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OCR MEI - S1 - 20th May 2015

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Original post by Qwertykeyboard15
Think I've lost marks for the given questions and I put the reason for the mode to be useful because the tourist office would know how many people is most likely would come in that day so they can get the right staff in etc 😂 totally different from what everyone else is saying... I didn't see the mean, so would have lost a mark there too 😣

I ran out of time with this exam, needed to have longer to check my answers


I tried to apply it to the tourist office situtation too lol, i thought when it said 'in this case' it meant the scenario rather than in this set of data.

I said it wasn't useful because all it tells them is that there are two hours in the day when the same number of people visit, which is no use to them.
The last question for anyone who is wondering :
you need these i think, where P(C) is prob of cherry
P(CC) + P(CCC) + P(CCCC) + P(CCCCC) +P(CCCCCC)
which would give you 3/30, which cancels to 1/10 or 0.1.
What was everyones opinion on the paper?
Original post by jpetersgill


James


Posted from TSR Mobile
This paper was definitely harder than last year!! i think the grade boundaries will be around 56/57
Last answer is 1/57.
To get more than 2 he doesn't like before he gets one he likes he has to pick 3 cherries in a row.
This can be calculated by (6/20)*(5/19)*(4/18) or by 1 - (14/20) - (6/20)*(14/19) - (6/20)*(5/19)*(4/18).
(edited 8 years ago)
Original post by Mathbomb
Last answer is 1/57.
To get more than 2 he doesn't like before he gets one he likes he has to pick 3 cherries in a row.
This can be calculated by (6/20)*(5/19)*(4/18) or by 1 - (14/20) - (6/20)*(14/19) - (6/20)*(5/19)*(4/18).


this is what i got but noone else seems to have gotten it
Original post by Mathbomb
Last answer is 1/57.
To get more than 2 he doesn't like before he gets one he likes he has to pick 3 cherries in a row.
This can be calculated by (6/20)*(5/19)*(4/18) or by 1 - (14/20) - (6/20)*(14/19) - (6/20)*(5/19)*(4/18).


It said 'He has to pick more than 2 before he gets one he likes' The only way this can happen is by getting a cherry first and second = 6/20 * 5/19 = 3/38.
Original post by corey7695
this is what i got but noone else seems to have gotten it

It's very easy to misinterpret the question and miss the significance of before, for example. That lead people to calculate (6/20)*(5/19) or 1 - (14/20) - (6/20)*(14/19), which would be true if before wasn't in the question.
i did 1 - (14/20*6/19) because i thought the prob of getting more than 2 cherries is 1-(prob of choosing 1 cherry), which by the looks it is wrong
Original post by Mathbomb
It's very easy to misinterpret the question and miss the significance of before, for example. That lead people to calculate (6/20)*(5/19) or 1 - (14/20) - (6/20)*(14/19), which would be true if before wasn't in the question.


Was that the wrong method though?
Original post by Mathbomb
It's very easy to misinterpret the question and miss the significance of before, for example. That lead people to calculate (6/20)*(5/19) or 1 - (14/20) - (6/20)*(14/19), which would be true if before wasn't in the question.


i like your thinking good sir
Hi how u use the probability table for the hypothesis testing???
Original post by Ritesh Jethwa
Hi how u use the probability table for the hypothesis testing???


You didn't use it in this exam because the specified probability of 0.78 isn't in the table.
Original post by Mathbomb
Last answer is 1/57.
To get more than 2 he doesn't like before he gets one he likes he has to pick 3 cherries in a row.
This can be calculated by (6/20)*(5/19)*(4/18) or by 1 - (14/20) - (6/20)*(14/19) - (6/20)*(5/19)*(4/18).


It did not say to get 'more than 2 he doesn't like'. It said 'that he has to select more than 2 before he finds one he likes'. (See pictures earlier on in thread).

Therefore 3/38 is the correct answer.
Original post by Connorbwfc
It did not say to get 'more than 2 he doesn't like'. It said 'that he has to select more than 2 before he finds one he likes'. (See pictures earlier on in thread).

Therefore 3/38 is the correct answer.


1- (P(has to select none(cherries) before he finds one he likes)+(has to select 1(cherries) before he finds one he likes) + (has to select 2 (cherries) before he finds one he likes)
Original post by Connorbwfc
It did not say to get 'more than 2 he doesn't like'. It said 'that he has to select more than 2 before he finds one he likes'. (See pictures earlier on in thread).

Therefore 3/38 is the correct answer.

Still seems like before would suggest you don't include finding the one he likes as one of the more than 2 times selected? I might be wrong :P
Original post by Mathbomb
Still seems like before would suggest you don't include finding the one he likes as one of the more than 2 times selected? I might be wrong :P


I think you are but the wording of that question was really bad.

You'll probably get 1/2 marks if you're answer is wrong because the process is correct.
for the one where it says is the mode useful, I said it is useful because the data is negatively skewed. is that enough for the one mark
http://www.thestudentroom.co.uk/showthread.php?t=3345445
I've just created a poll to rate the difficulty of this paper. It would be greatly appreciated if people could take time to choose an option, thank you:smile:
Original post by Kichelle


Both Kichelle and James

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