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Setler79
3 months ago
8

The Weibull distribution is widely used in statistical problems relating to aging of solid insulating materials subjected to agi

ng and stress. Use this distribution as a model for time (in hours) to failure of solid insulating specimens subjected to AC voltage. The values of the parameters depend on the voltage and temperature; suppose α = 2.6 and β = 220. (a) What is the probability that a specimen's lifetime is at most 250? Less than 250? More than 300? (Round your answers to four decimal places.) at most 250 less than 250 more than 300 (b) What is the probability that a specimen's lifetime is between 100 and 250? (Round your answer to four decimal places.) (c) What value is such that exactly 50% of all specimens
Mathematics
1 answer:
AnnZ [12.3K]3 months ago
7 0

Answer:

a

P(X \le 250 ) = 0.7564 [/tex], P(X < 250 ) = 0.7564,

P(X < 300 ) = 0.09922

b

P(100 < X < 250 ) =0.644

c

x = 192.1

Step-by-step explanation:

As per the question, we know that

The value for \alpha = 2.6

The value for \beta = 220

In general, the Weibull distribution function is expressed mathematically as

F( x, \alpha, \beta ) = \left \{ 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x < 0} \atop { 1- e^{-(\frac{x}{\beta } )^{\alpha } }}\ \ \ \ \ \ x \ge 0} \right

Typically, the probability for a specimen's lifetime being no more than 250 is mathematically illustrated as

P(X \le 250 ) = F(250, 2.7, 220 )

P(X \le 250 )=1 - e^{- (\frac{250}{220} )^{2.7}}

P(X \le 250 ) = 1 - 0.2436

P(X \le 250 ) = 0.7564

Usually, the probability that a specimen's lifespan is below 250

P(X < 250 ) = F(250, 2.7, 220 )

[texP(X < 250 ) =1 - e^{- (\frac{250}{220} )^{2.7}}[/tex]

P(X < 250 ) = 1 - 0.2436

P(X < 250 ) = 0.7564

Usually, the probability that a specimen's life is more than 300

P(X > 300 ) = 1- p(X \le 300 )

P(X > 300 ) = 1- F(300, 2.7, 220 )

[texP(X < 300) =1- [1 - e^{- (\frac{300}{220} )^{2.7}}][/tex]

P(X < 300 ) = 0.09922

Generally, the likelihood that a specimen's lifespan falls between 100 and 250 is

P(100 < X < 250 ) = P(X < 250) - P(X < 100)

=> P(100 < X < 250 ) =F(250, 2.7, 220 ) - F(100, 2.7, 220 )

=> P(100 < X < 250 ) =(1 - e^{-(\frac{250}{220})^{2.7}}) - (1 - e^{-(\frac{100}{220})^{2.7}})

=> P(100 < X < 250 ) = (1 - 0.244 ) - (1- 0.888)

=> P(100 < X < 250 ) =0.644

Generally, this value is such that precisely 50% of all specimens

P(X > x) = 1-P(X < x) = 0.50

=> P(X > x) = 1- (1 - e^{- (\frac{x}{220}) ^{2.7}}) = 0.50

=> P(X> x ) = e^(- \frac{x}{20})^{2.7} = 0.50

=> P(X> x ) = (- \frac{x}{20})^{2.7} = ln0.50

=> P(X> x ) = \frac{x}{20} =[ -ln0.50 ] ^{frac{1}{2.7}}

=> x = 220[ -ln0.50 ] ^{frac{1}{2.7}}

=> x = 192.1

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Answer and Explanation:

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3) The likelihood that the average lifespan of the sample falls between

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P

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−

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<

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Copy N 3

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Copy N 4

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Copy N 5

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Copy N 6

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Copy N 7

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The answer to Part b) is

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