Respuesta :
Answer:
(a) P(X<5) = 0.2237
(b) P(X=5) = 0.1454
(c) P(X≥5) = 0.7763
(d) P(X=either 4 or 5) = 0.2572
Step-by-step explanation:
We will consider this question as a Poisson process and use the following formula to determine the answers:
P(X=x) = λˣe^-λ / x!
Where λ = Mean number of chip parts per cookie
x = Number of chip parts found
(a) For calculating the probability of less than 5 chip parts being found, we need to consider all the values less than 5 i.e. 0,1,2,3,4 and calculate the probability for these values.
P(X < 5) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)
= [tex]6.5^{0} e^{-6.5}[/tex]/0! + [tex]6.5^{1} e^{-6.5}[/tex]/1! + [tex]6.5^{2} e^{-6.5}[/tex]/2! + [tex]6.5^{3} e^{-6.5}[/tex]/3! + [tex]6.5^{4} e^{-6.5}[/tex]/4!
P(X<5) = 0.0015 + 0.0098 + 0.0318 + 0.0688 + 0.1118
P(X<5) = 0.2237
The probability that in any particular cookie being inspected, less than 5 chip parts will be found is 0.2237
(b) P(X=5) = [tex]6.5^{5} e^{-6.5}[/tex]/5!
P(X=5) = 0.1454
The probability that in any particular cookie being inspected, exactly 5 chip parts will be found is 0.1454.
(c) To calculate the probability of more than or equal to 5 chip parts being found, we will consider the probability P(X<5) which we have already found in part (a) and subtract it from the total probability (i.e. 1) to get P(X≥5).
P(X≥5) = 1 - P(X<5)
= 1 - 0.2237
P(X≥5) = 0.7763
The probability that in any particular cookie being inspected 5 or more chip parts will be found is 0.7763.
(d) To calculate the probability that either 4 or 5 chip parts will be found, we must add P(X=4) and P(X=5). We know the value of P(X=5) from part (b) and the value of P(X=4) from part (a). Hence,
P(X=either 4 or 5) = P(X=4) + P(X=5)
= 0.1118 + 0.1454
P(X=either 4 or 5) = 0.2572
The probability that in any particular cookie being inspected, either 4 or 5 chip parts will be found is 0.2572.
The solution to the problem can be found using the Poisson distribution.
Given to us
The mean number of chip parts per cookie is 6.5.
What is Poisson distribution?
According to Poisson distribution,
[tex]P(X=x) = \dfrac{\lambda^x \times e^{-\lambda}}{x!}[/tex]
where λ is the mean.
For λ = 6.5 the graph of the Poisson distribution is plotted below.
What is the probability of any particular cookie being inspected less than five chip parts will be found?
We know about the Poisson distribution, therefore,
x < 5
λ = 6.5
As we need the probability of the cookie being inspected less than five chip parts will be found,
P(x<5) = P(x=0) + P(x=1) + P(x=2) + P(x=3) + P(x=4)
[tex]=\dfrac{6.5^0 \times e^{-6.5}}{0!} + \dfrac{6.5^1 \times e^{-6.5}}{1!}+\dfrac{6.5^2 \times e^{-6.5}}{2!} +\dfrac{6.5^3 \times e^{-6.5}}{3!} + \dfrac{6.5^4 \times e^{-6.5}}{4!}[/tex]
P(x<5) = 0.22366
What is the probability that in any particular cookie being inspected exactly five chip parts will be found?
We know about the Poisson distribution, therefore,
x = 5
λ = 6.5
P(x=5) = [tex]=\dfrac{6.5^5 \times e^{-6.5}}{5!}[/tex]
P(x=5) = 0.14537
What is the probability that in any particular cookie being inspected five or more chip parts will be found?
We know about the Poisson distribution, therefore,
x = 5
λ = 6.5
P(x≥5) = 1 - P(x<5)
= 1 - 0.22366
= 0.77634
What is the probability that in any particular cookie being inspected either four or five chip parts will be found?
P(x = 4 or 5) = P(x=4) + P(x = 5)
= 0.11182 + 0.14537
= 0.25719
Learn more about Poisson distribution:
https://brainly.com/question/5673802
