Question

question
given the following list of values, is the mean or the median likely to be a better measure of the center of the data set?
31, 32, 31, 34, 30, 30, 31, 31, 30, 31

select the correct answer below:
○ mean
○ median
feedback more instruction submit
content attribution

Explanation:

Step1: Analyze data distribution

The data set is: 30, 30, 30, 31, 31, 31, 31, 31, 32, 34. Most values are around 30 - 31, with one value (34) slightly higher.

Step2: Compare mean and median sensitivity

Mean is sensitive to outliers (or extreme values), but here 34 is not a strong outlier, and the data is relatively symmetric around 31. However, in cases with minimal skew (like this, since most values cluster), mean and median are close, but mean is still a good measure here as there's no extreme outlier distorting it. Wait, actually, let's check the values again. The data has 30 (three times), 31 (five times), 32, 34. The mean would be calculated as (330 + 531 + 32 + 34)/10 = (90 + 155 + 32 + 34)/10 = (90+155=245; 245+32=277; 277+34=311)/10 = 31.1. The median: since there are 10 values, the median is the average of the 5th and 6th values. Sorted data: 30,30,30,31,31,31,31,31,32,34. 5th is 31, 6th is 31, so median is 31. The mean is 31.1, very close to median. But the key is: when data has no strong outliers (extreme values that pull mean away), mean is a good measure. Here, 34 is not an extreme outlier (only 3 more than 31), so mean is a better measure (or equally good, but since the options are mean or median, and the data is not skewed heavily, mean is appropriate). Wait, maybe I made a mistake. Wait, the data: let's check the spread. The values are 30,30,30,31,31,31,31,31,32,34. The mean is 31.1, median 31. The difference is small. But in general, for symmetric or near - symmetric data (which this is, as most values are around 31, with 30s and 32,34 not too far), mean is a good measure of center. So the answer should be Mean.

Answer:

Mean