# Probability distribution of the mean of a discrete random variable

While self-studying a probability book, I came up with a modified version of a problem that I don't know how to solve. In the original statement we have a salesman that visites houses selling books, and his probability of selling a book in a visit is 0.3. The problem asks what his sales average per day is, if he does a fixed amount of 20 visits per day.

This is very easy, since the variable $$X$$ defined as the amount of successful sales on a day where he does 20 visits is distributed as a binomial, that is, $$X\sim Bi(n=20, p=0.3)$$. So the average we are looking for is $$\mu_X=np=20\cdot 0.3=6$$.

However, what if we asked the following... The salesman will have to work on the weekends if his average of successful sales for the week is less than 10. What is the probability of him working on the weekend?

I have no idea how to tackle this even though it sounds simple. Writing all the possible combinations of sales for the 5 days of the week that would result in an average of less than 5 would only be possible with a computer or with a lot of patience. Any advice?

• The salesman makes $5\times 20=100$ visits a week. You are asking for the probability that he makes at least $50$ sales. Doesn't matter what days he makes those sales.
– lulu
Commented Mar 19 at 11:12
• Should say: I'm not sure what you mean by "average". I took it to mean "average daily sales" but if that's the case, the poor man's task is effectively impossible. The weekly process has a mean of $30$ with $\sigma \sim 4.58$, so he needs a $6\sigma$ week just to have a day off.
– lulu
Commented Mar 19 at 11:15
• @lulu Wait, isn't the variance additive? For one day, $\sigma=20\cdot 0.3\cdot 0.7=4.2$, so $\sigma=4.2\cdot 5=21$ for the whole week. Am I wrong here? Commented Mar 19 at 13:33
• The variance is indeed $21$, but I was speaking of the standard deviation, hence $\sigma=\sqrt {21}\sim 4.58$.
– lulu
Commented Mar 19 at 13:54

You have initally random variable X, which is the amount of successful sales on a day. Next you need to define random variable $$X'$$ which is the amount of successful sales on a week.
Assuming that X for every day is independent of each other, we can consider $$X'$$ as a sum of Binomial distributions for 5 days. Thus:
$$X' \sim Bi(\sum n_i; p) \sim Bi(20 * 5 = 100; 0.3)$$
$$P(X' \le 9) = \sum_{i=0}^{9} {100\choose i} \cdot0.3^i\cdot0.7^{100-i}$$