You need some help with part A too.
A.
A box and whiskers plot indicates five points, the five quartile division. That's:
- the minimum, tip of the left whisker.
- the 25th percentile, left edge of the box
- the 50th percentile, aka median, line in the middle of the box
- the 75th percentile, right edge of the box
- the maximum, tip of the right whisker.
For us that's a min: 40, 25th percentile 45, median 55.5, 75th percentile 60, max 65.
Your plot looks good but you need to add the whiskers.
B.
We have to the left of the median more spread out data than to the right. It's not definitive, but it does hint that the average of the lower 5 data points will be less than the median by an amount greater than the average of the top five points is above the median. In other words, the mean of all ten points is likely less than the median 55.5.
C. We have two additional data points now. We're told the max increases by 2, so one of them must be the new maximum, and that must be 67.
Without the last point the data is
40 45 45 45 55 56 60 60 60 65 67
We're told when we add the last point the minimum and interquartile points doesn't change from the original 10. Maybe the median changes; we're not told one way or another.
When we have to interpolate for the percentiles the way it work is for percentile p (as a fraction) of N points numbers 1 to N, we need the value of the point with index
i = (1 - p) 1 + p N
Let's work it out for p=0, .25, .5, .75, 1, N=10
That's i=: 1, 3.25, 5.5, 7.75, 10
So the median i=5.5 is the average of points 5 and 6. The 25th percentile is the weighed average of points 3 and 4, 75% on 3, 25% on 4, etc.
With N=12 this works out to
i = 1, 3.75, 6.5, 9.25, 12
We we add the last point the 1st, 3.75th, and 9.25th positions must still be 40, 45 and 60 respectively.
Potential places for the last point:
45 60
a 40 45 | 45 45 55 | 56 60 60 | 60 65 67
40 b 45 | 45 45 55 | 56 60 60 | 60 65 67
40 45 45 | 45 c 55 | 56 60 60 | 60 65 67
40 45 45 | 45 55 d | 56 60 60 | 60 65 67
40 45 45 | 45 55 56 | e 60 60 | 60 65 67
40 45 45 | 45 55 56 | 60 60 60 | f 65 67
40 45 45 | 45 55 56 | 60 60 60 | 65 g 67
We don't need one after the max because we know the max is 67. By that token the min doesn't change, so we can rule out a (and save b=40 for b).
The bars indicate the points where the percentile is counted; we want the quartiles to stay 45 and 60. So we can rule out f (e can cover the 60 case) and g.
But everything else is ok for the last point, which we see must be between 40 (line b) and 60 (line C) inclusive.
Answer: One photograph took exactly 67 photos and the other between 40 and 60 inclusive.