We have the following scores:
52%, 85%, 89%, 90%, 83%, 89%
To calculate the median, we sort the scores from least to greatest:
52%, 83%, 85%, 89%, 89%, 90%.
As we have 6 scores, the median will be the average between the score that is in the third place and the score in the fourth place:
[tex]\text{Median}=\frac{x_3+x_4}{2}=\frac{85+89}{2}=\frac{174}{2}=87[/tex]The median is then 87%.
We can now calculate the mean as:
[tex]\begin{gathered} M_{}=\dfrac{1}{n}\sum ^n_{i=1}\, x_i \\ M=\dfrac{1}{6}(52+85+89+90+83+89) \\ M_{}=\dfrac{488}{6} \\ M_{}\approx81.3 \end{gathered}[/tex]The mean is 81.3%.
As 52% is an outlier that has a bigger effect in the mean than in the median, the median gives a better (higher) overall score for Nora's performance.
Also, the median is more representative of the set of data, as 52% can be considered an outlier.
Answer:
Median = 87%
Mean = 81.3%
The median gives a better overall score.
