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# Helppppp

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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?
Median: _______________
Mean: ______________

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?

Dec 16, 2020

#1
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The median is 12.3 and the avarage is 12498682.1052 so the answer is 153733789.894

Dec 16, 2020
#2
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Put these in order from least to greatest

9.9 , 10.8, 11.5, 12.3, 12.8,13.1, 34.5

a)

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

Dec 16, 2020