Questions: The accompanying data represent the actual amount (in mL) poured into a short, wide glass for individuals asked to pour 1.5 ounces (44.3 mL). 89.1, 68.9, 32.8, 37.2, 39.8, 46.6, 66.1, 79.3, 66.3, 52.3, 47.2, 64.1, 53.6, 63.1, 46.1, 63.0, 92.2, 57.6 (a) Calculate the values of the mean and standard deviation (in mL). (Round your answers to four decimal places.) mean 61.7444 mL standard deviation -118.0580

Calculate the mean of the poured amounts in mL.

Mean calculation formula.

The mean \( \mu \) is calculated using the formula: \[ \mu = \frac{\sum_{i=1}^N x_i}{N} \] where \( \sum_{i=1}^N x_i = 1065.3 \) and \( N = 18 \).

Substituting the values into the formula.

Substituting the values gives: \[ \mu = \frac{1065.3}{18} = 59.1833 \]

The mean is \( \boxed{59.1833} \).

Calculate the standard deviation of the poured amounts in mL.

Variance calculation formula.

The variance \( \sigma^2 \) is calculated using the formula: \[ \sigma^2 = \frac{\sum (x_i - \mu)^2}{n-1} \] where \( \sigma^2 = 279.1768 \).

Standard deviation calculation.

The standard deviation \( \sigma \) is the square root of the variance: \[ \sigma = \sqrt{279.1768} = 16.7086 \]

The standard deviation is \( \boxed{16.7086} \).

The mean of the poured amounts is \( \boxed{59.1833} \) and the standard deviation is \( \boxed{16.7086} \).