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Thread: Binomial Distribution [strange past exam question]

  1. #1
    Oct 2008

    Binomial Distribution [strange past exam question]

    A salesman makes 50 calls during a particular week. You may assume that independently for each house visited, the probability of a sale is 0.2.
    a] Find the probability that during this week, he makes
    i) exactly 12 sales,
    ii) between 10 and 14 (both inclusive) sales,
    iii) his first sale on the third house visited. [9]
    b] At the end of the week, he is paid 100 plus a commission of 50 for every sale. Find the mean and standard deviation of his total pay for this week. [5]

    For a]i), I used the binomial formula and got 0.1033 (to 4dp)
    For ii) I did P(X \leq 10) - P(X \leq 15) and taking these from my table values, I got 0.4956.
    For iii) I got confused - but I think it would be done by failure x failure x success which is 0.8 x 0.8 x 0.2 which is 0.128. Can't say I'm too confident about this.

    But what about part b? I've never known a question like this and hope it doesn't come up in the exam. How would you answer this?
    Thanks if you can help
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  2. #2
    MHF Contributor
    Nov 2008

    Let X be the random variable that counts the number oh houses that he sells per week.

    P(X=k) = \binom{50}{k}\:0.2^k\:0.8^{50-k}

    Let Y be the random variable of his pay.
    He is paid 100 + 50 for each house sold.

    P(X=k) = P(Y = 100 + 50 k) = \binom{50}{k}\:0.2^k\:0.8^{50-k}

    The expected value of this pay is

    100+50k)\:\binom{50}{k}\:0.2^k\:0.8^{50-k}" alt="\sum_{k=0}^{50}\100+50k)\:\binom{50}{k}\:0.2^k\:0.8^{50-k}" />

    which you can find equal to 600 by splitting the sum into 2 parts.
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