If we assume that these data represent measurement along a continuum...
But it won't make any difference in conclusions even if we use the formula: Median = L + W x (N/2 - C)/F
where:
L: the lower class boundary of median class
W: the width of median class
N/2: the sample size divided by 2
F: the frequency of median class
C: the previous cumulative frequency of the median class
Joubert PE median = 8.31 = Hanyu PE median
Joubert CH median = 8.31 = Hanyu CH median
Joubert IN median = 8.45 > Hany IN median = 8.33
Joubert median total > Hany median total

(Yes, I should have said Joubert got a higher median than Hanyu in IN and tied with Hanyu in the other two categories).
Was there a French Judge on the panel?
No. The judges were: Evgeni ROKHIN (UZB), Karin EHRHARDT (Austria), Ekaterina SEROVA (BLR), Eddy WU (TPE), Hailan JIANG (CHN), Ebru ANILDI (TUR), Sung-Hee KOH (KOR), Elena FOMINA (RUS), Inger ANDERSSON (SWE).
That 9.50 that Joubert got for interpretation stands out like a sore thumb. It is 2.34 standard deviations above the mean.
Are you assuming a normal distribution in scores since the concept of standard deviation and mean is used in your identification of outliers? If the judges were calibrated to apply similar criteria in assessing the performance, and if no more than, say, 1-point discrepancy among judges was the threshold of their training qualification, then it was Hanyu's 9.50 (i.e., an absolute deviation of 1.25 from the median) that would be considered a rogue score. It was likely contaminated because the judge apparently did not apply the same criteria as other judges and it had a "gaming" effect in a small sample size--able to pull up the favorite by saving the adjacent high score from trimming (if the highest and the lowest are trimmed) or by inflating the non-robust standard deviation so that some high scores would fall below the cutoff (if the trimming is determined by standard deviation). Even if we assume a normal distribution of the scores and ignore the limitation of the sample size and the unknown distribution of measurement error, why do we use μ + 2σ (95.45% confidence interval) rather than μ + 3σ (99.73% confidence interval, which is employed by the International Tchaikovsky Competition as the criterion for identifying rogue scores)? Roughly 5% of the scores (or 1 in 20 observed scores) will fall outside the 2-standard deviation cutoff, but those "outliers" may still very well be part of the "normal" distribution. Why do we automatically reject a score greater than 2SD rather than take it as a possibly legitimate opinion from the judge and winsorize it to the nearest acceptable score?
Of course, if we assume a normal distribution and use Dixon's Q test, the 9.50 that Joubert received would look even worse: Q = gap/range = 1/1.5 = 0.67 > Q99% = 0.598
