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the data below are the frequency of cremation burials found in 17 archa…

Question

the data below are the frequency of cremation burials found in 17 archaeological sites.
a. obtain the mean, median, and mode of these data.
b. which measure of center do you think works best here?
85 63 45 48 513 33 32 289 2391
45 368 28 86 416 50 237 125

a. the mean is 285.5 (round to one decimal place as needed.)
the median is 85.0 (round to one decimal place as needed.)
select the correct choice below and fill in any answer boxes within your choice.
a. the mode is \boxed{}. (round to one decimal place as needed.)
b. there is no mode.

Explanation:

Response
Part a: Mode Calculation

To find the mode, we check the frequency of each data point. The data set is: \(85, 63, 45, 48, 513, 33, 32, 289, 2391, 45, 368, 28, 86, 416, 50, 237, 125\).

Looking at each value:

  • \(28\): appears 1 time
  • \(32\): appears 1 time
  • \(33\): appears 1 time
  • \(45\): appears 2 times
  • \(48\): appears 1 time
  • \(50\): appears 1 time
  • \(63\): appears 1 time
  • \(85\): appears 1 time
  • \(86\): appears 1 time
  • \(125\): appears 1 time
  • \(237\): appears 1 time
  • \(289\): appears 1 time
  • \(368\): appears 1 time
  • \(416\): appears 1 time
  • \(513\): appears 1 time
  • \(2391\): appears 1 time

The value \(45\) appears twice, which is more frequent than any other value (all others appear once).

Brief Explanations

The data set has extreme values (like \(2391\) and \(513\)) which can skew the mean. The median is less affected by outliers, but here we also have a mode (45) which appears twice. However, looking at the data, there are outliers (very large values) that pull the mean up. The median is a better measure here? Wait, no—wait, the mean was calculated as 285.5, median as 85.0. Wait, actually, let's re - evaluate. The data has a few very large values (2391, 513, 416, 368, 289, 237). These are outliers. The mean is sensitive to outliers, the median is not. But also, the mode is 45, which is a low value. Wait, but let's check the data distribution. The data points: 28, 32, 33, 45, 45, 48, 50, 63, 85, 86, 125, 237, 289, 368, 416, 513, 2391. When we order them, the middle value (9th value) is 85 (median). The mean is pulled up by the large values (2391, 513 etc.). The mode is 45 (appears twice). But which measure works best? Since there are extreme outliers (like 2391 which is much larger than most other values), the mean is distorted. The median is a better measure of center because it is resistant to outliers. Alternatively, the mode? Wait, no, the mode is 45, but most values are above 45. Wait, maybe the median. Wait, let's think again. The mean is 285.5, median 85, mode 45. The data has a few very large values, so the mean is not representative. The median is better because it's not affected by the extreme values. Or is the mode appropriate? No, because the mode is just the most frequent, but it's a low value and only appears twice. So the median is a better measure of center here because it is resistant to the influence of the extreme outliers (like 2391) that skew the mean.

Answer:

A. The mode is \(45.0\) (Round to one decimal place as needed.)

Part b: Best Measure of Center