(i)
First we need to know n for each country, the number of insects measured.
Fortunately we're given the leaf count for each stem.
For X, n=10+18+16+16+11 = 71
We index them, meaning we number them from smallest to largest, 1 to 71. The middle one, the median, is .5*1 + .5*71 = 36 (that's its index). The 25 percentile corresponds to .75*1 + .25*71 = 18.5 and the 75th percentile is .25*1+.75*71=53.5.
So the median is the 36th number. We get 10+18=28 in the first two rows, so is the 8th number in the third row (from the smallest). That's a median of 0.825.
When the index isn't an integer we can interpolate (that's what we do for the median) or just take the first data point in the interquartile from 25%-75%, which is what I'll do here. That's index 19, rounding up 18.5. We count the 10 in the first row and nine more in the second row, so 25%ile is 0.815.
By symmetry we can count 19 from the top, 11 in row 84, so 8 more in, landing on .833 for the 75%ile.
That's an IQR of 0.833 - 0.815 = 0.018
Answer: Median 0.825 cm, IQR 0.018 cm
(ii)
This is a finicky one; q can be 3 or 4, meaning 0.823 or 0.824 and r can be 2 or 3, meaning 0.852 or 0.853. Without determining n, I'll guess these are in the right spot to be the ends of the IRQ calculation, so we need two whose difference is 0.028, so 0.852 - 0.824.
Answer: q=4, r=2
(iii)
Ugh if I had read ahead and knew that I had to plot I would have skipped this. I'll generate the data and see.
For Y we'll assume we got the quartiles, and now the median. n=83 so index 84/2=42, 13+15+17=45, so three back gives 0.837
X:
min: 0.802
25%: 0.815
median: 0.825
75%: 0.833
max: 0.848
Y:
min: 0.811
25%: 0.824
median: 0.837
75%: 0.852
max: 0.869
Those are the various points on the box and whisker plots. I may make a figure after I post, but let's get this answered.
(iv)
Comparing this figures, Y's bugs are bigger at every category. On average they're about 0.012 cm larger. They're sizes are also more spread out; a larger standard deviation, a larger variation in size.
