Answer to Question #151356 in Statistics and Probability for Hay

Question

Question 3

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

Here we have:

n = 40,

p = 1/4 = 0.25,

x number of customers to buy life insurance.

Here x follows the binomial distribution

$$P(X=x)=C(n,x)\cdot p^x\cdot (1-p)^{n-x}$$

i) $P(X=5)=C(40,5)\cdot 0.25^5\cdot (1-0.25)^{40-5}=\frac{40!}{5!35!}\cdot0.25^5\cdot0.75^{35}=0.02723$

ii)

$P(X\le10)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+$

$+P(X=4)+P(X=5)+P(X=6)+P(X=7)+P(X=8)+$

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

$P(X=0)=C(40,0)\cdot 0.25^0\cdot 0.75^{40}=0.00001$

$P(X=1)=C(40,1)\cdot 0.25^1\cdot 0.75^{39}=0.00013$

$P(X=2)=C(40,2)\cdot 0.25^2\cdot 0.75^{38}=0.00087$

$P(X=3)=C(40,3)\cdot 0.25^3\cdot 0.75^{37}=0.00368$

$P(X=4)=C(40,4)\cdot 0.25^4\cdot 0.75^{36}=0.01135$

$P(X=5)=C(40,5)\cdot 0.25^5\cdot 0.75^{35}=0.02723$

$P(X=6)=C(40,6)\cdot 0.25^6\cdot 0.75^{34}=0.05295$

$P(X=7)=C(40,7)\cdot 0.25^7\cdot 0.75^{33}=0.08573$

$P(X=8)=C(40,8)\cdot 0.25^8\cdot 0.75^{32}=0.11788$

$P(X=9)=C(40,9)\cdot 0.25^9\cdot 0.75^{31}=0.13971$

$P(X=10)=C(40,10)\cdot 0.25^10\cdot 0.75^{30}=0.14436$

$P(X\le10)=0.00001+0.00013+0.00087+0.00368+0.01135+0.02723+0.05295+0.08573+0.11788+0.13971+0.14436=0.5839$

iii)

$P(X\ge20)=1-P(X<20)$

$P(X=11)=C(40,11)\cdot 0.25^{11}\cdot 0.75^{29}=0.13124$

$P(X=12)=C(40,12)\cdot 0.25^{12}\cdot 0.75^{28}=0.10572$

$P(X=13)=C(40,13)\cdot 0.25^{13}\cdot 0.75^{27}=0.07590$

$P(X=14)=C(40,14)\cdot 0.25^{14}\cdot 0.75^{14}=0.04879$

$P(X=15)=C(40,15)\cdot 0.25^{15}\cdot 0.75^{25}=0.02819$

$P(X=16)=C(40,16)\cdot 0.25^{16}\cdot 0.75^{24}=0.01468$

$P(X=17)=C(40,17)\cdot 0.25^{17}\cdot 0.75^{23}=0.00690$

$P(X=18)=C(40,18)\cdot 0.25^{18}\cdot 0.75^{22}=0.00294$

$P(X=19)=C(40,19)\cdot 0.25^{19}\cdot 0.75^{21}=0.00114$

$P(X\ge20)=1-0.5839-0.13124-0.10572-0.0759-0.04879-0.02819-0.01468-0.0069-0.00294-0.00114=0.0006$

iv)

$E(X)= np=40\cdot0.25=10$

$SD(X)=\sqrt{np(1-p)}=\sqrt{40\cdot0.25\cdot(1-0.25)}=2.7386$

Answer:

i) 0.02723

ii) 0.5839

iii) 0.0006

iv) 10, 2.7386