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# Ocr mei statistics 1 exam thread (24/05/13) Watch

1. did you sneak it out of exam hall or are you a teacher :O
itd be better if u post the whole thing
2. (Original post by Attz_09)
hoa post picture or email them me i need to find out how i dun can get my bro to do the maths and then feel bad about how **** i dun
they're up mate
3. I need the tree diagram if you dont mind
4. (Original post by Attz_09)
did you sneak it out of exam hall or are you a teacher :O
itd be better if u post the whole thing
posted the whole thing just in 3 different posts by accident haha. Nope our school lets us take them out
5. do it landscape plz i have to tilt my head sideways and the shade makes it difficult to read black ink pleasee
6. (Original post by Attz_09)
UNOFFICIAL MARK SCHEME

1i) mean = 249.4, standard deviation = 14.5 [3]
ii) new mean = 209.46, new standard deviation = 13.0 [3]

2i) P(2 women, 1 man) = (5C2x5C1)/10C3 = 5/12 [4]
ii) P(four evenings with 2 women, 1 man) = (5/12)^4 + 4C3(5/12)^3(7/12) = 0.1989 [4]

3i) P(5underweight bags) = 50C5(0.1)^5(0.9)^45 = 0.1849 [3]
ii) P(at least one underweight bag in 20) = 1-(0.9)^20 = 0.8784 [2]
iii) E(X) = np = 48x0.8784=42.2 [2]

4i) 3! x 1/6 x 1/5 x 1/4 = 1/20 [2]
ii) E(X)= sum xP(X=x) = £1507.50, Var(X)=E(X^2)-E(X)^2 = £445511.25 [5]

5i) Because the researcher has assumed that the probability that a randomly selected person makes a correct identification is 0.5, as she assumes that there is a 50:50 chance of guessing the difference between tap and bottled water [2]
ii) Because the point of the test is to determine if the probability that a randomly selected person makes a correct identification is in fact greater than 0.5 [1]
iii) P(X≥13)=1-P(X≤12)=1-0.8684=0.1316>0.05 Therefore, 13 is not a member of the critical region and it is not significant. We accept H0. There is insufficient evidence from the data to suggest the probability that a randomly selected person makes a correct identification is greater than 0.5 [5]

6i) Median=3.32, LQ=(2.82+2.84)/2=2.83, UQ=(3.70+3.72)/2=3.71. IQR=0.88 [3]
ii) Draw box and whisker plot [3]
iii) 1step = 1.5xIQR=1.32. UQ+1.32=5.03, therefore no outliers above. LQ-1.32=1.51. Therefore 1.39 is an outlier below. It shouldn't be removed as there is no evidence to suggest that it forms a separate pool of data or is not a genuine data item. [4]
iv) Median=3.5. UQ=3.84, LQ=3.13. IQR=0.71 [3]
v) Males have a higher average weight (higher median). Males have less variation in birth weight (lower interquartile range). [2]
vi) (8/200)x(7/199). You can probably use any values from 7-10/200 (it depends how you read the graph). [3]

7i) Draw probability tree diagram [4]
iiA) 1-0.9x0.95x0.95=0.1878 (4 d.p.) [3]
iiB) 0.1x0.8x0.95+0.9x0.05x0.8+0.9x0. 95x0.05=0.1548 (4 d.p.) [4]
iii) 0.1548/0.1878=0.8242 [3]
iv) P(Hit,Hit,Hit) + P(Miss,Miss,Miss,Hit,Hit,Hit) = 0.0056 (4 d.p.) [4]

bold is the things am unsure/ lost marks on

im not sure if you did 3ii right because if i remember correctly it said number of bags in box 0.1 faulty. and 48 boxes in crate. so i did 20 times 0.1 which =2
then done 2 time 48
Nope i'm sorry, you needed to find the probability that it's greater than 1, which is 1-P(0 underweight bags)
7. (Original post by hoa123)
posted the whole thing just in 3 different posts by accident haha. Nope our school lets us take them out
thats wicked i wouldnt be waiting for an unnofficial mark scheme if i had the paper. do u live in liverpool by any chance lol
8. (Original post by Attz_09)
cbaaaaaaaa
9. (Original post by Attz_09)
thats wicked i wouldnt be waiting for an unnofficial mark scheme if i had the paper. do u live in liverpool by any chance lol
nope, newcastle haha
10. No 7 tree diagram page? Thank you.
11. (Original post by hoa123)
Nope i'm sorry, you needed to find the probability that it's greater than 1, which is 1-P(0 underweight bags)
i meant iii sorry it said 20 bags in a box 48 box in crate
12. (Original post by Attz_09)
i meant iii sorry it said 20 bags in a box 48 box in crate
yeah so you have P(≥1 underweight bag in a box)=0.8784 from ii

