contestada

A process that fills packages is stopped whenever a package is de- tected whose weight falls outside the specification. Assume that each package has probability 0.01 of falling outside the specification and that the weights of the packages are independent.
A) Find the mean number of packages that will be filled beforethe process is stopped.B) Find the variance of the number of packages that will befilled before the process is stopped.C) Assume that the process will not be stopped until fourpackages whose weight falls outside the specification are detected. Find the mean and variance of the number of packages thatwill be filled before the process is stopped.

Respuesta :

MikeS101

We have the following:

P(fail) = 0.01 and P(success) = 0.99

Parts (A) and (B) of this question uses the Geometric distribution.

Solution to part (A):

The mean of the Geometric distribution is:

P = 0.01; μx = 1/p = 1/0.01 = 100

Thus, the mean number of packages that will be filled before the process is stopped is 100

Solution to part (B):

The Variance of the Geometric is:

[tex]\frac{(1 - P)}{P^{2} } = \frac{0.99}{0.0001} = 9900[/tex]

Thus, the variance of the number of packages that will be filled before the process is stopped is 9900

Part (C) uses the negative Binomial distribution.

we have that:

P = 0.01 and r = 4

Thus, the mean is:

μx = r(1 - P) / P = 4 × 0.99 / 0.01 = 396

Also, the variance is:

[tex]\frac{r(1 - P)}{P^{2} } =\frac{4(0.99)}{0.0001} = 39600[/tex]

Therefore, the mean and variance of the number of packages that will be filled before the process is stopped is a mean of 396 and a variance of 39600.

Otras preguntas