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

Solution :

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

= $0.2 \times 5$

= 1 error

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

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

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 :

$P(X>1)$

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

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

$=1- \left( \frac{e^{-1} 1^0}{0!} + \frac{e^{-1} 1^0}{1!} \right)$

$=1-(2 \times e^{-1})$

$=1-( 2 \times 0.367879)$

$=1-0.735759$

$=0.264241$

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