A life insurance salesman sells on the average 3 life insurance policies per week. Use Poisson's law to calculate the probability that in a given week he will sell
a. Some policies
b. 2 or more policies but less than 5 policies.
c. Assuming that there are 5 working days per week, what is the probability that in a given day he will sell one policy?
Solution
Let $X$ denote the number of life insurance policies sold per week. The average number of life insurance policies sold per week is 3, i.e., $E(X)=\lambda = 3$.
$X\sim P(3)$.
The probability mass function of Poisson distribution with $\lambda =3$ is
$$ \begin{aligned} P(X=x) &= \frac{e^{-3}(3)^x}{x!},\\ &\qquad \; x=0,1,2,\cdots \end{aligned} $$
a. The probability that in a given week he will sell some policies is
$$ \begin{aligned} P(X\geq1) &= 1- P(X=0)\\ &= 1- \frac{e^{-3}3^{0}}{0!}\\ &= 1-0.0498\\ &= 0.9502 \end{aligned} $$
b. The probability that in a given week he will sell 2 or more policies but less than 5 policies is
$$ \begin{aligned} P(2\leq X < 5) &=P(2\leq X\leq 4)\\ &=P(X=2)+P(X=3)+P(X=4)\\ &= \frac{e^{-3}3^{2}}{2!}+\frac{e^{-3}3^{3}}{3!}+\frac{e^{-3}3^{4}}{4!}\\ &= 0.224+0.224+0.168\\ &= 0.6161 \end{aligned} $$
c. Assuming 5 working days per week, the average number of policies sold per day = $3/5 =0.6$. The probability that in a given day he will sell one policy is
$$ \begin{aligned} P(X=1) &= \frac{e^{-0.6}0.6^{1}}{1!}\\ &= 0.3293 \end{aligned} $$
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
