Questions: Checkout times. The checkout times (in minutes) for 12 randomly selected customers at a large supermarket during the store's busiest time are: 5.3, 5.6, 11.6, 6.7, 8.6, 13.1, 13.8, 7.2, 7.6, 5.8, 13.9, 13.9 a. Find the mean of the checkout times. 9.43 mins (Type an integer or a decimal. Round to two decimal places.) b. Find the standard deviation of the checkout times. mins (Type an integer or a decimal. Round to two decimal places.)

Solution Steps

To find the mean of the checkout times, we use the formula:

\[ \mu = \frac{\sum_{i=1}^N x_i}{N} \]

where \( N \) is the number of customers and \( x_i \) are the checkout times. For our data:

\[ \mu = \frac{113.1}{12} = 9.43 \]

Thus, the mean checkout time is \( 9.43 \) mins.

To find the standard deviation, we first calculate the variance using the formula:

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

Substituting the values, we find:

\[ \sigma^2 = 12.6 \]

The standard deviation is then calculated as:

\[ \sigma = \sqrt{12.6} \approx 3.55 \]

Thus, the standard deviation of the checkout times is \( 3.55 \) mins.

Final Answer