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    \mathrm{ Let\ \xi_1 ,\ \ldots \ ,\xi_15 \ be\ a\ random\ sample\ for\ a\ random\ variable\ \xi \ with\ values\ as\ listed\ below}

     6, 5, 5, 3, 5, 5, 5, 6, 4, 4, 5, 4, 4, 4, 6

    \mathrm{ Given\ that\ the\ random\ variable\ \xi\ has\ a\ Binomial\ distribution\ with\ parameters\ n\ and\ p} ;

    \mathrm{ (i)\ calculate\ the\ sample\ mean\ and\ the\ sample\ variance}
    \mathrm{ (ii)\ estimate\ p\ and\ n}

    I have calculated the mean as 71/15

    I calculated the sample variance as 82/105

    But then to calculate  n and  p , I know that:

     \dfrac{Var(\xi)}{\mathbb{E(\xi)}  } = 1-p

    Giving my value of p  = \dfrac{415}{497}

    And I can then work out \mathrm{n} as:

    \mathrm{n} = \dfrac{\mathbb{E(\xi)}}{p} = \dfrac{( \dfrac{71}{15})}{( \dfrac{415}{497})} = \dfrac{35287}{6225} = 5.668594378

    Which really doesn't seem right.

    I spoke to someone on my course, and he got:

    \mathrm{ n\ = 6,\ p = \dfrac{71}{90}}

    But I'm really not sure how.

    Can anyone point out where I've gone wrong?

    Cheers
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    Okay, I can see how he got  p = \dfrac{71}{90}

     \dfrac{\mathbb{E(\xi)}}{n} = \dfrac{(\dfrac{71}{15})}{6} = \dfrac{71}{90}

    But this still doesn't explain to me how he got  n = 6

    Obviously, my Variance is wrong, but I calculated the sample variance with the correct method:

    \mathrm{ Sample\ Variance} = \dfrac{1}{n-1} \sum_{i=1}^n (\xi_i - \bar{\xi_n})^2

    Really stuck on this, and can't see where :/
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    The only way I can see for this is that we know that n must be integer, so having calculated it as 5.66 we round to 6, and then use p = mean/n. This gives the (n,p)=(6,71/80) result. (Your calculations are correct.)
 
 
 
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