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.

Solution :

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

The number of errors thus follows the Poisson distribution.

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

=

= 1 error

Let X = the number of the errors in the hours

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

So the required probability is :

So the that the will be misdiagnosed as "" on the basis of the test whose result is 0.264241

there is no question content here, but you for the free points! : d

n=9

step-by-step explanation:

7 times a number = 7n

decreased by 13 = 7n-13

same as is =

5 times the number = 5n

increased by 5 = 5n+5

7n-13 = 5n+5

2n=18

n=9