Answer to Question #151351 in Statistics and Probability for Eric Koomson

Question

In a typical month, an insurance agent presents life insurance plans to 40 customers. Historically, one in four such customers chooses to buy life insurance from this agent. Based on the relevant information, answer the following questions:
(i) What is the probability that exactly 5 customers will buy life insurance from this agent in the coming month?
(ii) What is the probability that no more than 10 customers will buy life insurance from this agent in the coming month?
(iii) What is the probability that at least 20 customers will buy life insurance from this agent in the coming month?
(iv) Determine the mean and standard deviation of the number of customers who will buy life insurance from this agent in the coming month.

Answer 1

Let $X$ = the number of clients who will buy life insurance from this agent in the coming month:$X\sim Bin(n, p)$

$$P(X=x)=\dbinom{n}{x}p^x(1-p)^{n-x}$$

Given $n=40, p=0.25$

(i)

$$P(X=5)=\dbinom{40}{5}(0.25)^5(1-0.25)^{40-5}\approx0.027232$$

(ii)

$$P(X\leq10)=P(X=0)+P(X=1)+P(X=2)$$

$$+P(X=3)+P(X=4)+P(X=5)+P(X=6)$$

$$+P(X=7)+P(X=8)+P(X=9)+P(X=10)$$

$$\approx0.583904$$

(iii)

$$P(X\geq20)=1-P(X<20)\approx0.000572$$

(iv)

$$\mu=np=40(0.25)=10$$

$$\sigma=\sqrt{np(1-p)}=\sqrt{40(0.25)(1-0.25)}=\sqrt{7.5}$$

$$\approx2.7386$$