The question is :
There are $n$ jobs and $n$ machines. $T_i$ is the random variable for machine $i$ to complete its execution. Let $X= \max\{T_1 , \dots , T_n \}$ and let $Y = \min\{T_1 , \dots , T_n \}$ $T_i - \exp(\lambda_i)$.
Let $n=5$ and let the average time for job $n$ to be executed by $20$(ms). What is the probability that the total time is below $15$ms for $X$ and $Y$ respectively?
I need to know how I can use average time in this question to get to the result. It seems to me that I am missing how expected value can be used in this kind of question. Any advice is highly appreciated.
In general, if $Z_1,\ldots,Z_n$ are i.i.d. random variables, denote \begin{align} Z_\max &:= \max\{Z_1,\ldots,Z_n\},\\ Z_\min &:= \min\{Z_1,\ldots,Z_n\}.\\ \end{align}
For any $t\in\mathbb R$ we have $$ \{Z_\max \leqslant t\} = \bigcap_{i=1}^n \{Z_i\leqslant t\}. $$ It follows from independence that $$ \mathbb P(Z_\max \leqslant t) = \mathbb P\left(\bigcap_{i=1}^n \{Z_i\leqslant t\}\right) =\prod_{i=1}^n \mathbb P(Z_i\leqslant t) $$ and from being identically distributed that $$ \mathbb P(Z_\max \leqslant t) = \mathbb P(Z_1\leqslant t)^n = F_Z(t)^n, $$ where $F_Z$ is the distribution function of $Z_1$. Similarly, $$ \{Z_\min > t\} = \bigcap_{i=1}^n\{Z_i > t\}, $$ so that \begin{align} \{Z_\min > t\} &= \mathbb P\left(\bigcap_{i=1}^n\{Z_i > t\}\right)\\ &= \prod_{i=1}^n \mathbb P(Z_i>t)\\ &= \mathbb P(Z_1>t)^n\\ &= (1 -F_Z(t))^n, \end{align}
so that $\mathbb P(Z_\min \leqslant t) = 1 - (1-F_Z(t))^n$. Now let $T_1,\ldots,T_n$ be independent exponentially distributed random variables with parameter $\lambda$. Then \begin{align} \mathbb P(T_\max \leqslant t) &= (1-e^{-\lambda t})^n,\\ \mathbb P(T_\min \leqslant t) &= 1 - e^{-n\lambda t}. \end{align}
Note that $T_\min$ itself is exponentially distributed, with parameter $n\lambda$. To find the answer for your specific question, plug in $\lambda=1/20$, $n=5$, and $t=15$.