Question

The article "Expectation Analysis of the Probability of Failure for Water Supply Pipes" (J. of Pipeline Systems Engr. and Practice, May 2012: 36-46) pro-posed using the Poisson distribution to model the number of failures in pipelines of various types. Suppose that for cast iron pipe of a particular length, the expected number of failures is 1 (very close to one of the cases considered in the article). Then X, the number of failures, has a Poisson distribution with μ=1 A. Obtain P(X≤5) B. Determine P(X=2) C. Determine P(2≤X≤4) D. Find E(X) and V(X).

Answer 1

The probabilities for the Poisson distribution with μ=1 are calculated using the formula P(X=x) = (μ^x * e^-μ) / x!, summing them where necessary. The expected value and variance for a Poisson distribution are both equal to the mean μ, which is 1 in this case.

Step-by-step explanation:

The student has provided data indicating that for cast iron pipe of a particular length, the expected number of failures is 1, fitting a Poisson distribution with μ=1. Let's calculate the probabilities asked, using the Poisson distribution properties.

A. Obtain P(X≤5)

The probability that X is less than or equal to 5, P(X≤5), is calculated by summing the probabilities of X=0 through X=5. We use the formula P(X=x) = (μ^x * e^-μ) / x! where x is the number of occurrences, μ is the mean number of occurrences, and e is the base of the natural logarithm:

P(X=x) = (1^x * e^-1) / x!

Summing these probabilities from X=0 to X=5 gives us P(X≤5).

B. Determine P(X=2)

To find P(X=2), we plug x=2 into the formula, resulting in:

P(X=2) = (1^2 * e^-1) / 2! = (1 * e^-1) / 2 = 0.1839 (approximate)

C. Determine P(2≤X≤4)

P(X=2) + P(X=3) + P(X=4)

E(X) = μ = 1

V(X) = μ = 1

In this case, both the expected value and the variance are 1.

Answer 2

For a Poisson distribution with
`\(\mu = 1\)`:

a.
`\(P(X \leq 5) = 0.999\).`

b.
`\(P(X=2) \approx 0.1839\)` from formula and table.

c.
`\(P(2 \leq X \leq 4) \approx 0.2605\).`

d. Probability
`\(X\)` exceeds its mean by more than one standard deviation:
`\(P(X > 2) = 0.081\).`

For a Poisson distribution with a mean
`\(\mu = 1\)`:

a. To find
`\(P(X \leq 5)\)` using Table A.2 (Poisson cumulative distribution table), locate the row for
`\(\mu = 1\)` and find the corresponding value for
`\(X \leq 5\)`. The value in the table represents
`\(P(X \leq 5)\)`, which is approximately 0.999.

b. To determine
`\(P(X = 2)\)` using the probability mass function (pmf) formula for the Poisson distribution:

`\[P(X = k) = \frac{{e^(-\mu) \cdot \mu^k}}{{k!}}\]`

For
`\(k = 2\)` and
`\(\mu = 1\)`:

`\[P(X = 2) = \frac{{e^(-1) \cdot 1^2}}{{2!}} = \frac{{e^(-1)}}{{2}} \approx 0.1839\]`

You can also verify this value using Table A.2, where the row for
`\(\mu = 1\)` and column for
`\(X = 2\)` gives approximately
`\(0.1839\).`

c. To find
`\(P(2 \leq X \leq 4)\)`, you can sum the individual probabilities
`\(P(X = 2)\)`,
`\(P(X = 3)\)`, and
`\(P(X = 4)\)` using the Poisson pmf formula:

`\[P(X = 3) = \frac{{e^(-1) \cdot 1^3}}{{3!}} = \frac{{e^(-1)}}{{6}} \approx 0.0613\]`

`\[P(X = 4) = \frac{{e^(-1) \cdot 1^4}}{{4!}} = \frac{{e^(-1)}}{{24}} \approx 0.0153\]`

Therefore,
`\(P(2 \leq X \leq 4) = P(X = 2) + P(X = 3) + P(X = 4) \approx 0.1839 + 0.0613 + 0.0153 = 0.2605\).`

d. The standard deviation of a Poisson distribution is
`\(√(\mu)\)`. For
`\(\mu = 1\)`, the standard deviation
`(\(\sigma\))` is
`\(√(1) = 1\)`.

The probability that \(X\) exceeds its mean value by more than one standard deviation is the probability of
`\(X > \mu + \sigma = 1 + 1 = 2\)`. This is the same as finding
`\(P(X > 2)\)`.

`\[P(X > 2) = 1 - P(X \leq 2)\]`

Using Table A.2 for
`\(\mu = 1\)` and
`\(X = 2\)`,
`\(P(X \leq 2) \approx 0.9197\).`

So
`, \(P(X > 2) = 1 - 0.9197 = 0.0803\).`

complete the question

The article "Expectation Analysis of the Probability of Failure for Water Supply Pipes" (J. of Pipeline Systems Engr. and Practice, May 2012: 36-46) pro- posed using the Poisson distribution to model the num- ber of failures in pipelines of various types. Suppose that for cast-iron pipe of a particular length, the expected number of failures is 1 (very close to one of the cases considered in the article). Then X, the number of failures, has a Poisson distribution with μ = 1.

a. Obtain P(X ≤5) by using Appendix.

b. Determine P(X=2) first from the pmf formula and then from Appendix Table A.2.

c. Determine P(2 ≤ X ≤ 4).

d. What is the probability that X exceeds its mean value by more than one standard deviation?