El espesor de un borde de un componente de una aeronave está distribuido de manera uniforme entre 0.05 y 1.05 milimetros.Obtenga la función de distribución acumulada del espesor del borde.Calcule la proporción de bordes cuyo espesor es mayor que 1.02 milimetros.Qué espesor estEa excedido por el 90% de los bordes.Calcule la media y la varianza del espesor.

Answers

Answer 1

Answer:

a) [tex]F(x)\begin{cases} & 0\text{ if } x \leq 0.05 \\ \\ &\dfrac{x - 0.05}{1.05 - 0.05} = x - 0.05 \text{ if } 0.05 \leq x \leq 1.05 \\ \\ & 1 \text{ if } x \geq 1.05 \end{cases}[/tex]

b) 0.03

c) 0.15

d) La media es 0.55

La varianza es 8.[tex]\overline 3[/tex]

Step-by-step explanation:

La variación del espesor del componente de la aeronave = 0.05 a 1.05

El pdf de la variable 'X' se da de la siguiente manera;

[tex]f(x) = \dfrac{1}{1.05 - 0.05} = 1[/tex] por 0.05 < x < 1.05

a)

[tex]F(x)\begin{cases} & 0\text{ si } x \leq 0.05 \\ \\ &\dfrac{x - 0.05}{1.05 - 0.05} = x - 0.05 \text{ si } 0.05 \leq x \leq 1.05 \\ \\ & 1 \text{ si } x \geq 1.05 \end{cases}[/tex]

b) Para la proporción de los bordes que con un espesor superior a 1.02 milímetros, obtenemos;

P(X > 1.02) = 1 - F(1.02) = 1 - (1.02 - 0.05) = 0.03

P(X > 1.02) = 0.03

La proporción de bordes que tienen un espesor mayor que 1.02 = 0.03

c) El espesor, 'x', excedido en el 90% de los bordes, se encuentra de la siguiente manera;

P(X > x) = 1 - F(x)

∴ 90/100 = 0.9 = 1 - F(x) = 1 - (x - 0.05)

x - 0.05 = 1 - 0.9 = 0.1

x = 0.1 + 0.05 = 0.15

∴ El espesor excedido en el 90% de los bordes, x = 0.15

d) Para una variable aleatoria uniforme continua, a ≤ x ≤ b, tenemos;

[tex]El \ significado, \mu = E(X) = \dfrac{(a + b)}{2}[/tex]

[tex]La \ varianza, \sigma^2 = V(X) = \dfrac{(b - a)^2}{12}[/tex]

Para la distribución de espesor de borde dada, tenemos;

0.05 < x < 1.05

[tex]\therefore El \ significado, \mu = E(X) = \dfrac{(0.05 + 1.05)}{2} = 0.55[/tex]

El significado E(X) = 0.55

[tex]La \ varianza, \sigma^2 = V(X) = \dfrac{(1.05 - 0.05)^2}{12} = 1/12 = 8.\overline 3[/tex]

La varianza, V(X) = 8.[tex]\overline 3[/tex]


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