Let X be the number of material anomalies occurring in a particular region of an aircraft...

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Let X be the number of material anomalies occurring in a particular region of an aircraft...

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 Multiple-Anomaly Materials"� proposes a Poisson distribution for X. Suppose that ? = 4. (Round your answers to three decimal places.) (a) Compute both P(X ? 4) and P(X < 4). (b) Compute P(4 ? X ? 5). (c) Compute P(5 ? X). (d) What is the probability that the number of anomalies does not exceed the mean value by more than one standard deviation?

math Statistics-And-Probability

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Answer 1

Answer #1

Concepts and reason

The Poisson distribution occurs when there are events which do not occur as outcomes of a definite number of trials of an experiment but which occur at random points of time and space wherein our interest liners only in the number of occurrences of the event, not in its non-occurrences.

Fundamentals

The probability function for the Poisson distribution with parameter is,

tra=(x= x)d

The cumulative probability function the Poisson distribution with parameter is,

»*()-ܛܼܲ..

(a)

Let X be the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk.

The random variable, X - Poisson( 2 =4)

Since the random variable X follows Poisson distribution with parameter 4, use the following probability function for finding the required probabilities.

P(x = x) = - *
,
x=0,1,2,.....

Cumulative probability function is,

F() - ΣΡ(x =k)

Compute the probability, P(X 34)

P(X 54)= Fx (4)
é 4°
0!
e 4
1!
e 4²
2!
e 43 e 4²
3!*4!

= 0.0183 +0.0733 +0.1465 +0.1954 +0.1954
=0.6289

Compute the probability, P(X <4)

P(X <4)= P(X 33)
= Fx (3)
pe*4*
e*40
*4!

= 0.0183 + 0.0733 +0.1465 +0.1954
=0.4335

(b)

Compute the probability, P(45 X 58)

P(45 X 58)=P(X = 4)+ P( X = 5) + P(X = 6) + P ( X = 7)+P ( X = 8)
€444 445 446 447 448
4! 5! 6! 7! 8!
= 0.1954 +0.1563 +0.104

(c)

Compute the probability, P(55X)

P(5<X)=l-P(X<5)
=1-P(X<4)
=l-F;(4)
-4eܪܽ-I=

e*4!
1!
*42
2!
€ 4
3!
€ *4
4!
e *4º
| 0!
= 1-0.6288
= 0.3712

(d)

Compute the probability that the number of anomalies does not exceed the mean value by more than one standard deviation

Expected number and standard deviation of anomalies:

ܝܠܢ
ܠܛ

Compute the required probability.

P(X Su+o)=P(x +4+2)
= P(x 36)
= Fx (6)
2x!
je 41

0! 1! * 2!+ 3!
Le4 e*4 *46
1* 4! +5 +61
= 0.8893

Ans: Part a

The probability value when X is less than or equal to 4 with Poisson mean 4 is 0.6289

The probability value when X is less than 4 with Poisson mean 4 is 0.4335

Part b

The probability value when X is between 4 and 8 with Poisson mean 4 is 0.5452

Part c

The probability value when X is greater than or equal to 5 with Poisson rate 4 is 0.3712

Part d

Probability that the number of anomalies does not exceed the mean value by more than one standard deviation is 0.8893.

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Answer 2

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