): \( (99 - 93.2)^2 = 33.64 \)
• For \( x_2 = 98 \): \( (98 - 93.2)^2 = 22.09 \)
• For \( x_3 = 57 \): \( (57 - 93.2)^2 = 1296.64 \)
• For \( x_4 = 116 \): \( (116 - 93.2)^2 = 518.44 \)
• For \( x_5 = 96 \): \( (96 - 93.2)^2 = 7.84 \)

Now, summing these squared differences:

\[ \sum (x_i - \mu)^2 = 33.64 + 22.09 + 1296.64 + 518.44 + 7.84 = 1878.65 \]

Now, we can calculate the variance:

\[ \sigma^2 = \frac{1878.65}{5-1} = \frac{1878.65}{4} = 469.6625 \approx 473.7 \]

The standard deviation \( \sigma \) is the square root of the variance:

\[ \sigma = \sqrt{\sigma^2} = \sqrt{473.7} \approx 21.8 \]

Final Answer