Answer
a) $Q_1=76; Q_2=80.5; Q_3=101$
b) $IQR=25$
Work Step by Step
a) We are given the data:
54, 63, 68, 76, 78, 79, 80, 81, 84, 96, 101, 110, 121, 154
Determine $Q_2$: there are 14 numbers, $Q_2$ is the average between the 7th and the 8th numbers:
$Q_2=\dfrac{80+81}{2}=80.5$
Determine $Q_1$: there are 7 numbers on the left of $Q_2$, their median in the 4th number:
$Q_1=76$
Determine $Q_3$: there are 7 numbers on the right of $Q_2$, their median in the 11th number:
$Q_3=101$
b) Determine $IQR$:\\
$IQR=Q_3-Q_1=101-76=25$
Unlike the range, the IQR is not affected by outliers, therefore IQR is more useful than the range to compare the variability for distributions that are very highly skewed or that have severe outliers.
$1.5\times IQR=1.5\times 25=37.5$
$Q_1-1.5\times IQR=76-37.5=38.5$
$Q_3+1.5\times IQR=101+37.5=138.5$
This means 154 is an outlier, therefore the IQR is useful therefore IQR is more useful than the range to compare the variability for the distribution.
