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?
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 binomial distribution describes the probability of obtaining a certain number of successes in a fixed number of independent trials, given a probability of success for each trial.
In this case, the salesperson performs 15 demonstrations, and the probability of purchase after each demonstration is 0.30. We can find the probability of getting at least four orders by calculating the cumulative probability from four to fifteen demonstrations.
To calculate the probability of at least four orders, we can use the cumulative distribution function (CDF) of the binomial distribution. The formula to calculate the CDF is:
P(X ≥ r) = ∑ (nCr * p^r * q^(n-r))
Where:
P(X ≥ r) is the probability of getting at least r successes,
n is the number of trials (15 demonstrations),
r is the number of successes (4 or more orders),
p is the probability of success (0.30 in this case),
q is the probability of failure (1 - p = 0.70).
Now we can calculate the probability of at least four orders:
P(X ≥ 4) = ∑ (15C4 * 0.30^4 * 0.70^(15-4) ) + ∑ (15C5 * 0.30^5 * 0.70^(15-5) ) + ... + ∑ (15C15 * 0.30^15 * 0.70^(15-15) )
Similarly, to find the probability of between 4 and 8 (inclusive) orders, we need to calculate the individual probabilities for each number of orders from 4 to 8 and sum them up:
P(4 ≤ X ≤ 8) = P(X=4) + P(X=5) + P(X=6) + P(X=7) + P(X=8)
Now, to answer the next question, if the salesperson wants to be at least 90 percent confident of getting a satisfactory evaluation, she needs to determine the number of demonstrations that give her a probability of success greater than or equal to 0.90.
To find this number, the salesperson can use the inverse cumulative distribution function (also known as the quantile function or percent-point function) of the binomial distribution. This function will give her the minimum number of trials required for a given probability of success.
For example, to find the minimum number of demonstrations needed to be at least 90 percent confident of getting a satisfactory evaluation, she can use the following formula:
n ≥ qbinom(p, n, r)
Where:
qbinom is the inverse cumulative distribution function,
p is the desired probability of success (0.90 in this case),
n is the number of trials (unknown in this case),
r is the number of successes (at least 4 orders).
Finally, let's consider the situation where the probability of success increases to 0.35 due to training. To calculate the updated probabilities, we can use the same formulas as before but with the new probability of success. For example, to find the probability of at least four orders, we would now use a probability of 0.35 instead of 0.30 in the formula.