Mathematics, 28.08.2020 01:01 jacckiie4069

Suppose nonconforming parts are produced with a rate = 8 / hour (i. e. on average, 8 nonconforming parts will be made during a production period of one hour). (a) If x denotes the number of nonconforming parts made during one hour, what distribution does x follow? Please define the parameter(s) needed for defining the distribution based on the information provided.
(b) What is the probability that exactly 5 nonconforming parts are produced during a 1-hour period? What is the probability that at least 5 nonconforming parts are produced during a 1-hour period?
(c) What are the expected value and standard deviation of the number of nonconforming parts produced during a 120-min period?
(d) What is the probability that at least 20 nonconforming parts produced during a 2.5-hour period? That at most 10 nonconforming parts produced during this period?

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Answer from: anggar20

(a) x follows a poisson distribution with parameter λ = 8 / hour

(b) P( X = 5 ) = 0.09160

P( X ≥ 5 ) = 0.9004

(c) the expected value of the number of nonconforming parts produced during a 120-min period = 16 nonconforming parts per 120 minutes

The standard deviation = 4

(d) P(X ≥ 20) = 0.5297

P( X ≤ 10) = 0.0108

Step-by-step explanation:

Suppose nonconforming parts are produced with a rate = 8 / hour (i.e. on average, 8 nonconforming parts will be made during a production period of one hour).

(a) If x denotes the number of nonconforming parts made during one hour, then : x follows a poisson distribution with parameter λ = 8 / hour

(b)

What is the probability that exactly 5 nonconforming parts are produced during a 1-hour period? What is the probability that at least 5 nonconforming parts are produced during a 1-hour period?

the probability that exactly 5 nonconforming parts are produced during a 1-hour period can be computed as:

P( X = 5 ) = $e^{- \lambda } \lambda^x /x!$

P( X = 5 ) = $e^{- 8 }(8)^5 /5!$

P( X = 5 ) = 0.09160

the probability that at least 5 nonconforming parts are produced during a 1-hour period

P( X ≥ 5 ) = 1 - P( X ≤ 4)

$P( X \geq 5 ) = 1 - \sum \limits ^4_{x=0} e^{-\lambda} \lambda ^x/x!$

$P( X \geq 5 ) = 1 - (e^{-8} 8 ^4/4! + e^{-8} 8 ^3/3! + e^{-8} 8 ^2/2! +e^{-8} 8 ^1/1! + e^{-8} 8 ^0/0! )$

$P( X \geq 5 ) = 1 - 0.0996$

P( X ≥ 5 ) = 0.9004

c) What are the expected value and standard deviation of the number of nonconforming parts produced during a 120-min period?

the expected value of the number of nonconforming parts produced during a 120-min period $\lambda = 8 \times \dfrac {120}{60}$

= 16 nonconforming parts per 120 minutes

The standard deviation = $\lambda^{1/2}$

The standard deviation = $(16)^{1/2}$

The standard deviation = 4

(d) What is the probability that at least 20 nonconforming parts produced during a 2.5-hour period? That at most 10 nonconforming parts produced during this period?

During a 2.5 - hour period , the expected value = 8 × 2.5 = 20

As such, the probability that at least 20 nonconforming parts produced during a 2.5-hour period is:

P(X ≥ 20) = 1 - P(X ≤ 19)

$P( X \geq 20 ) = 1 - \sum \limits ^{19}_{x=0} e^{-\lambda} \lambda ^x/x!$

$P( X \geq 20 ) = 1 - (e^{-8} 8 ^{19}/19! + e^{-8} 8 ^{18}/{18}! + e^{-8} 8 ^{17}/17!+...+e^{-8} 8 ^2/2! +e^{-8} 8 ^1/1! + e^{-8} 8 ^0/0! )$

P(X ≥ 20) = 1 - 0.4703

P(X ≥ 20) = 0.5297

Probability that at most 10 nonconforming parts produced during this period is:

P( X ≤ 10) = $\sum \limits ^{10}_{x=0} e^{-\lambda} \lambda ^x/x!$

$P( X \leq 10) = e^{8} 8 ^{10}/{10}! + e^{8} 8 ^{9}/{9}! + e^{8} 8 ^{8}/{8}! + ...+ e^{8} 8 ^{2}/{2}!+ e^{8} 8 ^{1}/{1}! + e^{8} 8 ^{0}/{0}!$

P( X ≤ 10) = 0.0108

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