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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've done part a and for b got upto (13)-(1 divided by 2n)-(2) all divided by (square root 8.4 divided by square root n)=3.090
    How would i obtain a value for n?
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    for more clarity here's a picture
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    isnt  \displaystyle \bar{x} \sim (40, \sqrt{\frac{0.3}{n}}^2) and  \displaystyle P(\bar{x} &gt; X) = P(Z &gt; \frac{X - 40}{\sqrt\frac{{0.3}}{n}})

    (just checking I have this right before I go any further)
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    Wouldn't the variance be 8.4 over n, since mean for binomial approximated to normal is npq, so 40*0.3*0.7=8.4
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    (Original post by gooner1010)
    Wouldn't the mean be 8.4 over n, since mean for binomial approximated to normal is npq, so 40*0.3*0.7=8.4
    hmm as I feared, I cant remember how to do this :hide: sorry

    Edit: Is this S2? a binomial to normal approximation question? ill revise it and get back to you if you want?
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    (Original post by DylanJ42)
    hmm as I feared, I cant remember how to do this :hide: sorry

    Edit: Is this S2? a binomial to normal approximation question? ill revise it and get back to you if you want?
    Hi, yep it's a binomial approximation to normal, and help would be appreciated.
    Thanks
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    (Original post by gooner1010)
    Hi, yep it's a binomial approximation to normal, and help would be appreciated.
    Thanks
    sorry I honestly dont have a clue how to do this. Hopefully someone else can help you
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    (Original post by gooner1010)
    Hi, yep it's a binomial approximation to normal, and help would be appreciated.
    Thanks
    last try, is your answer to the first part 0.8869?

    Spoiler:
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    please say yes
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    (Original post by DylanJ42)
    last try, is your answer to the first part 0.8869?

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    please say yes

    Sorry. it's 0.0084
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    (Original post by gooner1010)
    Sorry. it's 0.0084
    ive got no idea then, sorry :laugh:

    iirc SeanFM knows his stats, maybe he can salvage this
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    (Original post by gooner1010)
    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've done part a and for b got upto (13)-(1 divided by 2n)-(2) all divided by (square root 8.4 divided by square root n)=3.090
    How would i obtain a value for n?
    So this is a normal approximation to the binomial type question. If you have a random variable X distributed as B(n,p) then it is approximated by a variable distributed as N(np, np(1-p)) (where the second parameter for the normal is given as the variance rather that standard deviation).

    So here, X is approximated by N(40 * 0.3, 40 * 0.3 * 0.7) = N(12, 8.4)

    If X is distributed as N(\mu, \sigma^2), then X-bar is distributed as N(\mu, \sigma^2/n) where n is the number of observations. So here we have the mean distributed as N(12, 0.1714286). So part (a) follows by looking at tables or using software.

    If n is unknown, then you need to work in reverse, try a few values of n and see what values you get for the probability - after a while you should be able to pinpoint the n you want.
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    (Original post by Gregorius)
    So this is a normal approximation to the binomial type question. If you have a random variable X distributed as B(n,p) then it is approximated by a variable distributed as N(np, np(1-p)) (where the second parameter for the normal is given as the variance rather that standard deviation).

    So here, X is approximated by N(40 * 0.3, 40 * 0.3 * 0.7) = N(12, 8.4)

    If X is distributed as N(\mu, \sigma^2), then X-bar is distributed as N(\mu, \sigma^2/n) where n is the number of observations. So here we have the mean distributed as N(12, 0.1714286). So part (a) follows by looking at tables or using software.

    If n is unknown, then you need to work in reverse, try a few values of n and see what values you get for the probability - after a while you should be able to pinpoint the n you want.
    The book has the answer 82, however I seem to be getting 81
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    (Original post by gooner1010)
    The book has the answer 82, however I seem to be getting 81
    I agree with your value of 81. I'm using the following R code

    N <- 81
    mm <- 12
    ss <- (40 * 0.3 * 0.7)/N
    1 - pnorm(13, mm, sqrt(ss))
 
 
 
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