Question

Suppose that the life distribution of an item has the hazard rate function$\lambda \left(t\right)=t^3,t>0$. What is the probability that

$\left(a\right)$the item survives to age$2?$

$\left(b\right)$the item’s lifetime is between$.4$and$1.4?$

$\left(c\right)$a $1-$year-old item will survive to age$2?$

Short answer

Therefore the,

$$\begin{gathered} \left(a\right)P(X\ge 2)\approx 0.0183 \\ \left(b\right)P(0.4\le X\le 1.4)\approx 0.6109 \\ \left(c\right)P(X\ge \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$X$}\right.\ge 1)\approx 0.0235 \end{gathered}$$

Step-by-step solution

Given information:

The life distribution of an item has the hazard rate function$\lambda \left(t\right)=t^3,t>0$.

Let $X$denote the life distribution. First, we compute the probability function of the life distribution.

$$\begin{gathered} F\left(t\right)=1-exp\left(-{\int }_0^t\lambda \left(u\right)du\right) \\ F\left(t\right)=1-exp\left(-\frac{1}{4}{t}^4\right) \end{gathered}$$

Part (a) step 2 Explanation:

It is to compute$P(X\ge 2)$

$$\begin{gathered} P(X\ge 2)=1-P(X<2) \\ =1-F\left(2\right) \\ =e^{-4}\approx 0.0183 \end{gathered}$$

Part (b) step 3 Explanation:

The probability is given by

$$\begin{gathered} P(0.4\le X\le 1.4)=P(X\le 1.4)-P(X\le 0.4) \\ =F(1.4)-F(0.4) \\ =(1-e^{-0.9604})-(1-e^{-0.0064}) \\ =e^{-0.0064}-e^{-0.9604}\approx 0.6109 \end{gathered}$$

Part (c) step 4 Explanation:

The probability that a $1-$year-old item will survive to age $2$is

$$\begin{gathered} P(X\ge \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$X$}\right.\ge 1)=\frac{P(X\ge 2)}{P(X\ge 1)} \\ =\frac{1-F\left(2\right)}{1-F\left(1\right)} \\ =\frac{e^{-4}}{e^{{\displaystyle \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}}=e^{\raisebox{1ex}{$-15$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\approx 0.0235 \end{gathered}$$