Mei Ocr s1 unofficial mark scheme

Someone relieve us of the answers for the last question :frown:

Reply 2

chocolatecookieM

6

this is what i got:
1i). For the stem and leaf diagram. Median = 29g
Q1= 27.8 g
Q2= 31.8g
IQR = 4g
ii) 1.5 x 4= 6
31.8 + 6= 37.8g
27.8-6= 21.8g
So, no outliers.

2i) The probability that team A does not lose is 0.5 * 0.5 * 0.5 = 0.125
ii) 0.125 + (0.2)^3 +(0.3)^3=
iii) 3(0.3*0.3*0.2) + 3(0.3*0.3*0.5) + 3(0.5*0.5*0.3) +3(0.5*0.5*0.2) etc you get the gist...

3i) 5!= 120
ii) 5P3= 60
iii) 5C3= 10

anyone wants to add to this so far?

Reply 3

chocolatecookieM

6

sure! the last question was

X~B( 20??? ,0.1)
i) P(X=3) = 20C3 * (0.1)^3 * (0.9)^17
ii) P(X less that or equal to 2)=
use the tables
iii) 0.1 * 20 =
iv)Ho : p= 0.1 H1 : p<0.1v) there would be no critical region for that value of 20. 20 is too small. 12.16 % would have to be the significance levelvi) X~B (65,0.1)2 people use the pin 1234P(X less than or equal to 2)= 0.0531 > 5 percent Accept the Ho. There's insuffient evidence to suggest number of people using 1234 as PIN lowered.anyone want to add to this?

Reply 4

AdamWallisjr

2

Yeah i got similar to that...Dont remember much...hopefully all good

Reply 5

malderson999

Original post by chocolatecookieM

this is what i got:
1i). For the stem and leaf diagram. Median = 29g
Q1= 27.8 g
Q2= 31.8g
IQR = 4g
ii) 1.5 x 4= 6
31.8 + 6= 37.8g
27.8-6= 21.8g
So, no outliers.

2i) The probability that team A does not lose is 0.5 * 0.5 * 0.5 = 0.125
ii) 0.125 + (0.2)^3 +(0.3)^3=
iii) 3(0.3*0.3*0.2) + 3(0.3*0.3*0.5) + 3(0.5*0.5*0.3) +3(0.5*0.5*0.2) etc you get the gist...

3i) 5!= 120
ii) 5P3= 60
iii) 5C3= 10

anyone wants to add to this so far?

Alright so pretty sure question 1 for me is wrong lol
But for your part 2, I don't think you've taken into account them drawing...I did 1-(0.2*0.2*0.2) = 0.992

Reply 6

Haneenamer_

For 2 i) did it not ask for the probability that the team doesn't lose at all in any of the 3 matches? So didn't it require more working out than that? As its the probabilities that it's win win draw or win draw draw etc.

Reply 7

Haneenamer_

And I think for question 3 i) and ii) they wanted probability's so the answer can't be bigger than 1

Reply 8

Daniel9998

15

Original post by chocolatecookieM

2i) The probability that team A does not lose is 0.5 * 0.5 * 0.5 = 0.125

no you have to include the probability to draw also

Reply 9

pointon.joel

2

Question 2 part i, the probability of not losing is 1-0.2=0.8, so not losing would be 0.8 cubed....

Reply 10

Original post by chocolatecookieM

sure! the last question was

X~B( 20??? ,0.1)
i) P(X=3) = 20C3 * (0.1)^3 * (0.9)^17
ii) P(X less that or equal to 2)=
use the tables
iii) 0.1 * 20 =
iv)Ho : p = 0.1 H1 : p<0.1v) there would be no critical region for that value of 20. 20 is too small. 12.16 % would have to be the significance levelvi) X~B (65,0.1)2 people use the pin 1234P(X less than or equal to 2)= 0.0531 > 5 percent Accept the Ho. There's insuffient evidence to suggest number of people using 1234 as PIN lowered.anyone want to add to this?

i) A) X~B (16,0.1), 16C3 * 0.1^3 * 0.9^13 = 0.1423

i) B) ATLEAST 3 so 1 - P(x<=2) = 1- 0.7892 = 0.2108

i) C) 0.1*16 = 1.6, 2

ii) H0 :tongue:

=0.1, H1<0.1, p is the probability of a person having 1234 as their PIN. H1 has this form as we are testing to see if the number of people using the PIN 1234 has reduced making it less likely to find someone with the PIN 1234

iii) A) X~B (20,0.1) using a significance level of 10% for a lower tail test will not work since P(X<=0) is greater than 10%, so even if no one uses 1234 we'd still be accepting H1

iv) P(X<=2)=0.0530 > 5% so accept Ho as 2 is not in the critical region (0 & 1) for the lower tail AND/OR accept H0 as there is no significant evidence to show there is a reduced amount of people using 1234 as their PIN so reject H1/accept H0
(for full marks I believe you'd need to APPLY it to the scenario as you've done since there is no other place for the 4 marks to be allocated)

Reply 11

Original post by malderson999

Alright so pretty sure question 1 for me is wrong lol
But for your part 2, I don't think you've taken into account them drawing...I did 1-(0.2*0.2*0.2) = 0.992

yeah...for part 2 instead of doing 3 multiplied by all outcomes, I did 1 - the probability of all 3 and 3 times the probability of getting each one once

Reply 12

Screen Shot 2016-05-25 at 16.14.56.png52.3KB

Original post by Haneenamer_

And I think for question 3 i) and ii) they wanted probability's so the answer can't be bigger than 1

I think the numbers were right but it was 1 over

so, it would be 1/60 and 1/10
I remember doubting myself but there is another method that gets you these numbers too

Reply 13

IXxi HIIDA ixXI

15

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Reply 14

For the last question, I looked up p(x=0) in the table when x~B (20, 0.1) and the number was bigger than 0.1, something like 0.16??, the number is in the formula book.
Because the probability of getting 0, so no people having a pin of 1234 was bigger than the 0.1 significant level.
I think that means that we couldn't necessarily reject the h0 at a 10% sig fig level????

Reply 15

Original post by chocolatecookieM

sure! the last question was

X~B( 20??? ,0.1)
i) P(X=3) = 20C3 * (0.1)^3 * (0.9)^17
ii) P(X less that or equal to 2)=
use the tables
iii) 0.1 * 20 =
iv)Ho : p= 0.1 H1 : p<0.1v) there would be no critical region for that value of 20. 20 is too small. 12.16 % would have to be the significance levelvi) X~B (65,0.1)2 people use the pin 1234P(X less than or equal to 2)= 0.0531 > 5 percent Accept the Ho. There's insuffient evidence to suggest number of people using 1234 as PIN lowered.anyone want to add to this?

13%, it asked for an integer

Reply 16

Original post by maarz

For the last question, I looked up p(x=0) in the table when x~B (20, 0.1) and the number was bigger than 0.1, something like 0.16??, the number is in the formula book.
Because the probability of getting 0, so no people having a pin of 1234 was bigger than the 0.1 significant level.
I think that means that we couldn't necessarily reject the h0 at a 10% sig fig level????

It means H0 would never have enough evidence to be rejected

Reply 17