Question

the ages (in years) of a random sample of shoppers at a gaming store are shown. determine the range, mean, variance, and standard deviation of the sample data set.

12, 21, 23, 14, 16, 18, 19, 17, 13, 16

the range is 11.
(simplify your answer.)

the mean is 16.9.
(simplify your answer. round to the nearest tenth as needed.)

the variance is
(simplify your answer. round to the nearest tenth as needed.)

Explanation:

Identify the given sample data

Using the Range Calculation and Mean Calculation knowledge points
The given sample data set of shopper ages is:

$$ \{12, 21, 23, 14, 16, 18, 19, 17, 13, 16\} $$

The sample size is \(n = 10\).

Calculate the sample mean

Using the Mean Calculation knowledge point

$$ \bar{x} = \frac{\sum x_i}{n} = \frac{12 + 21 + 23 + 14 + 16 + 18 + 19 + 17 + 13 + 16}{10} = \frac{169}{10} = 16.9 $$

Compute squared deviations from the mean

We calculate \((x_i - \bar{x})^2\) for each data point:

• \((12 - 16.9)^2 = (-4.9)^2 = 24.01\)
• \((21 - 16.9)^2 = (4.1)^2 = 16.81\)
• \((23 - 16.9)^2 = (6.1)^2 = 37.21\)
• \((14 - 16.9)^2 = (-2.9)^2 = 8.41\)
• \((16 - 16.9)^2 = (-0.9)^2 = 0.81\)
• \((18 - 16.9)^2 = (1.1)^2 = 1.21\)
• \((19 - 16.9)^2 = (2.1)^2 = 4.41\)
• \((17 - 16.9)^2 = (0.1)^2 = 0.01\)
• \((13 - 16.9)^2 = (-3.9)^2 = 15.21\)
• \((16 - 16.9)^2 = (-0.9)^2 = 0.81\)

Sum the squared deviations

We sum these squared differences:

$$ \sum (x_i - \bar{x})^2 = 24.01 + 16.81 + 37.21 + 8.41 + 0.81 + 1.21 + 4.41 + 0.01 + 15.21 + 0.81 = 108.9 $$

Calculate the sample variance

We divide the sum of squared deviations by \(n - 1\) to find the sample variance \(s^2\):

$$ s^2 = \frac{\sum (x_i - \bar{x})^2}{n - 1} = \frac{108.9}{9} = 12.1 $$

Answer:

The variance is <blank>12.1</blank>