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 calculate the probabilities in this scenario, we will use the binomial distribution formula:
P(x) = (nCx) * (p^x) * ((1-p)^(n-x))
Where:
- P(x) is the probability of obtaining exactly x successes
- n is the number of trials
- p is the probability of success in one trial (probability of purchase in this case)
- x is the number of successes
1. Probability of being satisfactory with 15 demonstrations:
To calculate the probability of getting at least four orders (satisfactory), we need to calculate the probability of getting exactly four, five, six, seven, eight, nine, ten, eleven, twelve, thirteen, fourteen, or fifteen orders and add those probabilities together.
P(satisfactory) = P(4) + P(5) + P(6) + P(7) + P(8) + P(9) + P(10) + P(11) + P(12) + P(13) + P(14) + P(15)
Let's calculate these probabilities one by one and then add them together:
P(4) = (15C4) * (0.30^4) * (0.70^11)
P(5) = (15C5) * (0.30^5) * (0.70^10)
P(6) = (15C6) * (0.30^6) * (0.70^9)
P(7) = (15C7) * (0.30^7) * (0.70^8)
P(8) = (15C8) * (0.30^8) * (0.70^7)
P(9) = (15C9) * (0.30^9) * (0.70^6)
P(10) = (15C10) * (0.30^10) * (0.70^5)
P(11) = (15C11) * (0.30^11) * (0.70^4)
P(12) = (15C12) * (0.30^12) * (0.70^3)
P(13) = (15C13) * (0.30^13) * (0.70^2)
P(14) = (15C14) * (0.30^14) * (0.70^1)
P(15) = (15C15) * (0.30^15) * (0.70^0)
Calculating each of these probabilities using the binomial coefficient formula (nCx = n! / (x! * (n-x)!)), and substituting the values, you will get:
P(satisfactory) = P(4) + P(5) + P(6) + P(7) + P(8) + P(9) + P(10) + P(11) + P(12) + P(13) + P(14) + P(15)
2. Probability of between 4 and 8 orders (inclusive):
To calculate the probability of between four and eight orders (inclusive), we need to calculate the probabilities of getting exactly four, five, six, seven, and eight orders and add those probabilities together.
P(4-8 orders) = P(4) + P(5) + P(6) + P(7) + P(8)
3. Number of demonstrations for at least 90% confidence:
To calculate the number of demonstrations needed for at least 90% confidence, we need to find the minimum number of demonstrations for which the probability of being satisfactory (getting at least four orders) is greater than or equal to 0.90.
Start with a reasonable number of demonstrations, say 100, and calculate the probability of being satisfactory. If it is less than 0.90, increase the number of demonstrations and recalculate until the probability reaches or exceeds 0.90.
4. Changes with increased probability of success:
If the probability of success (probability of purchase) increases to 0.35, repeat the calculations above with the new probability value. Replace the original value of "p" (0.30) with the new value of "p" (0.35) in all the formulas mentioned.