Probability for sold items A salesperson has a probability of 70% to make a sale. 
 A: The way to answer this is to use the Normal approximation to the binomial distribution. The number of sales is assumed to be distributed as $ X \sim Bin (500,0.7) $. The take home pay is $ Y = 50 + 10X $, which you can rearrange to $ X=\frac{Y-50}{10} $. You want to find the probability that $ Y> y $ (when $ y=4800$), so work out what that means in terms of the value of $ X $. Then compute the Normal approximation of $ X $, $ X^\prime \sim N (\mu,\sigma^2) $ and then calculate $\Pr(X^\prime > x) $ where $ x $ is the value of $ X $ for which $ Y=4800$.
A: I believe what you are interested in is the variance of the binomial distribution. Consider that you are essentially counting up a series of successes and failures (a sequence of independent Bernoulli trials). Since this is self study, I won't explicitly give the answer. 
Note: @tristan isn't wrong. However, with computers around, you can get a more accurate answer by using the true distribution (Binomial) than with a Normal approximation.
A: Let's say the salesperson sees 3 customers a day with 70% chance of a sale for each.


*

*Probability of 0 sales = 0.027

*Probability of 1 or more sales = 0.973

*Probability of 2 or more sales = 0.784

*Probability of 3 sales = 0.343
Based on this, expand to 500 customers a day. It helps to use the binomial distribution function.
