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# Official Edexcel S3 thread Wednesday 20 May 2015 Morning [6691] watch

1. (Original post by Damien_Dalgaard)
Any reason how come

Just for more practice, I will put links in the OP to papers etc.
sure
2. Can anyone please help with June 2013 R question 2. A step-by-step explanation would be much appreciated. I understand how to get the final answer as n=7, however my initial inequality is of the form < -1.6449 and not >1.6449.

3. (Original post by JaiGuruji)
Can anyone please help with June 2013 R question 2. A step-by-step explanation would be much appreciated. I understand how to get the final answer as n=7, however my initial inequality is of the form < -1.6449 and not >1.6449.

This is My solution,

X~N(40,9)
XBAR~(40,9/n)
P(XBAR>42)=P(Z>2/(3/rootn))
P(Z>2/(3/rootn))<5%
2/(3/rootn) greater than or equal to 1.6449
n is greater than or equal to 6.09
n = 7

Hope it isnt too ambiguous, just try and understand what each line means in terms of standard deviation, and which percentages we are looking to find. I often find drawing graphs and highlighting the section we are trying to find very useful
4. Hi there,

I dont understand how youve moved from <5% in one line to n is greater than or equal to 1.6449. I thought its meant to be less than or equal to -1.6449.

Thanks again

(Original post by shloke123)
This is My solution,

X~N(40,9)
XBAR~(40,9/n)
P(XBAR>42)=P(Z>2/(3/rootn))
P(Z>2/(3/rootn))<5%
2/(3/rootn) greater than or equal to 1.6449
n is greater than or equal to 6.09
n = 7

Hope it isnt too ambiguous, just try and understand what each line means in terms of standard deviation, and which percentages we are looking to find. I often find drawing graphs and highlighting the section we are trying to find very useful
5. Anyone? S3 June 2013 R Question 2 please?? Thanks!
6. (Original post by JaiGuruji)
Anyone? S3 June 2013 R Question 2 please?? Thanks!

X~N(40,9)

XBAR ~ (40, 9/n)

P(XBAR > 42 ) < 0.05 Is the same thing as 1 - P(XBAR < 42 ) < 0.05

Rearrange to get: P(XBAR < 42 ) > 0.95

then solve from there
7. (Original post by xKay)
X~N(40,9)

XBAR ~ (40, 9/n)

P(XBAR > 42 ) < 0.05 Is the same thing as 1 - P(XBAR < 42 ) < 0.05

Rearrange to get: P(XBAR < 42 ) > 0.95

then solve from there
Thanks for that! Much appreciated! Can you pls explain why just doing it as less than or equal to -1.6449 would be wrong from P(XBAR > 42 ) < 0.05.

Thanks again!
8. (Original post by JaiGuruji)
Thanks for that! Much appreciated! Can you pls explain why just doing it as less than or equal to -1.6449 would be wrong from P(XBAR > 42 ) < 0.05.

Thanks again!
if the probability of xBAR is GREATER than 42 is less than 0.05 then it is going to be on the right hand side of the normal distribution curve so its therefore the positive 1.6449

If it was xBAR is Less than 42 is less than 0.05 then it is on the right hand side of the distribution curve so we would use the negative -1.6449

9. (Original post by xKay)
if the probability of xBAR is GREATER than 42 is less than 0.05 then it is going to be on the right hand side of the normal distribution curve so its therefore the positive 1.6449

If it was xBAR is Less than 42 is less than 0.05 then it is on the right hand side of the distribution curve so we would use the negative -1.6449

Yeah it does! Thanks for that!
10. (Original post by JaiGuruji)
Yeah it does! Thanks for that!
11. i dont understand the mark scheme for Q 4 June 2014 (R)How do you do it... Many Thanks!
12. (Original post by Rkai01)
i dont understand the mark scheme for Q 4 June 2014 (R)How do you do it... Many Thanks!
i approached this question in a slightly different manner..
the thing is that you know that it is a one tail, 5% level, so the critical value is 1.6449,
next thing is you need to express the test statistic:

The test statistic, you need the mean, so you calculated the mean of the sample (the fifth value is n), and then the SD and sample size is given in the question, so you can express the test statistic.

Next step, in order to reject H0, you need test statistic greater than the critical value, so you express that (see the Mark scheme), and you solve it

Hope it helps
13. Hey! Can you guys help me out with Q 4 of the June 2014 paper. Thanks. I'm not sure how you get z = +-3.2905
A machine fills packets with X grams of powder where X is normally distributed with mean mu. Each packet is supposed to contain 1 kg of powder. To comply with regulations, the weight of powder in a randomly selected packet should be such that P(X 􀀝 mu – 30) = 0.0005
(a) Show that this requires the standard deviation to be 9.117 g to 3 decimal places. (3)
Hey! Can you guys help me out with Q 4 of the June 2014 paper. Thanks. I'm not sure how you get z = +-3.2905
A machine fills packets with X grams of powder where X is normally distributed with mean mu. Each packet is supposed to contain 1 kg of powder. To comply with regulations, the weight of powder in a randomly selected packet should be such that P(X 􀀝 mu – 30) = 0.0005
(a) Show that this requires the standard deviation to be 9.117 g to 3 decimal places. (3)
look at the formula booklet page 21, because you know that the probability is 0.0005, so the corresponding Z value is either + or - 3.2905, and then you can transform P(X 􀀝< mu – 30) = 0.0005
to P(Z < -30/ sigma) =0.0005, and you get -30/sigma = -3.2905 (because it is negative anyway)
15. Oh okay, I didn't realise it was .0005 and not .005. Silly me. Thank you DCMed96!
16. Hi do you have any more of these please.
17. (Original post by Damien_Dalgaard)
Hi do you have any more of these please.
(My stats section is not great but it will still be helpful)
18. (Original post by TeeEm)
(My stats section is not great but it will still be helpful)
19. (Original post by Damien_Dalgaard)
look in my profile because apparently I am not allowed to quote it in full
20. (Original post by TeeEm)
look in my profile because apparently I am not allowed to quote it in full
I was getting you confused with ten of them, you names are similar.

I think I did some of your papers before, great website sir!

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