91Ó°ÊÓ

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

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

Let $X_i$represents the lifetime of $i$th componentlocalid="1649784720827" $i=1,…,100$. 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${\mathrm{λ}}_i$. Then,

${\mathrm{μ}}_i=E\left[X_i\right]=\frac{1}{{\mathrm{λ}}_i}\text{and}{\mathrm{σ}}_i^2=Var\left(X_i\right)=\frac{1}{{\mathrm{λ}}_i^2},$

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

${\mathrm{μ}}_i=10+\frac{i}{10}\text{and}{\mathrm{σ}}_i^2=100+2i+\frac{i^2}{100}$

Let $X$denote the total lifetime of all componentslocalid="1649784745171" $X=\underset{1}{\overset{100}{∑}}{X}_i$.

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

$$\begin{gathered} E\left[X\right]=E\left[\underset{i=1}{\overset{100}{∑}}{X}_i\right]=\underset{i=1}{\overset{100}{∑}}{\mathrm{μ}}_i \\ =\underset{i=1}{\overset{100}{∑}}\left(10+\frac{i}{10}\right)=100\left(10\right)+\frac{1}{10}\underset{i=1}{\overset{100}{∑}}i\stackrel{(*)}{=}100\left(10\right)+\frac{100\left(101\right)}{10\left(2\right)} \\ =1505 \end{gathered}$$

and variance

$$\begin{gathered} Var\left(X\right)=Var\left(\underset{i=1}{\overset{100}{∑}}{X}_i\right)=\underset{i=1}{\overset{100}{∑}}{\mathrm{σ}}_i^2=\underset{i=1}{\overset{100}{∑}}\left(100+2i+\frac{i^2}{100}\right) \\ =100\left(100\right)+2\underset{i=1}{\overset{100}{∑}}i+\frac{1}{100}\underset{i=1}{\overset{100}{∑}}{i}^2 \end{gathered}$$

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

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:

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

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

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

The lifetime of components is uniformly distributed.

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

$i=1,…,100\text{.}$So,

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

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

In this case, the total lifetime of all componentslocalid="1649784800135" $X=\underset{1}{\overset{100}{∑}}{X}_{ir}$is a random variable with mean

$E\left[X\right]=E\left[\underset{i=1}{\overset{100}{∑}}{X}_i\right]=\underset{i=1}{\overset{100}{∑}}{\mathrm{μ}}_i=\underset{i=1}{\overset{100}{∑}}\left(10+\frac{i}{10}\right)=1505$

and variance

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

Using the central limit theorem we get the required probability:

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