The Weibull distribution is widely used in statistical problems relating to aging of solid insulating materials subjected to agi
1 answer:
Answer:
a
P(X \le 250 ) = 0.7564 [/tex], ,
b
c
Step-by-step explanation:
As per the question, we know that
The value for
The value for
In general, the Weibull distribution function is expressed mathematically as
Typically, the probability for a specimen's lifetime being no more than 250 is mathematically illustrated as
Usually, the probability that a specimen's lifespan is below 250
[texP(X < 250 ) =1 - e^{- (\frac{250}{220} )^{2.7}}[/tex]
Usually, the probability that a specimen's life is more than 300
[texP(X < 300) =1- [1 - e^{- (\frac{300}{220} )^{2.7}}][/tex]
Generally, the likelihood that a specimen's lifespan falls between 100 and 250 is
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Generally, this value is such that precisely 50% of all specimens
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Answer:
Step-by-step explanation:
To solve for v, reverse the operations performed on it, starting with the equation provided. Here, v is affected by:
- being multiplied by t
- subtracting gt^2 from the result
To start, we first need to add gt^2 back to the equation to counteract the subtraction:
h + gt^2 = vt
Next, we undo the multiplication by dividing the entire expression by the coefficient of v:
(h + gt^2)/t = v