What is the question asking us to do?
I'm finding the shortest distance (the pole will be perpendicular to the wall) at which it will hit when it falls.
By doing this I obtain answer E, however the answer is C.
So I think I may be misinterpreting the question.
Thanks in advance.
;also question 3 from the VCAA 2010 exam 1 (MC)
I've never seen a question stated like that before, how exactly do we calculate the standard deviation?
This one's a doosy - if you don't quite get what I'm saying, tell me and I'll re-explain with pictures.
You're right in that the maximum height will be then the poll falls the shortest distance. When we know the shortest this distance, we can use this to find the height via pythag.
So, the first thing you need to do is find this distance. The way I did this was to use the cosine-rule to find either angle made by the wall and the lines leading to the pole. With that, I used sin(x)=O/H to find the perpendicular distance between the pole and the wall. After that, I used c^2=a^2+b^2 to find the height, so in numbers:
Solve(5^2=6^2+4^2-2*4*6*cos(x),x), x=0.9734 (I will call this x for ease)
sin(x)=O/4 ---> 4*sin(x)=3.307 (hence O)
4^2-O^2=5.0625
sqrt(5.0625) = 2.25, closest answer is 2.3 (C)
For your standard deviation one, simply take approximate percentages from the mean and add them up. The mean (from the box-and-whisker plot) is 180. So, I add the two bars either side together, and get 40. I add the next two, and get 68. From this, I would assume that one standard deviation from the mean is 2 degrees. To confirm, I keep doing this until I hit 95% (or close to it, since guestimating the y-axis). If I hit 95% after another 2 degrees, then using the 68-95-99.7% rule, I would say that the standard deviation is 2 degrees.