Question

Suppose that in the production of 60-ohm radio resistors, nondefective items are those that have a resistance between 58 and 62 ohms and the probability of a resistor’s being defective is The resistors are sold in lots of 200, with the guarantee that all resistors are nondefective. What is the probability that a given lot will violate this guarantee? (Use the Poisson distribution.)

Answer 1

Answer:

The probability that a given lot will violate this guarantee is $1 - e^{-200x}$, 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:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

The probability of a resistor’s being defective is x:

This means that $\mu = nx$, in which n is the number of resistors.

The resistors are sold in lots of 200

This means that $n = 200$, so $\mu = 200x$

What is the probability that a given lot will violate this guarantee?

This is:

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = 0) = \frac{e^{-200x}*(200x)^{0}}{(0)!} = e^{-200x}$

So

$P(X \geq 1) = 1 - e^{-200x}$

The probability that a given lot will violate this guarantee is $1 - e^{-200x}$, in which x is the probability of a resistor being defective.