k = 3 Inter-Cluster Distances Cluster 1 Cluster 2 Cluster 3 Cluster 1 0 5.147 6.078 Cluster 2 5.147 0 5.432 Cluster 3 6.078 5.432 0 Within-Cluster Summary Size Average Distance Cluster 1 62 3.355 Cluster 2 65 3.999 Cluster 3 51 3.483 Total 178 3.627 k = 4 Inter-Cluster Distances Cluster 1 Cluster 2 Cluster 3 Cluster 4 Cluster 1 0 5.255 6.070 4.853 Cluster 2 5.255 0 5.136 4.789 Cluster 3 6.070 5.136 0 6.074 Cluster 4 4.853 4.789 6.074 0 Within-Cluster Summary Size Average Distance Cluster 1 56 3.024 Cluster 2 45 3.490 Cluster 3 49 3.426 Cluster 4 28 4.580 Total 178 3.498Please answer all questions completely 1. k-Means Clustering of Wines. Amanda Boleyn, an entrepreneur who recently sold her start-up for a multi- million-dollar sum, is looking for alternate investments for her newfound fortune. She is considering an investment in wine, similar to how some people invest in rare coins and fine art. To educate herself on the properties of fine wine, she has collected data on 13 different characteristics of 178 wines. Amanda has applied k-means clustering to this data for k = 1, ..., 10 and generated the following plot of total sums of squared deviations. After analyzing this plot, Amanda generates summaries for k = 2, 3, and 4. Which value of k is the most appropriate to categorize these wines? Justify your choice with calculations. Sum of WithinSS Over Number of Clusters Sum(WithinSS) Diff previous Sum(WithinSS) 500 1000 1500 2000 Sum of WithinSS -O - 0 - X -=>= X 4 6 8 10 k = 2 Inter-Cluster Distances Cluster 1 Cluster 2 Cluster 1 5.640 Cluster 2 5.640 0 Within-Cluster Summary Size Average Distance Cluster 1 87 4.003 Cluster 2 91 4.260 Total 178 4.134