Question

Let X be the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. The article "Methodology for Probabilistic Life Prediction of MultipleAnomaly Materials" (Amer. Inst. of Aeronautics and Astronautics J., 2006: 787793) proposes a Poisson distribution for X. Suppose that µ = 4. (a) Compute both P(X ≤ 4) and P(X < 4). (b) Compute P(4 ≤ X ≤ 8). (c) Compute P(8 ≤ X). (d) What is the probability that the number of anomalies exceeds its mean value by no more than one standard deviation

Answer 1

Answer:

a) $P(X \leq 4) = 0.6289$

$P(X < 4) = 0.4335$

b) $P(4 \leq X \leq 8) = 0.5452$

c) $P(X \geq 8) = 0.0511$

d) 0.2605

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

It is important to know that the variance has the same value as the mean in a Poisson distribution. The standard deviation is the square root of the variance.

In this problem, we have that:

$\mu = 4, \sigma = \sqrt{4} = 2$.

To help our solution, i am going to find each of P(X = x) from x = 0 to 8[/tex]

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-4}*4^{0}}{(0)!} = 0.0183$

$P(X = 1) = \frac{e^{-4}*4^{1}}{(1)!} = 0.0733$

$P(X = 2) = \frac{e^{-4}*4^{2}}{(2)!} = 0.1465$

$P(X = 3) = \frac{e^{-4}*4^{3}}{(3)!} = 0.1954$

$P(X = 4) = \frac{e^{-4}*4^{4}}{(4)!} = 0.1954$

$P(X = 5) = \frac{e^{-4}*4^{5}}{(5)!} = 0.1563$

$P(X = 6) = \frac{e^{-4}*4^{6}}{(6)!} = 0.1042$

$P(X = 7) = \frac{e^{-4}*4^{7}}{(7)!} = 0.0595$

$P(X = 8) = \frac{e^{-4}*4^{8}}{(8)!} = 0.0298$

(a) Compute both P(X ≤ 4) and P(X < 4)

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0183 + 0.0733 + 0.1465 + 0.1954 + 0.1954 = 0.6289$

---------

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0183 + 0.0733 + 0.1465 + 0.1954 = 0.4335$

(b) Compute P(4 ≤ X ≤ 8).

$P(4 \leq X \leq 8) = P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) = 0.1954 + 0.1563 + 0.1042 + 0.0595 + 0.0298 = 0.5452$

(c) Compute P(8 ≤ X)

That is $P(X \geq 8)$.

The sum of the probabilities must be 1 in decimal. Either X is greater or equal to 8, or X is lesser than 8.

So

$P(X < 8) + P(X \geq 8) = 1$

$P(X \geq 8) = 1 - P(X < 8)$

From what we have in a) and b)

$P(X < 8) = P(X < 4) + P(4 \leq X < 8) = 0.4335 + 0.1954 + 0.1563 + 0.1042 + 0.0595 = 0.9489$

So

$P(X \geq 8) = 1 - P(X < 8) = 1 - 0.9489 = 0.0511$

(d) What is the probability that the number of anomalies exceeds its mean value by no more than one standard deviation

One standard deviation is 2, and the mean is 4. So, this is:

$P(4 < X \leq 6) = P(X = 5) + P(X = 6) = 0.1563 + 0.1042 = 0.2605$