Question

An article suggests that a Poisson process can be used to represent the occurrence of structural loads over time. Suppose the mean time between occurrences of loads is 0.4 year.

(a) How many loads can be expected to occur during a 2-year period?
  loads

(b) What is the probability that more than eight loads occur during a 2-year period? (Round your answer to three decimal places.)
  

(c) How long must a time period be so that the probability of no loads occurring during that period is at most 0.2? (Round your answer to four decimal places.)
  yr

Answer 1

Concepts and reason

Poisson distribution: The Poisson distribution is a discrete probability distribution. This distribution is used when the number of occurrences for the given event is within a fixed interval of space or time.

Requirements for the Poisson probability distribution:

• The occurrences should be random.

• The occurrences should be independent to each other.

• For the given interval, the occurrences must be uniformly distributed.

Fundamentals

The Poisson probability distribution is,

$P\left( {X = x|\lambda } \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}\,\,\,;\,\,X = 0,1....$

Where,  is the expected number of non-conformities per area opportunity and x is the number of success per area opportunity in the sample.

The another formula to define the mean using duration (more than one unit) is,

$\lambda = t\mu$ , where, t represents the number of units.

(a)

The expected value of occurring time between loads for 2-year period is obtained as shown below:

The mean occurring time between loads is 0.4 per year. That is, $\lambda = \frac{1}{{0.4}}{\rm{ per year}}$ .

Let the random variable X is denoted as number of loads between the times.

The expected value is obtained below:

$\begin{array}{c}\\\lambda = t\mu \\\\ = 2\left( {\frac{1}{{0.4}}} \right)\\\\ = 5\\\end{array}$

(b)

The probability that more than eight loads occur during a 2-year period is obtained as shown below:

From part (a), the expected value for the number of loads occurring during a 2-year period is 5.

The required probability is,

$\begin{array}{c}\\P\left( {X > 8} \right) = 1 - P\left( {X \le 8} \right)\\\\ = 1 - \left[ {P\left( {X = 0} \right) + P\left( {X = 1} \right) + ... + P\left( {X = 7} \right) + P\left( {X = 8} \right)} \right]\\\\ = 1 - \left[ {\frac{{{e^{ - 5}}{{\left( 5 \right)}^0}}}{{0!}} + \frac{{{e^{ - 5}}{{\left( 5 \right)}^1}}}{{1!}} + ... + \frac{{{e^{ - 5}}{{\left( 5 \right)}^7}}}{{7!}} + \frac{{{e^{ - 5}}{{\left( 5 \right)}^8}}}{{8!}}} \right]\\\\ = 1 - \left[ \begin{array}{l}\\0.0067 + 0.0337 + 0.0842 + 0.1404 + 0.1755\\\\ + 0.1755 + 0.1462 + 0.1044 + 0.0653\\\end{array} \right]\\\end{array}$

$\begin{array}{c}\\ = 1 - 0.9319\\\\ = 0.068\\\end{array}$

(c)

The time period for which probability of no loads occurring during that period is at most 0.2 is obtained as shown below:

The required time period is,

Taking ln on both sides,

$\begin{array}{c}\\ - \frac{t}{{0.4}} = \ln \left( {0.2} \right)\\\\ - \frac{t}{{0.4}} = - 1.6094\\\\t = 0.6438\\\end{array}$

Ans: Part a

The expected number of loads occurring during a 2-year period is 5.

Part b

The probability that more than eight loads occur during a 2-year period is 0.068.

Part c

The time period for which probability of no loads occurring during that period is at most 0.2 is 0.6438 years.