A door-to-door sales person for a Household appliance has learned from her past experience that out of ten?

demonstrations of her appliance only four result in actual sales (long run average). This week she needs to make at least five sales. At least how many demonstrations does she need to perform to ensure that the probability of meeting her target is at least 90 percent?|||using a binomial calculator , for n = 17, P[%26gt;=5] = 0.8740, for n = 18, P[%26gt;=5] = 0.9058


ans: n = 18


------|||If she holds ten demonstration, she will have 4 sales. But she needs a fifth sale.


If she has fifteen demonstrations, she will have 2 more sales (4 out of 10 = 2 out of 5). So our answer is right between ten and fifteen. She can't have twelve and a half demonstrations. So does she need to do twelve or thirteen?


If she does 12 demonstrations, her success rate will be 4.8, or 4 and an 80% chance of the next sale.


That's not high enough for the word problem's conditions.


If she does 13, her success rate will be 5.2 - she will have (on average) five sales at least.