Question

Upgrading a certain software package requires installation of 68 new files. Files are installed consecutively. The installation time is random, but on the average, it takes 15 sec to install one file, with a variance of 11 sec2. (a) What is the probability that the whole package is upgraded in less than 12 minutes?

Answer 1

Answer:

The probability that the whole package is uppgraded in less then 12 minutes is 0,1271

Step-by-step explanation:

The mean distribution for the length of the installation (in seconds) of the programs will be denoted by X. Using the Central Limit Theorem, we can assume that X is normal (it will be pretty close). The mean of X is 15 and the variance is 15, hence, the standard deviation is √15 = 3.873.

We want to find the probability that the full installation process takes less than 12 minutes = 720 seconds. Then, in average, each program should take less than 720/68 = 10.5882 seconds to install. Hence, we want to find the probability of X being less than 10.5882. For that, we will take W, the standariation of X, given by the following formula

$W = \frac{X-\mu}{\sigma} = \frac{X-15}{3.873}$

We will work with $\phi$ , the cummulative distribution function of the standard Normal variable W. The values of $\phi$ can be found in the attached file.

$P(X < 10.5882) = P(\frac{X-15}{3.873} < \frac{10.5882-15}{3.873}) = P(W < -1,14)$

Since the density function of a standard normal random variable is symmetrical, then $\phi(-1.14) = 1-\phi(1.14) = 1-0.8729 = 0.1271$

Therefore, the probability that the whole package is uppgraded in less then 12 minutes is 0,1271.

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