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    The random variable X has a B(40. 0.3) distribution. The mean of a random sample of n observations of X is denoted by x̄. Find(a) P(x̄≥13) when n is 49,(b) the smallest value of n for which P(x̄≥13)<0.001.
    I got answer for part a which is 0.0084. However, I couldn't get answer for part b. Can someone help me? Thanks a lot.
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    (Original post by iceabbey)
    The random variable X has a B(40. 0.3) distribution. The mean of a random sample of n observations of X is denoted by x̄. Find(a) P(x̄≥13) when n is 49,(b) the smallest value of n for which P(x̄≥13)<0.001.
    I got answer for part a which is 0.0084. However, I couldn't get answer for part b. Can someone help me? Thanks a lot.
    What was your working out for part (b)? Badically, wherever you used 49 in part (a), use n instead and then work with that inequality. I'll only be able to help if you lay out your working for me.
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    (Original post by Zacken)
    What was your working out for part (b)? Badically, wherever you used 49 in part (a), use n instead and then work with that inequality. I'll only be able to help if you lay out your working for me.
    Thx a lot. This is my working.
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    (Original post by iceabbey)
    Thx a lot. This is my working.
    I'm not sure why you've done P(\bar{X} \geq 13 - \frac{1}{2n}) \leq 0.001?

    Remember that you can say \bar{X} \sim N(12, \frac{8.4}{n}), so what's P(\bar{X} \geq 13) in standardised terms?
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    (Original post by Zacken)
    I'm not sure why you've done P(\bar{X} &lt; 13 - \frac{1}{2n}) \leq 0.001?
    1/2n is the continuity correction.
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    (Original post by iceabbey)
    1/2n is the continuity correction.
    But you don't need a continuity correction if you invoke the CLT, do you? At least that's what I've learnt from Edexcel?
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    (Original post by Zacken)
    But you don't need a continuity correction if you invoke the CLT, do you? At least that's what I've learnt from Edexcel?
    I'm afraid in OCR we have to and that is how I got the right answer in part a. I tried to use 13 instead of 13- 1/2n. However, it did not give me the right answer.
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    (Original post by Zacken)
    But you don't need a continuity correction if you invoke the CLT, do you? At least that's what I've learnt from Edexcel?
    Binomial to normal. In Edexcel S3 we don't deal with explicitly converting discrete distributions to normal ones when using the CLT - at least, I've never seen it done!
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    (Original post by iceabbey)
    I'm afraid in OCR we have to and that is how I got the right answer in part a. I tried to use 13 instead of 13- 1/2n. However, it did not give me the right answer.
    P(X>=13) = P(X>=12.5) with cc, no?
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    (Original post by iceabbey)
    I'm afraid in OCR we have to and that is how I got the right answer in part a. I tried to use 13 instead of 13- 1/2n. However, it did not give me the right answer.
    Ah, okay. Fair enough. Let's work with the continuity correct then. Are you sure you need to be subtracting \frac{1}{2n} and not just \frac{1}{2}?
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    (Original post by tinkerbella~)
    P(X>=13) = P(X>=12.5) with cc, no?
    I don't think so. Because there is a unknown sample size n and will affect the continuity correction.
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    (Original post by aymanzayedmannan)
    Binomial to normal. In Edexcel S3 we don't deal with explicitly converting discrete distributions to normal ones when using the CLT - at least, I've never seen it done!
    Well, yes. But you're not really converting the binomial to a normal here, you're saying that X has a whatever distribution (in this case binomial but we don't need to care) so the sample mean is normally distributed by the CLT. We deal with sample means being normally distributed from not knowing the original distribution at all and we do so without continuity corrections, which is why I figured it wasn't needed here. :dontknow:

    Edit to add: some math.stack shows that continuity corrections are needed in certain scenarios and not in others. Apparently depends on the domain of the original distribution, I guess Edexcel just like to play nice with those. Oh well, good learning experience.
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    (Original post by Zacken)
    Ah, okay. Fair enough. Let's work with the continuity correct then. Are you sure you need to be subtracting \frac{1}{2n} and not just \frac{1}{2}?
    pretty sure about it.
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    (Original post by iceabbey)
    pretty sure about it.
    Just looked up some university probability theory notes for this and I'm quite confident that you need only subtract \frac{1}{2} and not \frac{1}{2n}. Try it.
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    Oh, and: Moved to maths.
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    (Original post by Zacken)
    Just looked up some university probability theory notes for this and I'm quite confident that you need only subtract \frac{1}{2} and not \frac{1}{2n}. Try it.
    The answer I got for that method is 302. the right answer should be 82. btw, I got part a right by using 13-1/2n therefore I don' t think it is 13-1/2.
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    (Original post by Zacken)
    Well, yes. But you're not really converting the binomial to a normal here, you're saying that X has a whatever distribution (in this case binomial but we don't need to care) so the sample mean is normally distributed by the CLT. We deal with sample means being normally distributed from not knowing the original distribution at all and we do so without continuity corrections, which is why I figured it wasn't needed here. :dontknow:
    That bit doesn't make sense, you're right about that - as per CLT, the mean of any distribution with a large enough sample size can be modelled by a normal distribution. However, on further research (with no solid evidence as to why), I've found that OCR/CIE does a continuity correction for sample means with an adjustment of 1/(2n). :confused:
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    (Original post by iceabbey)
    The answer I got for that method is 302. the right answer should be 82. btw, I got part a right by using 13-1/2n therefore I don' t think it is 13-1/2.
    Oh well, working as to what you say, we have: \displaystyle \frac{13 - \frac{1}{2n} - 12}{\frac{\sqrt{8.4}}{\sqrt{n}}} &gt; 3.0902

    But this gets me an answer around 320-ish again, just like it did when I used the non-continuity corrected version.

    Using +1/(2n) (as per Ayman's PDF) gets me another answer around 320 again? :dontknow:
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    (Original post by Zacken)
    ...
    Scroll down to sampling here. Do you believe it's incorrect?
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    (Original post by Zacken)
    I think your issue is doing 13 - \frac{1}{2n} when you should be doing 13 +\frac{1}{2n}?
    Sorry where does the 0.5n come from? I am confused.
 
 
 
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