Thread: Binomial Distribution [strange past exam question]
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Old June 6th, 2009, 01:03 PM
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Hi

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

\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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