Mathematics, 27.07.2021 02:50, Katlynnmarkle13

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.)

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answered: codie1103

a) the probability that the defective board was produced during the first hour of operation is $\frac{1}{10}$ or 0.1000

b) the probability that the defective board was produced during the last hour of operation is $\frac{1}{10}$ or 0.1000

c) the required probability is 0.2000

Step-by-step 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) = $\left \{ {{\frac{1}{b-a} }\\\ }} \right _0$ ; ( a ≤ y ≤ b ) $_{elsewhere$

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

f(y) = $\left \{ {{\frac{1}{10} }\\\ }} \right _0$ ; ( 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

= $\int\limits^1_0 {\frac{1}{10} } \, dy$

= $\frac{1}{10} [y]^1_0$

= $\frac{1}{10} [ 1 - 0 ]$

= $\frac{1}{10}$ or 0.1000

Therefore, the probability that the defective board was produced during the first hour of operation is $\frac{1}{10}$ 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

= $\int\limits^{10}_9 {\frac{1}{10} } \, dy$

= $\frac{1}{10} [y]^{10}_9$

= $\frac{1}{10} [ 10 - 9 ]$

= $\frac{1}{10}$ or 0.1000

Therefore, the probability that the defective board was produced during the last hour of operation is $\frac{1}{10}$ or 0.1000

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

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

$= (\frac{1}{10} [y]^{6}_5) / (\frac{1}{10} [y]^{10}_5)$

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

= 0.2000

Therefore, the required probability is 0.2000

answered: Guest

glide reflection.

its mainly because the pattern will be infinite and will never end. it will look like a gilder or like a filpbook when set to motion.

answered: Guest

rick has $1400 and sally has $1500

step-by-step explanation:

to find amount of money rick has after 10 years :

principal = $1000 , time = 10 years , rate = 4% annually

$\text{simple interest} = \frac{principal\times rate\times time}{100} \\\\\text{simple interest} = \frac{1000\times 4\times 10}{100}=\$400\\\\amount = principal + interest\\amount = 1000 + 400 = \$1400$

to find amount of money sally has after 10 years :

her friend gives to pay her $50 per year, therefore money she has after 10 years = 50 × 10 = $500

also, sally gets $1000 after 10 years,

therefore, total amount of money sally has after 10 years = 500 + 1000 = $1500

so, after 10 years : rick has $1400 and sally has $1500

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The number of defective circuit boards coming off a soldering machine follows a Poisson distribution...