Question

The number of defective circuit boards coming off a soldering machine follows a Poisson distribution. During a specific ten-hour period, one defective circuit board was found. (a) Find the probability that it was produced during the first hour of operation during that period. (Round your answer to four decimal places.) (b) Find the probability that it was produced during the last hour of operation during that period. (Round your answer to four decimal places.) (c) Given that no defective circuit boards were produced during the first five hours of operation, find the probability that the defective board was manufactured during the sixth hour. (Round your answer to four decimal places.)

Answer 1

a) the probability that the defective board was produced during the first hour of operation is
`(1)/(10)` or 0.1000

b) the probability that the defective board was produced during the last hour of operation is
`(1)/(10)` or 0.1000

c) the required probability is 0.2000

Explanation:

Given the data in the question;

During a specific ten-hour period, one defective circuit board was found.

Lets X represent the number of defective circuit boards coming out of the machine , following Poisson distribution on a particular 10-hours workday which one defective board was found.

Also let Y represent the event of producing one defective circuit board, Y is uniformly distributed over ( 0, 10 ) intervals.

f(y) =
`\left \{ {{(1)/(b-a) }\\\ }} \right _0`; ( a ≤ y ≤ b )
`_{elsewhere`

=
`\left \{ {{(1)/(10-0) }\\\ }} \right _0`; ( 0 ≤ y ≤ 10 )
`_{elsewhere`

f(y) =
`\left \{ {{(1)/(10) }\\\ }} \right _0`; ( 0 ≤ y ≤ 10 )
`_{elsewhere`

Now,

a) the probability that it was produced during the first hour of operation during that period;

P( Y < 1 ) =
`\int\limits^1_0 {f(y)} \, dy`

we substitute

=
`\int\limits^1_0 {(1)/(10) } \, dy`

=
`(1)/(10) [y]^1_0`

=
`(1)/(10) [ 1 - 0 ]`

=
`(1)/(10)` or 0.1000

Therefore, the probability that the defective board was produced during the first hour of operation is
`(1)/(10)` or 0.1000

b) The probability that it was produced during the last hour of operation during that period.

P( Y > 9 ) =
`\int\limits^(10)_9 {f(y)} \, dy`

we substitute

=
`\int\limits^(10)_9 {(1)/(10) } \, dy`

=
`(1)/(10) [y]^(10)_9`

=
`(1)/(10) [ 10 - 9 ]`

=
`(1)/(10)` or 0.1000

Therefore, the probability that the defective board was produced during the last hour of operation is
`(1)/(10)` or 0.1000

c)

no defective circuit boards were produced during the first five hours of operation.

probability that the defective board was manufactured during the sixth hour will be;

P( 5 < Y < 6 | Y > 5 ) = P[ ( 5 < Y < 6 ) ∩ ( Y > 5 ) ] / P( Y > 5 )

= P( 5 < Y < 6 ) / P( Y > 5 )

we substitute

`= (\int\limits^(6)_5 {(1)/(10) } \, dy) / (\int\limits^(10)_5 {(1)/(10) } \, dy)`

`= ((1)/(10) [y]^(6)_5) / ((1)/(10) [y]^(10)_5)`

= ( 6-5 ) / ( 10 - 5 )

= 0.2000

Therefore, the required probability is 0.2000