# Thread: Binomial Distribution [strange past exam question]

1. ## 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. 
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. 

For a]i), I used the binomial formula and got 0.1033 (to 4dp)
For ii) I did P(X $\displaystyle \leq$ 10) - P(X $\displaystyle \leq$ 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 2. Hi

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

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

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

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

The expected value of this pay is

$\displaystyle \sum_{k=0}^{50}\ 100+50k)\:\binom{50}{k}\:0.2^k\:0.8^{50-k}$

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

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