Respuesta :
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
[tex] W = \frac{X-\mu}{\sigma} = \frac{X-15}{3.873} [/tex]
We will work with [tex] \phi [/tex] , the cummulative distribution function of the standard Normal variable W. The values of [tex] \phi [/tex] can be found in the attached file.
[tex] P(X < 10.5882) = P(\frac{X-15}{3.873} < \frac{10.5882-15}{3.873}) = P(W < -1,14)[/tex]
Since the density function of a standard normal random variable is symmetrical, then [tex] \phi(-1.14) = 1-\phi(1.14) = 1-0.8729 = 0.1271 [/tex]
Therefore, the probability that the whole package is uppgraded in less then 12 minutes is 0,1271.
The probability of the entire package being upgraded in less than 12 minutes is 0,1271.
What is Normal Distribution?
A normal distribution is a probability distribution that is used to represent events with a default behaviour and accumulated potential departures from it.
Let X be the mean distribution for the length of time it takes to install the apps (in seconds). We may assume that X is normal using the Central Limit Theorem (it will be pretty close). Because X has a mean of 15 and a variation of 15, therefore, the standard deviation is √15 = 3.873.
We want to know what is the probability that the entire installation will take less than 12 minutes = 720 seconds. Therefore, each program should be installed in less than 720/68 = 10.5882 seconds on average. As a result, we're looking for the likelihood of X being smaller than 10.5882. For this, we'll use Z which is the standardization of x, and can be written as,
[tex]Z=\dfrac{X-\mu}{\sigma} = \dfrac{X-15}{3.873}[/tex]
We'll use the cumulative distribution function of the standard Normal variable Z as our tool. Thus, we can write,
[tex]P(X < 10.5882)=P(\dfrac{X-15}{3.873} < \dfrac{10.5882-15}{3.873})=P(Z < -1.14)[/tex]
Because a basic normal random variable's density function is symmetrical, therefore, we can write
[tex]\phi(-1.14)=1-\phi(-1.14)=1-0.8729 =0.1271[/tex]
Hence, the probability of the entire package being upgraded in less than 12 minutes is 0,1271.
Learn more about Normal distribution:
https://brainly.com/question/12421652
