Answer
$s=115~min$
$IQR=56.5~min$
IQR is resistant and the standard deviation is not.
Work Step by Step
In MINITAB, change the value of 346 by 0 and proceed just like in the previous problem to find the standard deviation.
$s=115$
New data in ascending order: 0, 345, 358, 372, 429, 437, 442, 442, 461, 466, 466, 470, 471, 480, 489, 490, 505, 515, 516, 549
The second quartile is equal to the median. Since we have 20 observations, the median is the mean between the 10th and the 11th observations in ascending order:
$Q_2=\frac{466+466}{2}=466$
The bottom half of the data: 0, 345, 358, 372, 429, 437, 442, 442, 461, 466. $Q_1$ is the median of these values. Since there are 10 observations, the median is the mean between the fifth and the sixth observations in ascending order:
$Q_1=\frac{429+437}{2}=433$
The top half of the data: 466, 470, 471, 480, 489, 490, 505, 515, 516, 549. $Q_3$ is the median of these values. Since there are 10 observations, the median is the mean between the fifth and the sixth observations in ascending order:
$Q_3=\frac{489+490}{2}=489.5$
$IQR=Q_3-Q_1=489.5-433=56.5$
