Question

Let X be the future lifetime random variable in years for a newborn (age 0). The probability density function of X is as follows:
f(x)= ke^-0.01x 0< x< 100

where k is a constant. Find the probability that a person age 20 will live for at least 20 more years.

Answer 1

Answer:

0

Step-by-step explanation:

First we need to find the value of constant k in equation. To find k we use the formula:

$\int\limits^{100}_0 {ke^{-0.01x}} \, dx =1\\Integrating:\\k\int\limits^{100}_0 {e^{-0.01x}} \, dx =1\\\frac{k}{-0.01}[e^{-0.01x}]_0^{100}=1\\-100k [e^{-0.01x}]_0^{100}=1\\-100k[e^{-0.01*100}-e^{-100*0}]=1\\-100k[e^{-1}-e^0]=1\\-100k(-0.632)=1\\63.2k = 1\\k=0.0158$

the probability that a person age 20 will live for at least 20 more years = P(40 ≤ x <∞).

The person would live for at least 40 years

Therefore:

P(40 ≤ x <∞) =

$\int\limits^{\infty}_{40} 0.0158e^{-0.01x}} \, dx \\Integrating:\\0.0158\int\limits^{\infty}_{40} {e^{-0.01x}} \, dx =\frac{0.0158}{-0.01}[e^{-0.01x}]_{40}^{\infty}\\=-1.58 [e^{-0.01x}]_{40}^{\infty}=-1.58[e^{-\infty}-e^{-100*40}]\\=-1.58[e^{-\infty}-e^{-4000}]=-1.58(0-0)=0$