Respuesta :
Answer:
(a) The probability of exactly three flaws in 150 m of cable is 0.21246
(b) The probability of at least two flaws in 100m of cable is 0.69155
(c) The probability of exactly one flaw in the first 50 m of cable, and exactly one flaw in the second 50 m of cable is 0.13063
Step-by-step explanation:
A random variable X has a Poisson distribution and it is referred to as Poisson random variable if and only if its probability distribution is given by
[tex]p(x;\lambda)=\frac{\lambda e^{-\lambda}}{x!}[/tex] for x = 0, 1, 2, ...
where [tex]\lambda[/tex], the mean number of successes.
(a) To find the probability of exactly three flaws in 150 m of cable, we first need to find the mean number of flaws in 150 m, we know that the mean number of flaws in 50 m of cable is 1.2, so the mean number of flaws in 150 m of cable is [tex]1.2 \cdot 3 =3.6[/tex]
The probability of exactly three flaws in 150 m of cable is
[tex]P(X=3)=p(3;3.6)=\frac{3.6^3e^{-3.6}}{3!} \approx 0.21246[/tex]
(b) The probability of at least two flaws in 100m of cable is,
we know that the mean number of flaws in 50 m of cable is 1.2, so the mean number of flaws in 100 m of cable is [tex]1.2 \cdot 2 =2.4[/tex]
[tex]P(X\geq 2)=1-P(X<2)\\P(X\geq 2)=1-(P(X=0)+P(X=1))[/tex]
[tex]P(X\geq 2)=1-p(0;2.4)-p(1;2.4)\\\\P(X\geq 2)=1-\frac{2.4^0e^{-2.4}}{0!}-\frac{2.4^1e^{-2.4}}{1!}\\\\P(X\geq 2)\approx 0.69155[/tex]
(c) The probability of exactly one flaw in the first 50 m of cable, and exactly one flaw in the second 50 m of cable is
[tex]P(X=1)=p(1;1.2)=\frac{1.2^1e^{-1.2}}{1!}\\P(X=1)\approx 0.36143[/tex]
The occurrence of flaws in the first and second 50 m of cable are independent events. Therefore the probability of exactly one flaw in the first 50 m and exactly one flaw in the second 50 m is
[tex](0.36143)(0.36143) = 0.13063[/tex]
Using the Poisson distribution, it is found that there is a :
a) 0.2125 = 21.25% probability of exactly three flaws in 150m of cable.
b) 0.3084 = 30.84% probability of at least two flaws in 100m of cable.
c) 0.1306 = 13.06% probability of exactly one flaw in the first 50m of cable, and exactly one flaw in the second 50m of cable.
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
The parameters are:
- x is the number of successes
- e = 2.71828 is the Euler number
- [tex]\mu[/tex] is the mean in the given interval.
Item a:
Mean number of flaws of 1.2 per 50m, thus, for 150m, [tex]\mu = 1.2\frac{150}{50} = 3.6[/tex]
The probability is P(X = 3), then:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 3) = \frac{e^{-3.6}(3.6)^{3}}{(3)!} = 0.2125[/tex]
0.2125 = 21.25% probability of exactly three flaws in 150m of cable.
Item b:
Mean number of flaws of 1.2 per 50m, thus, for 100m, [tex]\mu = 1.2\frac{100}{50} = 2.4[/tex]
The probability is:
[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]
In which
[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex]
Then
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-2.4}(2.4)^{0}}{(0)!} = 0.0907[/tex]
[tex]P(X = 1) = \frac{e^{-2.4}(2.4)^{1}}{(1)!} = 0.2177[/tex]
Then
[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.0907 + 0.2177 = 0.3084[/tex]
0.3084 = 30.84% probability of at least two flaws in 100m of cable.
Item c:
50m, thus, [tex]\mu = 1.2[/tex]
For each 50m, the probability of exactly one flaw is P(X = 1):
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 1) = \frac{e^{-1.2}(1.2)^{1}}{(1)!} = 0.3614[/tex]
The intervals are independent, thus:
[tex]p = 0.3614^2 = 0.1306[/tex]
0.1306 = 13.06% probability of exactly one flaw in the first 50m of cable, and exactly one flaw in the second 50m of cable.
A similar problem is given at https://brainly.com/question/16912674
