Question

Suppose Y follows a distribution with density function: f(y)={101846(100–y)? if(0<y<100) otherwise (Note: for this question you can enter your answer in decimals as well as fractions. You can round to two decimal places as well) 1. What must the value of k be so that f(x)is a probability density function? Submit Answer Tries 0/5 Based on this model, which event is more likely? A. a person dies between the ages of 70 and 80. B. a person lives beyond age 80. Submit Answer Tries 0/1 2. Given that a randomly selected person just celebrated his 60 th birthday, find the probability that he will live past age 80. Submit Answer Tries 0/5 3. Find the value y that maximizes f(y) (mode) Submit Answer Tries 0/2 4. Find the (average) life expectancy. Submit Answer Tries 0/5

Answer 1

The given PDF is $\fn_jvn f(y)=\frac{k}{10^{18}}y^6(100-y)^2,0<y<100$ .

1) The condition for valid PDF is

$\fn_jvn \int_{0}^{100}f(y)dy=1\\ \int_{0}^{100}\frac{k}{10^{18}}y^6(100-y)^2dy=1\\ \frac{k}{10^{18}}\int_{0}^{100}y^6(100-y)^2dy=1\\ \frac{k}{10^{18}}\int_{0}^{100}(100^2y^6-200y^7&plus;y^8)dy=1\\ \frac{k}{10^{18}}(100^2y^7/7-200y^8/8&plus;y^9/9]_{0}^{100}=1\\ {\color{Blue} k=252}$

The probability,

$\fn_jvn P(70<Y<80)=\int_{70}^{80}f(y)dy\\ P(70<Y<80)=\int_{70}^{80}\frac{k}{10^{18}}y^6(100-y)^2dy\\ P(70<Y<80)=0.2754$

and

$\fn_jvn P(Y>80)=\int_{80}^{100}f(y)dy\\ P(Y>80)=\int_{80}^{100}\frac{k}{10^{18}}y^6(100-y)^2dy\\ P(Y>80)=0.2618$

The probability that the person dies between 70 and 80  is higher.

2) The conditional probability,

$\fn_jvn P(Y>80|Y>60)=\frac{P(Y>80\cap Y>60)}{P(Y>60)}\\ P(Y>80|Y>60)=\frac{P(Y>80)}{P(Y>60)}\\ P(Y>80|Y>60)=\frac{0.2618}{\int_{60}^{100}\frac{k}{10^{18}}y^6(100-y)^2dy}\\ {\color{Blue} P(Y>80|Y>60)=0.3408}$

3) $\fn_jvn f(y)=\frac{k}{10^{18}}y^6(100-y)^2,0<y<100$ is maximized when

$\fn_jvn f'(y)=0\\ \frac{\mathrm{d} }{\mathrm{d} y}y^6(100-y)^2=0\\ 6y^5(100-y)^2-2y^6(100-y)=0\\ 6(100-y)-2y=0\\ {\color{Blue} y=75}$

4) The expected value of the distribution is

$\fn_jvn E(Y)=\int_{0}^{100}yf(y)dy\\ E(Y)=\int_{0}^{100}\frac{k}{10^{18}}y^7(100-y)^2dy\\ E(Y)= \frac{k}{10^{18}}\int_{0}^{100}y^7(100-y)^2dy=1\\ E(Y)= \frac{k}{10^{18}}\int_{0}^{100}(100^2y^7-200y^8&plus;y^9)dy\\ E(Y)=\frac{k}{10^{18}}(100^2y^8/8-200y^9/9&plus;y^{10}/10]_{0}^{100}\\ {\color{Blue} E(Y)=58.29 \textup{ years}}$