c) The number of flaws in a new brand of fibre optic cable is known to follow a Poisson distribution, with a mean of 2 flaws per 100 metres of cable. i. Use the output below to determine the probability of observing 5 flaws in 100 metres of cable, i.e. P(X = 5) dpois (5, lambda = 100) [1] 3.100063e-36 dpois (5, lambda = 2) [1] 0.03608941 dpois (2, lambda = 5) [1] 0.08422434 dpois (100, lambda = 5) [1] 5.695402e-91 ii. Use the output below to determine the probability of at least 3 flaws in 100 metres of cable, i.e. P(X > 3) sum (dpois (0:2, lambda = 2)) %3D [1] 0.6766764 1-sum (dpois (0:2, lambda = 2)) [1] 0.3233236 sum (dpois (0:2, lambda = 3)) [1] 0.4231901 1-sum (dpois (0:2, lambda = 3)) [1] 0.5768099
c) The number of flaws in a new brand of fibre optic cable is known to follow a Poisson distribution, with a mean of 2 flaws per 100 metres of cable. i. Use the output below to determine the probability of observing 5 flaws in 100 metres of cable, i.e. P(X = 5) dpois (5, lambda = 100) [1] 3.100063e-36 dpois (5, lambda = 2) [1] 0.03608941 dpois (2, lambda = 5) [1] 0.08422434 dpois (100, lambda = 5) [1] 5.695402e-91 ii. Use the output below to determine the probability of at least 3 flaws in 100 metres of cable, i.e. P(X > 3) sum (dpois (0:2, lambda = 2)) %3D [1] 0.6766764 1-sum (dpois (0:2, lambda = 2)) [1] 0.3233236 sum (dpois (0:2, lambda = 3)) [1] 0.4231901 1-sum (dpois (0:2, lambda = 3)) [1] 0.5768099
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![c) The number of flaws in a new brand of fibre optic cable is known to follow a Poisson
distribution, with a mean of 2 flaws per 100 metres of cable.
i. Use the output below to determine the probability of observing 5 flaws in 100 metres
of cable, i.e. P(X = 5)
dpois (5, lambda = 100)
[1] 3.100063e-36
dpois (5, lambda = 2)
[1] 0.03608941
dpois (2, lambda = 5)
[1] 0.08422434
dpois(100, lambda = 5)
[1] 5.695402e-91
ii. Use the output below to determine the probability of at least 3 flaws in 100 metres of
cable, i.e. P(X > 3)
sum (dpois (0:2, lambda
2))
%3D
[1] 0.6766764
1-sum (dpois (0:2, lambda = 2))
[1] 0.3233236
sum (dpois (0:2, lambda = 3))
[1] 0.4231901
1-sum (dpois (0:2, lambda = 3))
[1] 0.5768099](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe6e636e7-d926-467f-8992-f2860e2db250%2Fe84ed84c-3558-4958-9812-57a528216c50%2Ftbkdgdj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:c) The number of flaws in a new brand of fibre optic cable is known to follow a Poisson
distribution, with a mean of 2 flaws per 100 metres of cable.
i. Use the output below to determine the probability of observing 5 flaws in 100 metres
of cable, i.e. P(X = 5)
dpois (5, lambda = 100)
[1] 3.100063e-36
dpois (5, lambda = 2)
[1] 0.03608941
dpois (2, lambda = 5)
[1] 0.08422434
dpois(100, lambda = 5)
[1] 5.695402e-91
ii. Use the output below to determine the probability of at least 3 flaws in 100 metres of
cable, i.e. P(X > 3)
sum (dpois (0:2, lambda
2))
%3D
[1] 0.6766764
1-sum (dpois (0:2, lambda = 2))
[1] 0.3233236
sum (dpois (0:2, lambda = 3))
[1] 0.4231901
1-sum (dpois (0:2, lambda = 3))
[1] 0.5768099
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