Please help

The following scores were recorded on a 200-point final examination: 193,185,163,186,192,135,158,174,188,172,168,183,195,165,183.
A. Find the mean final examination score
B. Find the median final examination score.
C. Is the mean or median a more useful respentative of the final examination scores?

Here is a really good explanation -- not only how to find the mean and the median, but also which is a better (more useful) expression.

http://www.factmonster.com/ipka/A0001736.html

=)

A. To find the mean score, you need to sum up all the scores and divide it by the number of scores. Here's how you can do it:

1. Add up all the scores: 193 + 185 + 163 + 186 + 192 + 135 + 158 + 174 + 188 + 172 + 168 + 183 + 195 + 165 + 183 = 2722.
2. There are 15 scores, so divide the sum by 15: 2722 / 15 = 181.47.

So, the mean final examination score is approximately 181.47.

B. To find the median score, you need to arrange the scores in ascending order and find the middle value. Here's how you can do it:

1. Arrange the scores in ascending order: 135, 158, 163, 165, 168, 172, 174, 183, 183, 185, 186, 188, 192, 193, 195.
2. Since there are an odd number of scores (15 in this case), the median is the middle value. In this case, the 8th score is the median: 183.

So, the median final examination score is 183.

C. To determine whether the mean or median is a more useful representation of the final examination scores, consider the following:

- Mean: The mean gives you an overall average of the scores. It takes into account all the scores and calculates the average. However, in cases where there are extreme outliers, the mean can be greatly affected.
- Median: The median gives you the middle value of the scores, which can be useful to understand the central tendency of the data. It is less affected by outliers compared to the mean.

In this case, since the median is 183 and the mean is 181.47, there is not a significant difference between the two values. However, if there were extreme outliers or a skewed distribution, the median would be a more useful representation as it is less affected by those outliers.