Q 5.4 - A First Course in Probability

Question

The probability density function of X, the lifetime of a certain type of electronic device (measured in hours), is given by 

$f\left(x\right)=\left\{\begin{array}{ll}\frac{10}{x^2}& x>10\\ 0& x\le 10\end{array}\right.$

$\left(a\right)$Find$P\{X>20\}$

$\left(b\right)$What is the cumulative distribution function of $X?$

$\left(c\right)$) What is the probability that of$6$such types of devices, at least $3$will function for at least $15$hours? What assumptions are you making?

Step-by-Step Solution

Answer

$\left(a\right) P\{X>20\}=0.5$

$\left(b\right)$The cumulative distribution function (COD) of$X$ is$\left\{\begin{array}{ll}0& x\in (-\infty ,10]\\ 1-\frac{10}{x}& x\in (10,\infty )\end{array}\right.$

$\left(c\right)$$P\left(A_3\right)=\sum _{i=1}^6\left(\begin{array}{c}6\\ i\end{array}\right){\left(\frac{2}{3}\right)}^i{\left(\frac{1}{3}\right)}^{6-i}$

Part a Step 1 Given information.

The probability density function of $X$, the lifetime of a certain type of electronic device (measured in hours), is given by

$f\left(x\right)=\left\{\begin{array}{ll}\frac{10}{x^2}& x>10\\ 0& x\le 10\end{array}\right.$

Part a Step 2 explanation

We find as follows 

$$\begin{gathered} P\{X>20\}={\int }_{20}^{\infty }f\left(x\right)dx \\ =10{\int }_{20}^{\infty }{x}^{-2}dx \\ =10\underset{t\to \infty }{\lim }{\int }_{20}^t{x}^{-2}dx \\ =10\underset{t\to \infty }{\lim }{\left[\frac{-1}{x}\right]}_{20}^t \\ =10\underset{t\to \infty \left(\begin{array}{cc}\frac{-1}{t}+& \frac{1}{20}\end{array}\right)}{\lim } \\ =\frac{10}{20}=0.5 \end{gathered}$$

Part b Step 1 Explanation

We have that

$$\begin{gathered} F\left(x\right)= P(X\le x)={\int }_{-\infty }^{x}f\left(t\right)dt \\ =10{\int }_{10}^x{t}^{-2}dt \\ =10{\left[\frac{-1}{t}\right]}_{10}^x \\ =10\left(-\frac{1}{x}+\frac{1}{10}\right) \\ =1-\frac{10}{x} \end{gathered}$$

We can express the CDF AS $F\left(x\right)\left\{\begin{array}{ll}0& x\in (-\infty ,10]\\ 1-\frac{10}{x}& x\in (10,\infty )\end{array}\right.$

part c Step 1 Explanation.

Let $A$be the event that a single device works for at least $15$hours. We thus have that$$\begin{gathered} P\left(A\right)=1-F\left(15\right) \\ =1-\left(1-\frac{10}{15}\right) \\ =\frac{10}{15}=\frac{2}{3} \end{gathered}$$

If we let$A_3$ denote the event that at least three will work for at least $15$hours, the we have