Q 5.4 The probability density function... [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 5: Q 5.4 (page 212)

The probability density function of X, the lifetime of a certain type of electronic device (measured in hours), is given by
$f (x) = \{\begin{cases} \frac{10}{x^{2}} & x > 10 \\ 0 & x 鈮� 10 \end{cases}$
$(a)$ Find $P {X > 20}$
localid="1646589462481" $(b)$ What is the cumulative distribution function of localid="1646589521172" $X ?$
localid="1646589534997" $(c)$ ) What is the probability that of $6$ such types of devices, at least localid="1646589580632" $3$ will function for at least localid="1646589593287" $15$ hours? What assumptions are you making?

Short Answer

Expert verified

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

$(b)$ The cumulative distribution function (COD) of $X$ is $\{\begin{cases} 0 & x 鈭� (- 鈭�, 10] \\ 1 - \frac{10}{x} & x 鈭� (10, 鈭�) \end{cases}$

$(c)$ $P (A_{3}) = {鈭�}_{i = 1}^{6} (\begin{matrix} 6 \\ i \end{matrix}) {(\frac{2}{3})}^{i} {(\frac{1}{3})}^{6 - i}$

Step by step solution

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

$f (x) = \{\begin{cases} \frac{10}{x^{2}} & x > 10 \\ 0 & x 鈮� 10 \end{cases}$

Part a Step 2 explanation

We find as follows

$P {X > 20} = {鈭�}_{20}^{鈭�} f (x) d x = 10 {鈭�}_{20}^{鈭�} x^{- 2} d x = 10 \lim_{t 鈫� 鈭�} {鈭�}_{20}^{t} x^{- 2} d x = 10 \lim_{t 鈫� 鈭�} {[\frac{- 1}{x}]}_{20}^{t} = 10 \lim_{t 鈫� 鈭� (\begin{matrix} \frac{- 1}{t} + & \frac{1}{20} \end{matrix})} = \frac{10}{20} = 0.5$

Part b Step 1 Explanation

We have that

$F (x) = P (X 鈮� x) = {鈭�}_{- 鈭�}^{x} f (t) d t = 10 {鈭�}_{10}^{x} t^{- 2} d t = 10 {[\frac{- 1}{t}]}_{10}^{x} = 10 (- \frac{1}{x} + \frac{1}{10}) = 1 - \frac{10}{x}$

We can express the CDF AS $F (x) \{\begin{cases} 0 & x 鈭� (- 鈭�, 10] \\ 1 - \frac{10}{x} & x 鈭� (10, 鈭�) \end{cases}$

part c Step 1 Explanation.

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

If we let $A_{3}$ denote the event that at least three will work for at least $15$ hours, the we have

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Short Answer

Step by step solution

Part a Step 1 Given information.

Part a Step 2 explanation

Part b Step 1 Explanation

part c Step 1 Explanation.

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