Then iii asks for the number of boxes with ≥1underweight bags. So you do E(X)=np=48x0.8784
13. (Original post by hoa123)
cbaaaaaaaa
Hey any chance you could post the info for me to draw the tree diagram again?
14. (Original post by Blobar)
No 7 tree diagram page? Thank you.
15. (Original post by hoa123)
Sorry it's upside down haha
16. its great you posting all these pictures and everthing but am turning my laptop screen sideways and upside down to read it haha
17. (Original post by Blobar)
Hey any chance you could post the info for me to draw the tree diagram again?
probability of bull's-eye with first throw is 0.1, for each subsequent throw it's 0.2 if she hit with the previous dart and 0.05 if she missed with the previous dart
18. (Original post by Attz_09)
its great you posting all these pictures and everthing but am turning my laptop screen sideways and upside down to read it haha
sorry haha
19. (Original post by hoa123)
UNOFFICIAL MARK SCHEME

1i) mean = 249.4, standard deviation = 14.5 [3]
ii) new mean = 209.46, new standard deviation = 13.0 [3]

2i) P(2 women, 1 man) = (5C2x5C1)/10C3 = 5/12 [4]
ii) P(four evenings with 2 women, 1 man) = (5/12)^4 + 4C3(5/12)^3(7/12) = 0.1989 [4]

3i) P(5underweight bags) = 50C5(0.1)^5(0.9)^45 = 0.1849 [3]
ii) P(at least one underweight bag in 20) = 1-(0.9)^20 = 0.8784 [2]
iii) E(X) = np = 48x0.8784=42.2 [2]

4i) 3! x 1/6 x 1/5 x 1/4 = 1/20 [2]
ii) E(X)= sum xP(X=x) = £1507.50, Var(X)=E(X^2)-E(X)^2 = £445511.25 [5]

5i) Because the researcher has assumed that the probability that a randomly selected person makes a correct identification is 0.5, as she assumes that there is a 50:50 chance of guessing the difference between tap and bottled water [2]
ii) Because the point of the test is to determine if the probability that a randomly selected person makes a correct identification is in fact greater than 0.5 [1]
iii) P(X≥13)=1-P(X≤12)=1-0.8684=0.1316>0.05 Therefore, 13 is not a member of the critical region and it is not significant. We accept H0. There is insufficient evidence from the data to suggest the probability that a randomly selected person makes a correct identification is greater than 0.5 [5]

6i) Median=3.32, LQ=(2.82+2.84)/2=2.83, UQ=(3.70+3.72)/2=3.71. IQR=0.88 [3]
ii) Draw box and whisker plot [3]
iii) 1step = 1.5xIQR=1.32. UQ+1.32=5.03, therefore no outliers above. LQ-1.32=1.51. Therefore 1.39 is an outlier below. It shouldn't be removed as there is no evidence to suggest that it forms a separate pool of data or is not a genuine data item. [4]
iv) Median=3.5. UQ=3.84, LQ=3.13. IQR=0.71 [3]
v) Males have a higher average weight (higher median). Males have less variation in birth weight (lower interquartile range). [2]
vi) (8/200)x(7/199). You can probably use any values from 7-10/200 (it depends how you read the graph). [3]

7i) Draw probability tree diagram [4]
iiA) 1-0.9x0.95x0.95=0.1878 (4 d.p.) [3]
iiB) 0.1x0.8x0.95+0.9x0.05x0.8+0.9x0. 95x0.05=0.1548 (4 d.p.) [4]
iii) 0.1548/0.1878=0.8242 [3]
iv) P(Hit,Hit,Hit) + P(Miss,Miss,Miss,Hit,Hit,Hit) = 0.0056 (4 d.p.) [4]
With the last part, i got 0.0085.

P( Hit, Hit, Hit ) = 0.1 x 0.2 x 0.2 = 0.004

P ( Miss, Miss, Miss, Hit, Hit, Hit ) = 0.9 x 0.5 x 0.5 x 0.5 x 0.2 x 0.2 = 0.0045

0.004 + 0.0045 = 0.0085

?
How is it 0.0056 ?
20. (Original post by RomanMarkOliver)
With the last part, i got 0.0085.

P( Hit, Hit, Hit ) = 0.1 x 0.2 x 0.2 = 0.004

P ( Miss, Miss, Miss, Hit, Hit, Hit ) = 0.9 x 0.5 x 0.5 x 0.5 x 0.2 x 0.2 = 0.0045

0.004 + 0.0045 = 0.0085

?
How is it 0.0056 ?
Hit, hit, hit = 0.1x0.2x0.2 yes
Miss, miss, miss, hit, hit, hit = 0.9 x 0.95 x 0.95 x 0.05 x 0.2 x 0.2

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