Question

A quality product is one that is consistent and has very little variability in its characteristics. Controariability can be more difficult with agricultural products than with those that are manufactured. The following table gives the weights, in ounces, of the 25 potatoes sold in a 10-pound bag. 7.7 7.9 8.1 6.9 6.7 7.8 7.9 7.8 7.5 7.8 7.0 4.8 7.6 6.3 4.8 4.7 4.8 6.2 6.1 5.2 4.4 7.8 5.1 6.1 3.8 (a) Make a stemplot for these data. (Split each stem in two with leaves ranging from 0 to 4 on the first stem and 5 to 9 on the second. Enter numbers from smallest to largest separated by spaces. Enter NONE for stems with no values.) 6 1123 6 79 7 1567888899Find the numerical summary. (Round your answers to three decimal places.) 432 standard deviation 1.388 minimum Q. median 95 maximum (b) Do you think that your numerical summaries do an effective job of describing these data? Why or why not? O Yes: these data were given together so one set of numerical summaries is sufficient. O No: the range of these data is too large for one set of numerical summaries to be sufficient. ONo: these data are clustered so one set of numerical summaries is not sufficient O No: the standard deviation is too large for one set of numerical summaries to be sufficient. (c) There appear to be two distinct clusters of weights for these potatoes. Divide the sample into two subsamples based on the clustering. Summarize the data numerically. (Round your answers to three decimal places.) Smaller Weights Larger Weights tandard deviation tandard deviation Q1 median median maximum maximum Do you think that this way of summarizing these data is better than a numerical summary that uses all the data as a single sample? Give a reason for your answer

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Answer #1

(c)
Based on stem plot, data can be divided into 2 clusters.
1. All values below 5.5 (total 8 values)
2. All values above 5.5 (total 17 values)

For Smaller weights, the data is 3.8,4.4,4.7,4.8,4.8,4.8,5.1,5.2

Mean = 4.7

Standard deviation = 0.438

Minimum = 3.8

Q1 = 4.625

Median = 4.8

Q3 = 4.875

Maximum = 5.2

For Larger weights, the data is 6.1,6.1,6.2,6.3,6.7,6.9,7.0,7.5,7.6,7.7,7.8,7.8,7.8,7.8,7.9,7.9,8.1

Mean = 7.247

Standard deviation = 0.720

Minimum = 6.1

Q1 = 6.7

Median = 7.6

Q3 = 7.8

Maximum = 8.1

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