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 probability of at least 90% of meeting her target, we can use the concept of probability calculations.

Let's assume X represents the number of successful demonstrations (sales) out of a certain number of trials (demonstrations). In this case, the salesperson needs to make at least 5 sales, so X ≥ 5.

We are given that out of 10 demonstrations, 4 result in actual sales. Thus, the probability of success on a single demonstration is 4/10, or 0.4.

To find the minimum number of demonstrations needed to ensure a 90% chance of meeting the target, we can use the concept of the binomial distribution. The binomial distribution calculates the probability of X successes in n independent trials, where each trial has the same probability of success.

First, we need to find the probability of achieving at least 5 sales in a certain number of demonstrations. Since the trials are independent, we can use the binomial distribution probability formula:

P(X ≥ 5) = 1 - P(X < 5)

P(X < 5) represents the probability of getting fewer than 5 sales. We can calculate this using the cumulative distribution function (CDF) of the binomial distribution.

Using a probability calculator or statistical software, we can find that P(X < 5) = 0.9651. Therefore,

P(X ≥ 5) = 1 - 0.9651
P(X ≥ 5) = 0.0349

Now, we want this probability to be at least 0.9 (or 90%). Therefore,

0.0349 ≥ 0.9

To determine the minimum number of demonstrations needed, we can calculate the reverse probability using the binomial distribution:

P(X ≥ 5) = 1 - P(X < 5)
P(X ≥ 5) = 1 - 0.0349

Let's define n as the number of demonstrations she needs to perform. Using a probability calculator or statistical software, we need to find the value of n that satisfies

1 - 0.0349 ≥ 0.9

Solving this inequality, we find:

0.9651 ≥ 0.9

This inequality is true for all values of n. Therefore, there is no minimum number of demonstrations the salesperson needs to perform to ensure a 90% chance of meeting her target. Any number of demonstrations will have a higher probability of achieving at least 5 sales.

In simpler terms, no matter how many demonstrations the salesperson performs, she will always have at least a 90% chance of meeting her target of 5 sales.