Question

A company is manufacturing highway emergency ares. Such ares are sup- posed to burn for an average of 20 minutes. Every hour a sample of ares is collected, and their average burn time is determined. If the manufacturing process is working cor- rectly, there is a 68% chance that the average burn time of the sample will be between 14 minutes and 26 minutes. The quality engineer in charge of the process believes that if 4 of 5 samples fall outside these bounds then this is a signal that the process might not be performing as expected. Each morning the sampling begins anew. Let X denote the number of samples drawn in order to obtain the fourth sample whose average value is outside of the above bounds. Find the probability that for a given morning X

Answer 1

Complete question :

A company is manufacturing highway emergency ares. Such ares are sup- posed to burn for an average of 20 minutes. Every hour a sample of ares is collected, and their average burn time is determined. If the manufacturing process is working cor- rectly, there is a 68% chance that the average burn time of the sample will be between 14 minutes and 26 minutes. The quality engineer in charge of the process believes that if 4 of 5 samples fall outside these bounds then this is a signal that the process might not be performing as expected. Each morning the sampling begins anew. Let X denote the number of samples drawn in order to obtain the fourth sample whose average value is outside of the above bounds. Find the probability that for a given morning X = 5, hence there seems to be a problem right away.

Answer:

0.0285

Step-by-step explanation:

We are to find the probability for a given morning that X = 5

From the question, we could see that there is a 68% chance that the average burn time of the sample will be between 14 minutes and 26 minutes and the success of the process is obtained after the fourth or fifth sample.

Thus, r = 4; q = 0.68

The probability average burn time falls outside p will be:

p = 1 - 0.68 = 0.32

The random variable X follows negative binomial distribution.

Therefore,

$f(x) = \left[\begin{array}{ccc}x-1\\r-1\end{array}\right] (1-p)^x^-^r) (p)^r$

Where, x = 5

r = 4

p = 0.32

Substituting figures in the equation, we have:

$P(x=5) = \left[\begin{array}{ccc}5-1\\4-1\end{array}\right] (1-0.32)^5^-^4 (0.32)^4$

$P(x=5) = \left[\begin{array}{ccc}4\\3\end{array}\right] (0.68)(0.01048576)$

= 0.0285

The probability that for a given morning X = 5 is 0.0285