91影视

Create a random variable. The number $X$represents the machine's lifetime. It is assumed that the provided hazard function defines $X$.

We'll use the relationship between the hazard rate and the cumulative function provided by

$F\left(t\right)=1鈭�\exp 鈦�\left(鈭�{鈭�}_0^t鈥�位\left(s\right)ds\right)$

The task at hand is to locate $P(X>6)$. We can deduce from the above link that

$$\begin{gathered} P(X>6)=1鈭�P(X鈮�6) \\ =1鈭�F\left(6\right) \\ =\exp 鈦�\left(鈭�{鈭�}_0^6鈥�位\left(s\right)ds\right) \end{gathered}$$

Let's calculate ${鈭�}_0^6鈥�位\left(s\right)ds$. We have that

${鈭�}_0^6鈥�位\left(s\right)ds={鈭�}_0^2鈥�0.2ds+{鈭�}_2^5鈥�(0.2+0.3(s鈭�2\left)\right)ds+{鈭�}_5^6鈥�1.1ds$

$$\begin{gathered} =0.4+1.95+1.1 \\ =3.45 \end{gathered}$$

So we have that

$P(X>6)=e^{鈭�3.45}鈮�0.0317$

We are required to find $P(X鈮�8鈭�X>6)\text{.}$. It is equal to

$P(X鈮�8鈭�X>6)=\frac{P(6<X<8)}{P(X>6)}$

The probability in the denominator is equal to

Probability $P(X>8)$ is calculated similarly as in (a). We have that

${鈭�}_0^8鈥�位\left(s\right)ds={鈭�}_0^2鈥�0.2ds+{鈭�}_2^5鈥�(0.2+0.3(s鈭�2\left)\right)ds+{鈭�}_5^8鈥�1.1ds$

$$\begin{gathered} =0.4+1.95+3.3 \\ =5.65 \end{gathered}$$

So,

$P(X>8)=e^{鈭�5.65}鈮�0.003517$