Question

State Farm tnsurance studies show that in Colvade, 6S% of the auto merance dans santed mer po env ong net sinna by Renoa, a yat, sp damage claims Involving automobiles are selected at random prin (o) Let r be the number of diaims made by males under age 25. Make a histogram for the r-distribution probablites 025 015 e0s 015 0s

Answer 1

The probability that a randomly selected auto insurance claim is a claim made by male under age of 25 is 0.65

Out of 9 randomly selected claims, let R be the number of claims made by males under age of 25. R has a Binomial distribution with parameters, number of trials, n=9 and success probability p=0.65.

The probability that R=r claims are made by males under age of 25 out of n=9 claims is

$\begin{align*} P(R=r)&=\binom{n}{r}p^r(1-p)^{n-r}\\ &=\binom{9}{r}0.65^r(1-0.65)^{9-r}\\ &=\frac{9!}{r!(9-r)!}0.65^r(1-0.65)^{9-r}\\ \end{align*}$

R can take values from 0,1,...,9

We calculate the following values of the probabilities

$\begin{align*} P(R=0)&=\binom{9}{0}0.65^0(1-0.65)^{9-0}=0.00008\\ P(R=1)&=\binom{9}{1}0.65^1(1-0.65)^{9-1}=0.00132\\ P(R=2)&=\binom{9}{2}0.65^2(1-0.65)^{9-2}=0.00979\\ P(R=3)&=\binom{9}{3}0.65^3(1-0.65)^{9-3}=0.04241\\ P(R=4)&=\binom{9}{4}0.65^4(1-0.65)^{9-4}=0.11813\\ P(R=5)&=\binom{9}{5}0.65^5(1-0.65)^{9-5}=0.21939\\ P(R=6)&=\binom{9}{6}0.65^6(1-0.65)^{9-6}=0.27162\\ P(R=7)&=\binom{9}{7}0.65^7(1-0.65)^{9-7}=0.21619\\ P(R=8)&=\binom{9}{8}0.65^8(1-0.65)^{9-8}=0.10037\\ P(R=9)&=\binom{9}{9}0.65^9(1-0.65)^{9-9}=0.02071\\ \end{align*}$

If we plot a histogram of the following in excel using insert--->columns

R	P(R=r)
0	0.00008
1	0.00132
2	0.00979
3	0.04241
4	0.11813
5	0.21939
6	0.27162
7	0.21619
8	0.10037
9	0.02071

we get

This is similar to the 4th graph

a) ans: 4th graph

b) the probability that seven or more claims are made by males under age 25, is same as the probability that R>= 7

$\begin{align*} P(R\ge 7)&=P(R=7)+P(R=8)+P(R=9)\\ &= 0.21619 + 0.10037 + 0.02071 \\ &= 0.337 \end{align*}$

Ans: the probability that seven or more claims are made by males under age 25 is 0.337

c) Using the standard results for a Binomial distribution we know that

The expected value of R is

$\begin{align*} E(R)=np=9\times 0.65=5.85 \end{align*}$

The standard deviation of R is

$\begin{align*} SD(R)=\sqrt{np(1-p)}=\sqrt{9\times 0.65\times (1-0.65)}=1.43 \end{align*}$

ans:

$\begin{align*} \mu=5.85\\ \sigma=1.43 \end{align*}$