The following times, in seconds, were recorded for a mouse to run through a
maze: 34.5, 12.8, 10.8, 11.5, 12.3, 13.1, 9.9
a) What are the median and mean times?
b) Which measure of central tendency best represents the data?
c) Identify any possible outlier(s). Should the outlier(s) be removed from the
data set? Explain why or why not.
d) How would removing the outlier(s) affect the median and the mean?
The median is 12.3 and the avarage is 12498682.1052 so the answer is 153733789.894
Put these in order from least to greatest
9.9 , 10.8, 11.5, 12.3, 12.8,13.1, 34.5
Median = in an odd number of values, the median is the middle value = 12.3
Mean .add the values and divide by 7......this gives us ≈ 14.99
b) The median is better because the value of 34.5 tends to raise the average score
c) Outliers - 34.5 appears to be the outlier......this should be removed from the data set....it tends to exaggerate the mean
d) Without the outlier
Median = ( 11.5 + 12.3)/2 = 11.9
Mean ≈ 11.73
Note that the median isn't affected much, but the mean drops by more than 3