Answered: The first pair and the second pair look… | bartleby

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7]: fig, ax = plt.subplots()
data_d.plot.hist(density=False,
ax=ax, title='Histogram: Set1 and Setl samples vs. Set2 and Set2 samples', bins=40)
data.plot.hist (density=False, ax=ax, bins=40)
ax.set_ylabel('Count')
ax.grid(axis='y')
Count
Histogram: Set1 and Set1 samples vs. Set2 and Set2 samples
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Use boxplots to compare the four sets. Discuss their differences.
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8]: fig = plt.figure(figsize =(10, 7))
plt.boxplot ([set1, set1_s, set2, set2_s],1, '')
plt.show()
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The first pair and the second pair look similar while the two pairs look differnet, right? The question is how can we KNOW if two sets are truly (significantly) different or not?

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It is straightforward and efficient to compute the union of two sets using Boolean values. We may create a new union set by Oring the matching items of the two BitArrays since the union of two sets is a combination of the members of both sets. At other words, if the value in the corresponding place of either BitArray is True, a member is added to the new set.Computing the intersection of two sets is analogous to computing the union, except that the And operator is used instead of the Or operator Using the same technique we used to detect the difference, we can determine if one set is a subset of another. For example, if:setA(index) && !(setB(index))evaluates to False then setA is not a subset of setB.The BitArray Set ImplementationWrite The code for a CSet class based on a BitArray.
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A collection cannot be represented by a linked list. This is a set data structure that has absolutely no values. There are no similarities between the two sets. Sets are relevant to mathematical set functions like union and intersection.
Im relooking at the problem rn, im starting to think that maybe it should be preorder instead of post order. How would the coding implemention change?