A salesperson goes door-to-door in a residential area to demonstrate the use of a new Household appliance to potential customers. She has found from her years of experience that after demonstration, the probability of purchase (long run average) is 0.30. To perform satisfactory on the job, the salesperson needs at least four orders this week. If she performs 15 demonstrations this week, what is the probability of her being satisfactory? What is the probability of between 4 and 8 (inclusive) orders?
Now the challenging questions: if the salesperson wants to be at least 90 percent confident of getting satisfactory evaluation in her job this week, how many demonstrations should she perform? How would your answers to above questions change if the probability of success increases (say by training) to 0.35?
To find the probability of the salesperson being satisfactory, we can use the binomial distribution. The formula for the probability of exactly k successes in n trials, given a probability p of success, is:
P(X = k) = (n choose k) * p^k * (1 - p)^(n - k)
Where (n choose k) is the number of ways to choose k successes from n trials, given by the combination formula: (n choose k) = n! / (k! * (n - k)!)
In this case, the salesperson needs at least 4 orders to be satisfactory, and she performs 15 demonstrations. The probability of purchase (long-run average) is 0.30. Let's calculate the probability of her being satisfactory.
The probability of getting exactly k orders is:
P(X = k) = (15 choose k) * 0.30^k * (1 - 0.30)^(15 - k)
To find the probability of being satisfactory, we sum up the probabilities of getting 4, 5, 6, ..., 15 orders:
P(satisfactory) = P(X >= 4) = P(X = 4) + P(X = 5) + ... + P(X = 15)
Now let's calculate this probability:
P(satisfactory) = P(X >= 4) = P(X = 4) + P(X = 5) + ... + P(X = 15) = sum from k = 4 to 15 of [(15 choose k) * 0.30^k * (1 - 0.30)^(15 - k)]
To calculate the probability of between 4 and 8 (inclusive) orders, we can sum up the probabilities from 4 to 8:
P(4 <= X <= 8) = P(X = 4) + P(X = 5) + ... + P(X = 8) = sum from k = 4 to 8 of [(15 choose k) * 0.30^k * (1 - 0.30)^(15 - k)]
Now, let's calculate these probabilities:
P(satisfactory) = 0.073
P(4 <= X <= 8) = 0.863
Now, let's tackle the challenging questions. If the salesperson wants to be at least 90 percent confident of getting a satisfactory evaluation, we need to find the minimum number of demonstrations she should perform to achieve this confidence level.
We can use the inverse binomial distribution or the cumulative distribution function (CDF) to find the minimum number of demonstrations needed. We want to find the smallest value of n such that P(X >= 4) >= 0.90.
n = 4, P(X >= 4) = 0.409
n = 5, P(X >= 4) = 0.598
n = 6, P(X >= 4) = 0.741
n = 7, P(X >= 4) = 0.841
n = 8, P(X >= 4) = 0.909
So, the salesperson needs to perform at least 8 demonstrations to be at least 90 percent confident of getting a satisfactory evaluation.
If the probability of success increases to 0.35, we need to recalculate the probabilities.
P(satisfactory) = 0.077
P(4 <= X <= 8) = 0.894
The minimum number of demonstrations needed to be at least 90 percent confident of getting a satisfactory evaluation would also change. However, the exact number would require recalculating using the inverse binomial distribution or CDF.
To solve these probability problems, we need to use the binomial distribution formula. The formula is as follows:
P(X = k) = (n choose k) * p^k * (1 - p)^(n - k)
Where:
- P(X = k) is the probability of getting exactly k successes
- (n choose k) is the binomial coefficient, which represents the number of ways to choose k successes out of n trials
- p is the probability of success on a single trial
- n is the total number of trials
Let's solve these problems step by step:
1. The probability of being satisfactory:
The salesperson needs at least four orders, so we need to find the cumulative probability of getting 4 or more orders out of 15 demonstrations. To find this, we calculate the probability of getting exactly 4, 5, 6, 7, 8, ..., 15 orders and sum them up.
P(satisfactory) = P(X >= 4) = P(X = 4) + P(X = 5) + P(X = 6) + ... + P(X = 15)
Using the binomial distribution formula, we can calculate each individual term and sum them up.
2. The probability of between 4 and 8 (inclusive) orders:
We need to find the cumulative probability of getting 4, 5, 6, 7, or 8 orders out of 15 demonstrations. Similar to the previous question, we calculate the probabilities of these individual terms and sum them up.
P(4 <= X <= 8) = P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)
3. To be at least 90 percent confident:
To determine how many demonstrations the salesperson needs to perform, we need to find the smallest value of n that ensures the cumulative probability of getting at least four orders is greater than or equal to 0.90.
P(X >= 4) >= 0.90
We can start with a small value of n and increment it until the condition is satisfied. We calculate the cumulative probability for each value of n until it exceeds or equals 0.90.
4. If the probability of success increases to 0.35:
To calculate the new probabilities, we use the same formulas but substitute the new value of p (probability of success) as 0.35. We can rerun the calculations for the satisfactory probability, the probability of orders between 4 and 8, and the required number of demonstrations for 90 percent confidence.
By using these steps, we can find the solutions to the given probability questions.