Question

Gamma, Exponential, Weibull and Beta Distributions (Part 3)

1. The random variable X can modeled by a Weibull distribution with B = 1 and 0 = 1000. The spec time limit is set at x = 4000. What is the proportion of items not meeting spec? 2. Suppose that the response time X at a certain on-line computer terminal (the elapsed time between the end of a user's inquiry and the beginning of the system's response to that inquiry) has an exponential distribution with expected response time equal to 5 Sec. a. Find the probability that the response time is at most 10 sec? b. Find the Probability that the response time is between 5 and 12 sec? 3. To complete any particular activity once it has been started has a beta distribution with A= the optimistic time (if everything goes well) and B = the pessimistic time (if everything goes badly). Suppose that in constructing a single-family house, the time X (in days) necessary for laying the foundation has a beta distribution with A = 7, B = 9, a = 4, and B = 6. a. Compute the probability that it takes at most 6 days to backfill the site to allow consolidation before construction. b. Estimate the expectation or mean and variance of this project.

Answer 1

Answer:

Given that,

(1).

The random variable X can modeled by a Weibull distribution with $\beta$ = ½ and $\theta$ = 1000. The spec time limit is set at x = 4000.

What is the proportion of items not meeting spec:

The fraction of item is not meeting spec is,

P(X > 4000)=1-P(X $\leq$ 4000)

=1-F(4000)

Since foe weibull distribution.

F(t) = 1-

001 0001 = e

$=e^{-2}$

=0.1353

i.e, all but about 13.53% of items will not meet spec.

(2).

Suppose that the response time X at a certain on-line computer terminal (the elapsed time between the end of a user’s inquiry and the beginning of the system’s response to that inquiry) has an exponential distribution with expected response time equal to 5 Sec.

E(X) =

$\lambda =0.2$

(a).

Find the Probability that the response time is at most 10 sec:

$P(X\leq 10)=1-e^{-\lambda \times 10}$

$=1-e^{-0.2 \times 10}$

$=1-e^{-2}$

=1-0.1353

=0.8646

(b).

Find the Probability that the response time is between 5 and 12 sec:

$P(5\leq X\leq 12)=(F(12)-F(5))$

$=(1-e^{-0.2\times 12})-\left ( 1-e^{-0.2\times 5} \right )$

=(1-0.0907)-(1-0.9678)

=0.2771

(3).

To complete any particular activity once it has been started has a beta distribution with,

A= the optimistic time (if everything goes well) and

B = the pessimistic time (if everything goes badly).

Suppose that in constructing a single-family house, the time X (in days) necessary for laying the foundation has a beta distribution with,

A = 7, B = 9, α = 4, and β = 6.

(a).

Compute the probability that it takes at most 6 days to backfill the site to allow consolidation before construction:

The probability that it takes at most 6 days to lay the foundation.

$P(X\leq 6)=\int_{4}^{6}\frac{1}{9-7}\times \frac{9!}{3!\times 5!}\times \left ( \frac{x-7}{9-7} \right )^3\times \left ( \frac{9-x}{9-7} \right )^{6-1}dx$

$=\frac{504}{2}\int_{4}^{6}\left ( \frac{x-7}{2} \right )^3\times \left ( \frac{9-x}{2} \right )^{5}dx$

$=\frac{504}{512}\int_{4}^{6}\left ( x-7\right )^3\times \left ( 9-x \right )^{5}dx$

(b).

Estimate the expectation or mean and variance of this project:

$\frac{\alpha }{\alpha +\beta }=\frac{4}{10}=0.4$

$E(X)=A+(B-A)\frac{\alpha }{\alpha +\beta }$

$=7+(9-7)\frac{4 }{4+6 }$

$=7+(2)\frac{4 }{10}$

=7.8

$\sigma ^{2}=\frac{(B-A)^2\alpha \beta }{(\alpha +\beta )^2\left ( \alpha +\beta +1 \right )}$

$=\frac{(9-7)^2(4)(6) }{(4 +6 )^2\left ( 4 +6 +1 \right )}$

$=\frac{(9)(24)}{(100 )\left ( 11 \right )}$

=0.0873