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Thread: Binomial Distribution

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    Exclamation Binomial Distribution




    A company starts a fund of M dollars from which it pays $1000 to each employee who achieves
    high performance during the year. The probability of each employee achieving this goal is 0:10 and
    is independent of the probabilities of the other employees doing so. If there are n = 10 employees,
    how much should M equal so that the funds has a probability of at least 99% of covering those
    payments?


    Is this a binomial distribution question? I am pretty confused. All I could think of is

    10
    Σ 1000Ck (0.1)^k (0.9)^(1000-k)
    k=0

    Help pleaseeeeeeee

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    Re: Binomial Distribution

    Can you state the information you were given in an even more informative/organized fashion?

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    Re: Binomial Distribution

    cool question

    seems you're saying each person is an iid Bernoulli trial
    with
    N=10
    p=0.10
    x=number "successes"
    the binomial distribution is (x=0,...,10)
    0.3486784 0.7360989 0.9298092 0.9872048 0.9983651
    0.9998531 0.9999909 0.9999996 1.0000000 1.0000000 1.0000000
    So,
    Pr(x<=3)=0.987
    Pr(x<=4)=0.998
    meaning (if I'm understanding correctly) M=$4000

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    Re: Binomial Distribution


    also, if this is a real-life question, and there's real $$ to be lost, I'd consider very closely whether the "trials" are truly independent, and I'd probably seek some insight into how sensitive the answer is to the assumed model (which is iid Bernoulli trials, leading to binomial)

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