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

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

did you sneak it out of exam hall or are you a teacher :O
itd be better if u post the whole thing

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

posted the whole thing just in 3 different posts by accident haha. Nope our school lets us take them out

do it landscape plz i have to tilt my head sideways and the shade makes it difficult to read black ink pleasee

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 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

cbaaaaaaaa

No 7 tree diagram page? Thank you.

Hey any chance you could post the info for me to draw the tree diagram again?

its great you posting all these pictures and everthing but am turning my laptop screen sideways and upside down to read it haha

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 ?