Respuesta :
Answer:
(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 ) = [tex]e^{- \lambda } \lambda^x /x![/tex]
P( X = 5 ) = [tex]e^{- 8 }(8)^5 /5![/tex]
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)
[tex]P( X \geq 5 ) = 1 - \sum \limits ^4_{x=0} e^{-\lambda} \lambda ^x/x![/tex]
[tex]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! )[/tex]
[tex]P( X \geq 5 ) = 1 - 0.0996[/tex]
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 [tex]\lambda = 8 \times \dfrac {120}{60}[/tex]
= 16 nonconforming parts per 120 minutes
The standard deviation = [tex]\lambda^{1/2}[/tex]
The standard deviation = [tex](16)^{1/2}[/tex]
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)
[tex]P( X \geq 20 ) = 1 - \sum \limits ^{19}_{x=0} e^{-\lambda} \lambda ^x/x![/tex]
[tex]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! )[/tex]
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) = [tex]\sum \limits ^{10}_{x=0} e^{-\lambda} \lambda ^x/x![/tex]
[tex]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}![/tex]
P( X ≤ 10) = 0.0108
Following are the calculation to the given points:
For point a)
With parameter, x follows the poisson distribution is [tex]\lambda=\frac{8}{hour}[/tex]
For point b)
The probability that exactly 5 nonconforming parts are generated in a one-hour period.
[tex]\to P(X=5)=e^{-\lambda }\frac{\lambda ^{x}}{x!} =0.0916[/tex]
There is a good chance that at least 5 nonconforming parts exist.
[tex]\to P(X\geq 5) =1-P(X\leq 4)[/tex]
[tex]=1 -\sum_{x=0}^{4} \ e^{-\lambda } \frac{\lambda ^{x}}{x!} \\\\=1-0.0996\\\\=0.9004[/tex]
For point c)
Expected value for 120-minute period [tex]\lambda=8\times \frac{120}{60} =16[/tex] 120 minutes of nonconforming sections
[tex]\to \sigma ==(\lambda)^{\frac{1}{2} =(16)^{\frac{1}{2}} =4[/tex]
For point d)
Expected value for 2.5 hours
[tex]\to \lambda =8\times 2.5=20[/tex]
As a result, there is a good chance that at least 20 nonconforming items were manufactured.
[tex]\to P(X\geq 20)=1-P(X\leq 19)[/tex]
[tex]=1-\sum_{x=0}^{19}\ e^{-\lambda }\frac{\lambda^{x}}{x!}\\\\=1-0.4703\\\\=0.5297[/tex]
There is a chance that no more than 10 nonconforming parts were created.
[tex]\to P(X\leq 10)=\sum_{x=0}^{10}e^{-\lambda }\frac{\lambda ^{x}}{x!} =0.0108[/tex]
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