Q. 8.5 8.5 The amount of time that a ce... [FREE SOLUTION]

91影视

A First Course in Probability

Sheldon M. Ross

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 432 pages

ISBN: 9780321794772

Chapter 8: Q. 8.5 (page 393)

8.5 The amount of time that a certain type of component functions before failing is a random variable with probability density function
$f (x) = 2 x 0 < x < 1$
Once the component fails, it is immediately replaced by
another one of the same type. If we let denote the life-time of the $i$ th component to be put in use, then $S_{n} = {鈭�}_{i = 1}^{n} X_{i}$ represents the time of the $n$ th failure. The long-term rate at which failures occur, call it $r$ , is defined by
$r = \lim_{n 鈫� 鈭�} \frac{n}{S_{n}}$
Assuming that the random variables $X i, i 鈮� 1,$ are independent, determine $r$ .

Short Answer

Expert verified

The r is $\frac{3}{2}$ .

Step by step solution

Given information

A random variable with probability density function is $f (x) = 2 x 0 < x < 1 .$ Let $X_{i}$ denote the life-time of the $i$ th component , then $S_{n} = {鈭�}_{i = 1}^{n} X_{i}$ represents the time of the $n$ th failure. The long-term rate failures occur, is defined by $r = \lim_{n 鈫� 鈭�} \frac{n}{S_{n}}$ .

Explanation

Let $X_{i}$ denote the $i$ th component and understand that $X_{1}, X_{2}, . . .$ , is a sequence of independent and similarly distributed random variables, with finite mean.
$渭 = E [X_{i}] = {鈭�}_{0}^{1} x f (x) d x = {鈭�}_{0}^{1} 2 x^{2} d x = {2 \frac{x^{3}}{3}|}_{0}^{1} = \frac{2}{3}$
Then, $S_{n} = {鈭�}_{i = 1}^{n} X_{i}$ represents the time of $n$ th failure, with probability $1$ ,
$\frac{S_{n}}{n} 鈫� 渭,$ as $n 鈫� 鈭�$

Then,

$r = \lim_{n 鈫� 鈭�} \frac{n}{S_{n}} = \lim_{n 鈫� 鈭�} \frac{1}{\frac{S_{n}}{n}} = \frac{1}{渭} = \frac{3}{2}$

Hence, $r = \frac{3}{2}$

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