Answer
$Q_1=9.15~g/dL$
$Q_2=9.95~g/dL$
$Q_3=11.1~g/dL$
Work Step by Step
The values in ascending order: 5.7, 7.7, 7.8, 8.7, 8.9, 9.4, 9.5, 9.6, 9.6, 9.9, 10.0, 10.3, 10.6, 10.7, 11.0, 11.2, 11.7, 12.9, 13.0, 13.4
There are 20 observations. The second quartile (median) is the mean between the 10th and the 11th observations:
$Q_2=\frac{9.9+10.0}{2}=9.95$
The bottom half of the data: 5.7, 7.7, 7.8, 8.7, 8.9, 9.4, 9.5, 9.6, 9.6, 9.9. $Q_1$ is the median of these values:
$Q_1=\frac{8.9+9.4}{2}=9.15$
The top half of the data: 10.0, 10.3, 10.6, 10.7, 11.0, 11.2, 11.7, 12.9, 13.0, 13.4. $Q_3$ is the median of these values:
$Q_3=\frac{11.0+11.2}{2}=11.1$
