Branson Creek Middle School has decided to make fitness a key message to their students in the upcoming school year. As a result, they will be participating in a national fitness program. To participate, they must randomly select 15 students in the 5th grade and record their exercise time each day. The data (in minutes) are shown.
85, 80, 76, 78, 82, 88, 80, 80, 110, 85, 85, 82, 83, 88, 76
A. Describe the distribution of the data.
B. Determine the median and mean of the data.
C. Which measure better represents a typical value in the data set? Explain.
A. The distribution of the data appears to be roughly symmetric and unimodal, with a possible outlier at 110 minutes.
B. Median: To find the median, we first need to order the data from least to greatest: 76, 76, 78, 80, 80, 80, 82, 82, 83, 85, 85, 85, 88, 88, 110. The middle value is the 8th value (82), since there are an odd number of values. Therefore, the median exercise time is 82 minutes.
Mean: To find the mean, we add up all the exercise times and divide by the number of students: (85+80+76+78+82+88+80+80+110+85+85+82+83+88+76) / 15 = 83.2. Therefore, the mean exercise time is 83.2 minutes.
C. Both the median and mean can be used to represent a typical value in the data set, but it depends on the context. If we want a value that is not affected by outliers, the median would be a better choice. In this case, the outlier of 110 minutes could greatly affect the mean, making it higher than what most students typically exercise. However, if we want a value that takes into account all the data points, including the outlier, the mean would be a better choice. It is important to consider both measures and the context in which they are being used to determine which is more appropriate.