Mathematics: Suppose that an appliance is constructed in such a way that it requires that n independent electronic components are all functioning. Assume that the lifespan of each of these, Tj, is an exponential random variable with parameter Î»j. a. Let X be the random variable giving the lifespan of the appliance. Find the CDF and PDF for X. b. Find the expected value of X. c. Now find the median lifespan of the appliance (that is, the time t at which half of the appliances are likely to have broken and half

Suppose that an appliance is constructed in such

Ответ:

clevelandjaniya

18.04.2021

Step-by-step explanation:

$T_j \sim exp ( \lambda j) \ \ \ j = 1 (1) n \\ \\ f_{Tj}(tj) = \lambda j e^{-\lambda j tj } , tj0 \\ \\ P\Big[ Tj x\Big] = \int \limits ^{\infty}_{x} \lambda j e^{-\lambda j tj}\ dtj \\ \\ = e^{-\lambda j x}, x 0 \\ \\ \\ \\ a) F_x (x) \\ \\ = P[X \le x ] \\ \\ = 1 - P[X x] \\ \\ = 1 - \pi \limits ^{n}_{j =1} \Big\{ P[T_j x ] \Big \}$

$\text{This is because the appliance has the capacity to work for more than (x) }$ $\text{hours if and only if all the "n" components work more than (x) hours.}$

$\text{Then:}$

$= 1 - e \ \pi^{n}_{j=1} \ e^{-\lambda j x} \\ \\ = 1 - e^{- (\sum \limits ^{n}_{j=1} \lambda j)x}$

∴

$CDF = 1 - e^{- (\sum \limits ^{n}_{j=1} \lambda j)x}\ , \ x0$

$PDF =\Big( \sum \limits ^{n}_{j =1} \lambda j\Big) e^{- (\sum \limits ^{n}_{j=1} \lambda j)x}\ , \ x0$

$f_x(x) = \Big( \sum \limits ^{n}_{j =1} \lambda j\Big) e^{- (\sum \limits ^{n}_{j=1} \lambda j)x}\ , \ x0$

(b)

$E(x) = \int \limits ^{\infty}_{o }x f_x (x) \ dx \\ \\ = \int \limits ^{\infty}_{o }x \ \Big( \sum \limits ^{n}_{j =1} \lambda j\Big) e^{- (\sum \limits ^{n}_{j=1} \lambda j)x}\ , \ dx$

$= \dfrac{1}{\sum \limits ^n_{j=1} \lambda j} \ \ \int \limits ^{\infty}_{o} \Big [(\sum \limits ^n_{j=1} \lambda j )x \Big] e ^{-\Big ( \sum \limits ^{n}_{j=1} \lambda j \Big)x} \ \ d \Big( x \sum \limits ^n_{j=1} \lambda j \Big)$

$= \dfrac{1}{\sum \limits ^n_{j= 1} \lambda j} \int \llimits ^{\infty}_{o} t e^{-t} \ dt$

$= \dfrac{1}{\sum \limits ^n_{j= 1} \lambda j} \int \llimits ^{\infty}_{o} t e^{-t} \ dt \ \ \ \ \text{\Big[By transformation }t =( \sum \limits ^n_{j=1} \lambda j )x \Big]$

$= \dfrac{1}{\sum \limits ^n_{j= 1} \lambda j}$

(c)

$Let \ \ l_y_{\dfrac{1}{2}} \text{ be the median}$

∴

$F(l_y_{\dfrac{1}{2}}) = \dfrac{1}{2} \\ \\ 1 - e ^{-\Big (\sum \limits ^{n}_{j=1} \lambda j \Big)} {l_y}_{\frac{1}{2}} = \dfrac{1}{2} \\ \\ \dfrac{1}{2} = e^{-\Big (\sum \limits ^{n}_{j=1} \lambda j \Big) } {l_y}_{\frac{1}{2}} \\ \\ - In 2 = {-\Big (\sum \limits ^{n}_{j=1} \lambda j \Big)} {l_y}_{\frac{1}{2}} \\ \\ \\ \\ \mathbf{ {l_y}_{\frac{1}{2}} = \dfrac{ In \ 2}{\sum \limits ^{n}_{j=1} \lambda j }}$

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Ответ:

afosburgh20

30.05.2022

Mathematics

-6

Step-by-step explanation:

Hello!

Since we are finding f(-1) we put -1 in for x

2(-1)^2 + 8(-1)

Follow PEMDAS

2 * 1 + 8(-1)

2 * 1 + -8

2 + -8 = -6

The answer is -6

Hope this helps!

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