Exercise 2 a=c(21,36,42,24,25,36,35,200,32) mean(a)  tmean(a)  median(a)  35 Sample mean is not resistant as its value largely inflated with a change of a single value.
Exercise 3 The resistance of the 20% trimmed mean is 0.2, meaning that a change in more than 20% of the observations are required to generate a large change in its value. In this case n=9, so 9×0.2=1.8, rounded down to the nearest integer, =1. This means that a large change in the 20% trimmed mean value requires a change of at least 2 observations.
Exercise 4 The resistance of the Median is 0.5, meaning that a change in more than 50% of the observations are required to generate a large change in its value. In this case n=9, so 9×0.5=4.5. This means that a large change in the median requires a change of at least 5 observations.
Exercise 10 d=(21,36,42,24,25,36,35,102,32) winvar(d)  Winsorized variance remained the same.
Exercise 11 Yes, because we shift the extreme values closer to the mean. This reduces the dispersion in the data. The mean squared distances from the mean decreases accordingly.
Exercise 12 The variance has s sample breakdown point of 1/n, thus a single observation can render it value arbitrarily large or small.
Exercise 13 The sample breakdown point of the 20% Winsorized variance is 0.2. In the case of n=25, this would be 25×0.2= 5. Thus, we need at to change at least 6 observation to render the Winsorized variance arbitrarily large.
Exercise 22 3 outliers (none detected) 2 outliers detected (one is masked on graph) print(boxplot(g,plot=F)$out)  print(boxplot(g,plot=F)$out) numeric(0)
Exercise 23 The boxplot has a sample break down point of 0.25%. The number of outliers it detects does not exceed 25% of the sample. For example, when we had 3 outliers with n=10, all outliers disappeared.