Answer:

Now we can find the limits in order to determine outliers like this:


So for this case the left boundary would be 3, if a value is lower than 3 we consider this observation as an outlier
b. 3
Step-by-step explanation:
For this case we have the following summary:
represent the minimum value
represent the first quartile
represent the median
represent the third quartil
represent the maximum
If we use the 1.5 IQR we need to find first the interquartile range defined as:

Now we can find the limits in order to determine outliers like this:


So for this case the left boundary would be 3, if a value is lower than 3 we consider this observation as an outlier
b. 3