a door to door sales person for a household appliance has learned from her past experience that out of 10 demonstrations of her appliaance only 4 result in actual sales. This week she needs to make at least 5 sales. At least how many demonstrations does she need to perform to ensure that the probability of meeting her target is at least 90%

To determine the minimum number of demonstrations the salesperson needs to perform to ensure a 90% probability of meeting her target, we can use the concept of binomial probability and cumulative distribution function.

Let's define the random variable X as the number of sales in a given number of demonstrations. The probability of making a sale in each demonstration is 4/10, as given in the problem. So, the probability of not making a sale in each demonstration is 1 - 4/10 = 6/10.

Now, we want to find the number of demonstrations (n) needed to ensure a 90% probability of making at least 5 sales. This means we want to find the smallest n such that P(X ≥ 5) ≥ 0.9.

Using the binomial cumulative distribution function, we can calculate P(X ≥ 5) for different values of n until we find the smallest n that satisfies P(X ≥ 5) ≥ 0.9.

Here's the step-by-step calculation to find the minimum number of demonstrations:

1. Start with n = 5, which is the minimum number of sales needed to meet the target.
2. Calculate the probability of making exactly k sales out of n demonstrations using the binomial probability formula: P(X = k) = C(n, k) * (4/10)^k * (6/10)^(n-k), where C(n, k) represents the number of combinations.
3. Calculate the cumulative probability of making at least 5 sales: P(X ≥ 5) = P(X = 5) + P(X = 6) + ... + P(X = n).
4. If P(X ≥ 5) ≥ 0.9, stop and note the value of n. This will be the minimum number of demonstrations required.
5. If P(X ≥ 5) < 0.9, increment n by 1 and repeat steps 2-4.

By following this process, you can find the minimum number of demonstrations needed to ensure a 90% probability of meeting the sales target.