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

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.

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) = (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) = (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.