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 45 . (round to one decimal place as needed.)

b. there is no mode.

b. which measure of center works best here?

a. the mean works best. the mode is highly affected by the largest value and the median is well to the left of the distributions center.

b. each center of measure works equally well.

c. the mode works best. the median is highly affected by the largest value and the mean is well to the left of the distributions center.

d. the median works best. the mean is highly affected by the largest value and the mode is well to the left of the distributions center.

Explanation:

Part a

Mean Calculation

Step1: Sum all data points

The data points are: 85, 63, 45, 48, 513, 33, 32, 289, 2391, 45, 368, 28, 86, 416, 50, 237, 125.
Sum = \(85 + 63 + 45 + 48 + 513 + 33 + 32 + 289 + 2391 + 45 + 368 + 28 + 86 + 416 + 50 + 237 + 125\)
First, add step - by - step:
\(85+63 = 148\); \(148 + 45=193\); \(193 + 48 = 241\); \(241+513 = 754\); \(754+33 = 787\); \(787+32 = 819\); \(819+289 = 1108\); \(1108+2391 = 3499\); \(3499+45 = 3544\); \(3544+368 = 3912\); \(3912+28 = 3940\); \(3940+86 = 4026\); \(4026+416 = 4442\); \(4442+50 = 4492\); \(4492+237 = 4729\); \(4729+125 = 4854\)

Step2: Divide by the number of data points (n = 17)

Mean=\(\frac{4854}{17}\approx285.5\)

Median Calculation

Step1: Order the data

Ordered data: 28, 32, 33, 45, 45, 48, 50, 63, 85, 86, 125, 237, 289, 368, 416, 513, 2391

Step2: Find the middle value (n = 17, so the 9th value)

The 9th value is 85, so median = 85.0

Mode Calculation

Step1: Identify the most frequent value

The value 45 appears twice, and all other values appear once. So the mode is 45.

Part b

The data has an extreme outlier (2391). The mean is affected by extreme values (since \(\bar{x}=\frac{\sum x}{n}\), a large outlier will increase the mean). The mode is 45, which is on the left - hand side of the data distribution. The median is the middle value and is less affected by extreme outliers. So the median works best because the mean is highly affected by the largest value and the mode is well to the left of the distribution's center.

Answer:

a. Mean: \(285.5\), Median: \(85.0\), Mode: \(45\)
b. D. The median works best. The mean is highly affected by the largest value and the mode is well to the left of the distribution's center.