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ad-work [718]
3 years ago
13

El tiempo hasta que falle un sistema informático sigue una distribución Exponencial con media de 600hs. (Utilice 3 decimales par

a sus respuestas) Si el sistema ha estado funcionando sin fallas durante 400hs ¿cuál es la probabilidad de que funcione por lo menos 200hs más?
Engineering
1 answer:
Lesechka [4]3 years ago
7 0

Answer:

La probabilidad pedida es 0.717

Explanation:

Primero comencemos definiendo la variable aleatoria. Para nuestro problema, la variable aleatoria es la siguiente :

X: '' El tiempo (en horas) hasta que falle un sistema informático ''

La variable aleatoria X será entonces una variable aleatoria continua.

Sabemos que sigue una distribución exponencial con una media de 600 hs.

Esto se escribe :

X ~ ε ( λ ) (I)

En donde λ es igual a la inversa de la media. Esto se escribe :

λ =\frac{1}{E(X)}

En donde E(X) es la media de la variable. Por ende, si reemplazamos los datos del ejercicio obtenemos ⇒

λ =\frac{1}{E(X)} ⇒ λ =\frac{1}{600}

Si reemplazamos el valor de λ en (I) obtenemos :

X ~ ε (\frac{1}{600})

La función de distribución de X (por ser una variable aleatoria exponencial) es :

F_{X}(x)=P(X\leq x)=  1 - e ^ ( - λx) con x > 0 y F_{X}(x)=0 en caso contrario.

Si reemplazamos el valor de λ en la función de distribución de X obtenemos :

F_{X}(x)=P(X\leq x)=1-e^{-\frac{x}{600}}  

Dado que la variable aleatoria X se distribuye de manera exponencial, el hecho de saber que el sistema ha estado funcionando sin fallas durante 400 hs no nos aporta ninguna información sobre lo que ocurrirá después. Esta característica se conoce como propiedad de perdida de memoria de la variable aleatoria exponencial. Entonces, la probabilidad pedida se reduce a calcular :

P(X>200)    

Dado que saber que el sistema ha estado funcionando sin fallas durante 400 hs no nos dice nada sobre lo que ocurrirá instantes posteriores a esas 400 hs.

Calculamos entonces la probabilidad pedida :

P(X>200)=1-P(X\leq 200)=1-F_{X}(200)=1-(1-e^{-\frac{200}{600}})=e^{-\frac{1}{3}}=0.717

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A bakery wants to determine how many trays of doughnuts it should prepare each day. Demand is normal with a mean of 15 trays and
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The number of trays that should be prepared if the owner wants a service level of at least 95% is; 7 trays

<h3>How to utilize z-score statistics?</h3>

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Formula for z-score is;

z = (x' - μ)/σ

We want to find the value of x such that the probability is 0.95;

P(X > x) = P(z > (x - 15)/5) = 0.95

⇒ 1 -  P(z ≤ (x - 15)/5) = 0.95

Thus;

P(z ≤ (x - 15)/5) = 1 - 0.95

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Complete Question is;

A bakery wants to determine how many trays of doughnuts it should prepare each day. Demand is normal with a mean of 15 trays and standard deviation of 5 trays. If the owner wants a service level of at least 95%, how many trays should he prepare (rounded to the nearest whole tray)? Assume doughnuts have no salvage value after the day is complete. 6 5 4 7 unable to determine with the above information.

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2 years ago
A closed system of mass 10 kg undergoes a process during which there is energy transfer by work from the system of 0.147 kJ per
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Answer:

-50.005 KJ

Explanation:

Mass flow rate = 0.147 KJ per kg

mass= 10 kg

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Δu= -5 kJ/kg

ΔKE + ΔPE+ ΔU= Q-W

0.5×m×(30^2- 15^2)+ mgΔh+mΔu= Q-W

Q= W+ 0.5×m×(30^2- 15^2) +mgΔh+mΔu

= 1.47 +0.5×1/100×(30^2- 15^2)-9.7×50/1000-50

= 1.47 +3.375-4.8450-50

Q=-50.005 KJ

7 0
3 years ago
Read 2 more answers
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