A life insurance salesman sells on the average 3 life insurance policies per week. What type of this distribution is? (2 marks)
Let in a given week, calculate probability of the salesman will sell,
i. at least one policy. (2 marks)
ii. from two to four policies. (3 marks)
iii. Assume that there are 5 working days per week, what is the probability that in a given day he will sell one policy? (3 marks)
inadequate data.
To answer these questions, we need to understand the concept of probability distribution and specifically the distribution that applies to the given situation.
The distribution that describes the average number of life insurance policies sold per week is called the Poisson distribution. The Poisson distribution is commonly used to model the occurrence of events over a fixed interval of time or space given a known average rate of occurrence.
i. To calculate the probability that the salesman will sell at least one policy in a given week, we can use the complement rule. The complement of "at least one policy" is "no policies sold." The average number of policies sold per week is 3, so the average rate of occurrence of not selling a policy is also 3.
The probability of not selling a policy in a given week can be calculated using the Poisson distribution formula:
P(X = 0) = (e^(-λ) * λ^0) / 0!
Where λ is the average rate of occurrence (in this case, 3).
P(X = 0) = (e^-3 * 3^0) / 0! = 0.0498 (rounded to four decimal places)
The probability of selling at least one policy is the complement of not selling any policies:
P(X ≥ 1) = 1 - P(X = 0) = 1 - 0.0498 = 0.9502 (rounded to four decimal places)
ii. To calculate the probability that the salesman will sell from two to four policies in a given week, we need to calculate the individual probabilities of selling two, three, and four policies and sum them.
P(X = 2) = (e^-3 * 3^2) / 2! = 0.2240 (rounded to four decimal places)
P(X = 3) = (e^-3 * 3^3) / 3! = 0.2240
P(X = 4) = (e^-3 * 3^4) / 4! = 0.1680 (rounded to four decimal places)
P(X = 2 to 4) = P(X = 2) + P(X = 3) + P(X = 4) = 0.2240 + 0.2240 + 0.1680 = 0.6160 (rounded to four decimal places)
iii. Assuming there are 5 working days per week, to calculate the probability that the salesman will sell one policy in a given day, we can use the average rate of occurrence per day. The average number of policies sold per week is 3, so the average number of policies sold per day is 3 divided by 5 (number of working days).
Average number of policies sold per day = 3 / 5 = 0.6
The probability of selling one policy in a given day can be calculated using the same Poisson distribution formula:
P(X = 1) = (e^-0.6 * 0.6^1) / 1! = 0.3293 (rounded to four decimal places)