Question

An electronic switching device occasionally malfunctions, but the device is considered satisfactory if it makes, on average, no more than 0.20 error per hour. A particular 5-hour period is chosen for testing the device. If no more than 1 error occurs during the time period, the device will be considered satisfactory.

(a) What is the probability that a satisfactory device will be considered unsatisfactory on the basis of the test? Assume a Poisson process.
(b) What is the probability that a device will be accepted as satisfactory when, in fact, the mean number of errors is 0.25? Again, assume a Poisson process.

Answer 1

The probability that a satisfactory device will be considered unsatisfactory on the basis of the test is 0.7358 or 73.58%.

Step-by-step explanation:

In this problem, we are dealing with a Poisson process, where the number of errors occurring in a given time period follows a Poisson distribution. We are given that a satisfactory device makes, on average, no more than 0.20 errors per hour. Therefore, the mean number of errors during the 5-hour period is 5 * 0.20 = 1. Using the Poisson distribution formula, we can find the probability of no more than 1 error occurring. Let's calculate:

1. Probability(X = 0) = (e^(-1) * 1^0) / 0! = 0.3679
2. Probability(X = 1) = (e^(-1) * 1^1) / 1! = 0.3679
3. Probability(X ≤ 1) = Probability(X = 0) + Probability(X = 1) = 0.3679 + 0.3679 = 0.7358

Therefore, the probability that a satisfactory device will be considered unsatisfactory on the basis of the test is 0.7358 or 73.58%.

Answer 2

It is given that the device works satisfactorily if it makes an average of no more than
`0.2` errors per hour.

The number of errors thus follows the Poisson distribution.

It is given that in
`5` hours test period, the number of the errors follows is

=
`0.2 * 5`

= 1 error

Let X = the number of the errors in the
`5` hours

`$X \sim \text{Poisson } (\lambda = 0.2 * 5 =1)$`

Now that we want to find the
`\text{probability}` that a
`\text{satisfactory device}` will be misdiagnosed as "
`\text{unsatisfactory}`" on the basis of this test. We know that device will be unsatisfactory if it makes more than
`1` error in the test. So we will determine probability that X is greater than
`1` to get required answer.

So the required probability is :

`P(X>1)`

`$=1-P(X \leq 1)$`

`$=1-[P(X=0)+P(X=1)]$`

`$=1- \left( (e^(-1) 1^0)/(0!) + (e^(-1) 1^0)/(1!) \right) $`

`$=1-(2 * e^(-1))$`

`$=1-( 2 * 0.367879)$`

`$=1-0.735759$`

`=0.264241`

So the
`\text{probability}` that the
`\text{satisfactory device}` will be misdiagnosed as "
`\text{unsatisfactory}`" on the basis of the test whose result is 0.264241