Please help me solve the following problem
Time of production of one electronic component is given with
exponential distribution with parameter λ. If the process lasts less than 3 hours, the component is working, otherwise it is defective. Events that component is working or that is defective have equally probability. Find:
- Parameter λ
- Probability that the process will last less than 1 hour.
Best Answer
Let $X$ be the random variable that measures length of life, in hours. We are told that $X$ has density function $\lambda e^{-\lambda x}$ (for $x\gt 0)$.
We are also told that $\Pr(X\lt 3)=\Pr(X\ge 3)$. Each is therefore $\frac{1}{2}$.
But $\Pr(X\gt x)$, by integration, is equal to $e^{-\lambda x}$. It follows that $$e^{-3\lambda}=\frac{1}{2}.$$ Now we can solve for $\lambda$, by taking logarithm to the base $e$ of both sides.
And now that we have $\lambda$, we can find the probability the thing lasts less than $1$ hour, since $\Pr(X\le x)=1-e^{-\lambda x}$ for $x\gt 0$.
Remarks: $1.$ We were asked to find $\lambda$, so we did. But note that the second problem can be solved without finding $\lambda$ explicitly. For we had $e^{-3\lambda}=\frac{1}{2}$. Taking cube roots, we find that $e^{-\lambda}=2^{-1/3}$, and therefore $\Pr(X\lt 1)=1-2^{-1/3}$.
$2.$ The calculations of the problem are closely related to half-life calculations that you have undoubtedly done several times in the past. Individual atoms of a radioactive isotope have a lifetime that is exponentially distributed. The median lifetime is called the half-life of the substance.