Answer:
The probability that a given lot will violate this guarantee is
, in which x is the probability of a resistor being defective.
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
![P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20%5Cfrac%7Be%5E%7B-%5Cmu%7D%2A%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
In which
x is the number of sucesses
e = 2.71828 is the Euler number
is the mean in the given interval.
The probability of a resistor’s being defective is x:
This means that
, in which n is the number of resistors.
The resistors are sold in lots of 200
This means that
, so ![\mu = 200x](https://tex.z-dn.net/?f=%5Cmu%20%3D%20200x)
What is the probability that a given lot will violate this guarantee?
This is:
![P(X \geq 1) = 1 - P(X = 0)](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%201%29%20%3D%201%20-%20P%28X%20%3D%200%29)
In which
![P(X = 0) = \frac{e^{-200x}*(200x)^{0}}{(0)!} = e^{-200x}](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20%5Cfrac%7Be%5E%7B-200x%7D%2A%28200x%29%5E%7B0%7D%7D%7B%280%29%21%7D%20%3D%20e%5E%7B-200x%7D)
So
![P(X \geq 1) = 1 - e^{-200x}](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%201%29%20%3D%201%20-%20e%5E%7B-200x%7D)
The probability that a given lot will violate this guarantee is
, in which x is the probability of a resistor being defective.