Ran out of time when I realised how the histogram is done goodbye A
Omg same Was looking at it worked it out after staring at it forever and started writing and they tell us to put our pens down... I was pissed off because I know I nwo got 0 in that
this is most probably a silly question, sorry, but does that mean I wouldn't get the mark for 17?
Probably not I'm afraid. The way they intended for the question to be answered is either by finding the value of t where p=16 on the graph by inspection, or subbing 16 into the formula.
Definitely 16. You could argue that at 16.0000001, and so on is the point at which it is above 16 and therefore will cause the death at that moment, so as it is so close to 16, you would simply just take it as 16. They may give you the marks for 17 though, who knows.
5e - yeah prob, i thought it was odd that it was quite close to being symmetrical.
For 5e could you not argue that you should use the mean because this includes cars speeding and is therefore a better measure of average speed, because the median would end up completely ignoring such individuals?
I didn't revise much (I know it's not really an excuse) so in the rush of the exam I thought that P(BnT)=P(B)P(T|B) implied that they're independent which is obviously nonsense.
Ah, that's an impressive proof! Really unfortunate though, unlucky. I can say goodbye to my Cambridge application if I keep dropping marks like this... Looking at 95 UMS for this hopefully. I'm off to bed now, night!
Hey two questions, if anyone can answer i'd appreciate it a lot:
1. For question 4a i gave two examples which were (B n W) and (T n W) . Each one is correct but would i be marked down for giving two? i got the explanation correct btw...
2. For question 5e would i get away by saying that the data is almost symmetrical and therefore it would be appropriate to use the mean? Also for question 5d i said the distribution was symmetrical for the same reasons above, again would i get the marks?
Thank you
1) Yeah what Arsey said, you'll get the marks. 2) I did the same as you! I thought to myself, in real life, it's so unlikely to get a perfectly symmetrical distribution so if median is almost the same as the mean then surely we would use normal distribution to represent it! Oh well, apart from that, wasn't too bad. My friend put negative skew but put the reasons for positive skew!