The 11 members of the swim team completed the following numbers of laps at today's practice. \[ 76,76,78,80,81,82,83,84,85,87,90 \] Identify all values that are outliers. If there is more than one outlier, separate them with commas. If there are no outliers, click on "None".

We first identify the median ( \(Q_2\) ), first quartile ( \(Q_1\) ), and third quartile ( \(Q_3\) ) of the dataset. The data in sorted order is: \[ 76,\; 76,\; 78,\; 80,\; 81,\; 82,\; 83,\; 84,\; 85,\; 87,\; 90 \] Since there are 11 data points, the median is the \(6^{\text{th}}\) term: \[ Q_2 = 82 \] The lower half (not including the median) is: \[ 76,\; 76,\; 78,\; 80,\; 81 \] So, the first quartile is the median of these \(5\) values: \[ Q_1 = 78 \] The upper half (not including the median) is: \[ 83,\; 84,\; 85,\; 87,\; 90 \] So, the third quartile is the median of these \(5\) values: \[ Q_3 = 85 \] The interquartile range (IQR) is: \[ \text{IQR} = Q_3 - Q_1 = 85 - 78 = 7 \] Next, we calculate the fences for outliers. An observation is an outlier if it is below: \[ Q_1 - 1.5 \times \text{IQR} = 78 - 1.5 \times 7 = 78 - 10.5 = 67.5 \] or above: \[ Q_3 + 1.5 \times \text{IQR} = 85 + 1.5 \times 7 = 85 + 10.5 = 95.5 \] Now, we check the dataset: all values lie between \(67.5\) and \(95.5\). Therefore, there are no outliers. None