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OCR (Not MEI) S1 Wednesday 8th June 2016

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Reply 180
Avatar for Ab9009
Ab9009
Original post by samh2301
Any predictions for grade boundaries?

57/58
Original post by metasysta
Let's do another community MS. Please press the button to request permission to contribute.
https://docs.google.com/document/d/1uL5DuXry9Kc-sc2DE1QXz7e2FqW9JsV0TmVhI79VAGI/edit?usp=sharing


does anyone remember roughly the marks for each part?
were we supposed to compare the 2 different graphs for spearmans rank?
Original post by samh2301
Any predictions for grade boundaries?


Probably around the same as last year, maybe slightly lower as this paper seemed a little harder.
Original post by student199919
does anyone remember roughly the marks for each part?


Question numbers and questions would be useful too
0.0864 for very last question
Original post by metasysta
Let's do another community MS. Please press the button to request permission to contribute.
https://docs.google.com/document/d/1uL5DuXry9Kc-sc2DE1QXz7e2FqW9JsV0TmVhI79VAGI/edit?usp=sharing
Original post by LukeBlizard
0.0864 for very last question


I got 1.37
Original post by sen99
wouldn't it be 12 bc it said less than 140

This is probably wrong, but I got that Q1 was (something like) the 13.5th value, meaning that there are 13 values below the lower quartile
It was 12 as 13 was the lower quartile at 140 and it wanted values lower than 140
Original post by torilee
This is probably wrong, but I got that Q1 was (something like) the 13.5th value, meaning that there are 13 values below the lower quartile


I did too
Original post by Btec certified
I got 1.37

Lol how can a probability be more than one?
Reply 191
Avatar for sen99
sen99
7
Original post by Ozil5
Haha no worries, you had to use their position on the x and y axis as ranks, so they were 12345 and 12354. I think the final answer was 0.9


I cant believe how simple it actually was :frown:((. well there goes 5 marks. I used the wrong formula for variance but for the next part of the question i used the right method to get em2. Do you think I'll get any marks?
How's the BARBARA question 4/35. You already take in to account the repetitions in the first part, so you just have 1/210 as the probability.
Original post by Vanilla Poison
How's the BARBARA question 4/35. You already take in to account the repetitions in the first part, so you just have 1/210 as the probability.


this.
Reply 194
Avatar for ytfy
ytfy
Mark Scheme: Kind of.
1)
Spearman (Given)= -1
Spearmen (Calculated)= 0.9
2)
Last Q)
P(X=3) = 0.128
P(9<X<20) = P(X≤19)-P(X≤9) = 0.120 (3sf)
3)
Marie Wins:
(0.2X0.2X0.7)+(0.2X0.2X0.3) =0.04
Nadine Wins:0.0864
4)
Perm/com
A) 210
B)1/210
C)1/70
5)
Group of 12 into 3: 27,720
6)
PMCC: 0.924 Therefore more then 0.92
7) Mean N: 6.5
Mean Z: About 20.9
Total output: 20.9x12= 251
8) E(X)=2.7
VAR (X)= Forgot it but it was something obvious
9)
Coded Data
Mean: 146.5
Var= 12.8
In the range: 13 apples
Outliers: None (Marks for calculating both and saying no outliers)
Sum of M^2: 11107676.1 )Something like that, forgot it, because it's a long ****ing long number)
(edited 7 years ago)
Reply 195
Avatar for toosie
toosie
1
Can someone explain to me why there aren't any outliers? What I did was find the iqr (15?) then multiply that by 1.5 then add it to the uq which was 155? So then I got that 177.5-180 were outliers. Is that wrong and why?
What do we think that 69/72 would be approximately UMS wise?
Original post by Buymoria
What do we think that 69/72 would be approximately UMS wise?


Probably the full 100 ums
Reply 198
Avatar for ytfy
ytfy
The highest Value was 176 on the box and whisker plot, so there couldn't be any outliers.
Guys would I lose marks for using 147 and 21.8 to find the sum of msquared instead of 146.5 and 21.75?

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