Question

We have $100$components that we will put to use in a sequential fashion. That is, the component $1$is initially put in use, and upon failure, it is replaced by a component$2$, which is itself replaced upon failure by a componentlocalid="1649784865723" $3$, and so on. If the lifetime of component i is exponentially distributed with a mean $10+i/10,i=1,...,100$estimate the probability that the total life of all components will exceed$1200$. Now repeat when the life distribution of component i is uniformly distributed over$(0,20+i/5),i=1,...,100$.

Short answer

- the lifetime of the component is exponentially distributed$.9767$.

- the lifetime of the component is uniformly distributed$.9997$.

Step-by-step solution

Given Information.

Given $100$components that we will put in use in a sequential fashion. That is, the component $1$is initially put in use, and upon failure, it is replaced by a component$2$, which is itself replaced upon failure by the component$3$, and so on. If the lifetime of component i is exponentially distributed with mean

$10+i/10,i=1,...,100$.

Explanation.

Let $X_i$represents the lifetime of $i$th componentlocalid="1649784720827" $i=1,\dots ,100$. Assume that the components are used one at a time, whereby the failed component is replaced immediately new one. Also, let these lifetimes be exponential random variables with parameters${\lambda }_i$. Then,

${\mu }_i=E\left[X_i\right]=\frac{1}{{\lambda }_i}\text{and}{\sigma }_i^2=Var\left(X_i\right)=\frac{1}{{\lambda }_i^2},$

and since it is given that the lifetime is variable with mean$10+i/10,i=1,\dots ,100$, we have that${\lambda }_i=10/(100+i)$. So,

${\mu }_i=10+\frac{i}{10}\text{and}{\sigma }_i^2=100+2i+\frac{i^2}{100}$

Let $X$denote the total lifetime of all componentslocalid="1649784745171" $X=\sum _1^{100}{X}_i$.

Then, since lifetimes are obviously independent random variables, using the corresponding properties of expectation and variance we get: the random variable $X$is a gamma random variable with mean

$$\begin{gathered} E\left[X\right]=E\left[\sum _{i=1}^{100}{X}_i\right]=\sum _{i=1}^{100}{\mu }_i \\ =\sum _{i=1}^{100}\left(10+\frac{i}{10}\right)=100\left(10\right)+\frac{1}{10}\sum _{i=1}^{100}i\stackrel{(*)}{=}100\left(10\right)+\frac{100\left(101\right)}{10\left(2\right)} \\ =1505 \end{gathered}$$

and variance

$$\begin{gathered} Var\left(X\right)=Var\left(\sum _{i=1}^{100}{X}_i\right)=\sum _{i=1}^{100}{\sigma }_i^2=\sum _{i=1}^{100}\left(100+2i+\frac{i^2}{100}\right) \\ =100\left(100\right)+2\sum _{i=1}^{100}i+\frac{1}{100}\sum _{i=1}^{100}{i}^2 \end{gathered}$$

$(*),(**)=100\left(100\right)+100\left(101\right)+\frac{100\left(101\right)\left(201\right)}{100\left(6\right)}=23483.5$

Explanation.

The probability that the total life of all components will exceed $1200$is

$P\{X>1200\}.$

To approximate this probability we use the central limit theorem and in that case, we get:

$$\begin{gathered} P\{X>1200\}=1-P\{X\le 1200\}=1-P\left\{\frac{X-E\left[X\right]}{\sqrt{Var\left(X\right)}}\le \frac{1200-E\left[X\right]}{\sqrt{Var\left(X\right)}}\right\} \\ =1-P\left\{\frac{X-1505}{\sqrt{23483.5}}\le \frac{1200-1505}{\sqrt{23483.5}}\right\}\approx 1-\Phi (-1.99) \\ \stackrel{\Phi (-z)=1-\Phi \left(z\right)}{=}\Phi (1.99) \end{gathered}$$

Explanation.

$(*)$the sum of first $n$natural numbers:$\sum _{i=1}^{n}i=\frac{n(n+1)}{2}$

$(**)$the sum of the squares of first $n$natural numbers: $\sum _{i=1}^n{i}^2=\frac{n(n+1)(2n+1)}{6}$

Explanation.

The lifetime of components is uniformly distributed.

Now, assume that the lifetime $X_i$is uniformly distributed over$(0,20+i/5)$,

$i=1,\dots ,100\text{.}$So,

${\mu }_i=E\left[X_i\right]=\frac{(20+i/5)+0}{2}=10+\frac{i}{10}$

${\sigma }_i^2=Var\left(X_i\right)=\frac{\left(\right(20+i/5)-0{)}^2}{12}=\frac{100}{3}+\frac{2}{3}i+\frac{1}{300}{i}^2$

In this case, the total lifetime of all componentslocalid="1649784800135" $X=\sum _1^{100}{X}_{ir}$is a random variable with mean

$E\left[X\right]=E\left[\sum _{i=1}^{100}{X}_i\right]=\sum _{i=1}^{100}{\mu }_i=\sum _{i=1}^{100}\left(10+\frac{i}{10}\right)=1505$

and variance

$$\begin{gathered} Var\left(X\right)=Var\left(\sum _{i=1}^{100}{X}_i\right)=\sum _{i=1}^{100}{\sigma }_i^2=\sum _{i=1}^{100}\left(\frac{100}{3}+\frac{2}{3}i+\frac{1}{300}{i}^2\right) \\ =\frac{100}{3}\left(100\right)+\frac{2}{3}\sum _{i=1}^{100}i+\frac{1}{300}\sum _{i=1}^{100}{i}^2 \\ \stackrel{(*),(*)}{=}\frac{10000}{3}+\frac{1}{3}100\left(101\right)+\frac{100\left(101\right)\left(201\right)}{300\left(6\right)}=7827.83 \end{gathered}$$

Using the central limit theorem we get the required probability:

$$\begin{gathered} P\{X>1200\}=1-P\{X\le 1200\}=1-P\left\{\frac{X-E\left[X\right]}{\sqrt{Var\left(X\right)}}\le \frac{1200-E\left[X\right]}{\sqrt{Var\left(X\right)}}\right\} \\ =1-P\left\{\frac{X-1505}{\sqrt{7827.83}}\le \frac{1200-1505}{\sqrt{7827.83}}\right\}\approx 1-\Phi (-3.45) \\ \underset{\Phi (-z)=1-\Phi \left(z\right)}{=}\Phi (3.45)\stackrel{\text{Table 5.1 (textbook, Chapter 5)}}{=}.9997 \end{gathered}$$