Respuesta :
Answer:
(a) 0.7054
(b) 0.9713
(c) 0.2196
(d) The amount of web traffic that will require Smiley’s People to purchase additional server capacity is 3.72.
Step-by-step explanation:
Let X = number of visitors to the web site of Smiley’s People, Inc.
It is provided that [tex]X\sim N(\mu = 4.56 mn, \sigma = 0.82mn)[/tex]
(a)
Determine the value of P (X < 5) as follows:
[tex]P (X <5)=P(\frac{X-\mu}{\sigma}< \frac{5-4.56}{0.82})\\ =P(Z<0.5366)\\=0.7054[/tex]
Use a z-table to determine the probability.
Thus, the probability that the web site has fewer than 5 million visitors in a single day is 0.7054.
(b)
Determine the value of P (X ≥ 3) as follows:
[tex]P (X \geq 3)=P(\frac{X-\mu}{\sigma}\geq \frac{3-4.56}{0.82})\\ =P(Z\geq -1.9024)\\=P(Z<1.9024)\\=0.9713[/tex]
Thus, the probability that the web site has 3 million or more visitors in a single day is 0.9713.
(c)
Determine the value of P (3 ≤ X ≤ 4) as follows:
[tex]P 93\leq X\leq 4)=P(X\leq 4) - P(X\leq 3)\\=P(\frac{X-\mu}{\sigma}\leq \frac{4-4.56}{0.82})-P(\frac{X-\mu}{\sigma}\leq \frac{3-4.56}{0.82}) \\=P(Z\leq -0.683)-P(Z\leq -1.9024)\\=0.2483-0.0287\\=0.2196[/tex]
Thus, the probability that the web site has between 3 million and 4 million visitors in a single day is 0.2196.
(d)
The probability that Smiley's web server will require to purchase additional server capacity is = 1 - 0.85 = 0.15.
That is, P (X > x) = 0.15.
Determine the value of x as follows;
[tex]P(X<x)=0.15\\P(\frac{X-\mu}{\sigma}<\frac{x-4.56}{0.82})=0.15\\ P(Z<z)=0.15[/tex]
Using the z tables the value of z such that P (Z < z) = 0.15 is -1.03.
Then the value of x is:
[tex]\frac{x-4.56}{0.82}=-1.03\\ x=4.56-(1.03\times0.82)\\=3.7154\\\approx3.72[/tex]
Thus, the amount of web traffic that will require Smiley’s People to purchase additional server capacity is 3.72.
Using the normal distribution, we have that:
a) There is a 0.7054 = 70.54% probability that the web site has fewer than 5 million visitors in a single day.
b) 0.0287 = 2.87% probability that the web site has 3 million or more visitors in a single day.
c) 0.2196 = 21.96% probability that the web site has between 3 million and 4 million visitors in a single day.
d) For a traffic of 5.41 million visitors and above, additional server capacity has to be purchased.
In normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It measures how many standard deviations the measure is from the mean.
- After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of measure X.
In this problem:
- Mean of 4.56 million, thus [tex]\mu = 4.56[/tex]
- Standard deviation of 820,000 visitors, our unit is millions, thus [tex]\sigma = 0.82[/tex]
Item a:
This probability is the p-value of Z when X = 5, thus:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{5 - 4.56}{0.82}[/tex]
[tex]Z = 0.54[/tex]
[tex]Z = 0.54[/tex] has a p-value of 0.7054.
0.7054 = 70.54% probability that the web site has fewer than 5 million visitors in a single day.
Item b:
This probability is 1 subtracted by the p-value of Z when X = 3, thus:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{3 - 4.56}{0.82}[/tex]
[tex]Z = -1.9[/tex]
[tex]Z = -1.9[/tex] has a p-value of 0.0287.
0.0287 = 2.87% probability that the web site has 3 million or more visitors in a single day.
Item c:
This probability is the p-value of Z when X = 4 subtracted by the p-value of Z when X = 3, so:
X = 4:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{4 - 4.56}{0.82}[/tex]
[tex]Z = -0.68[/tex]
[tex]Z = -0.68[/tex] has a p-value of 0.2483.
X = 3:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{3 - 4.56}{0.82}[/tex]
[tex]Z = -1.9[/tex]
[tex]Z = -1.9[/tex] has a p-value of 0.0287.
0.2483 - 0.0287 = 0.2196
0.2196 = 21.96% probability that the web site has between 3 million and 4 million visitors in a single day.
Item d:
This is traffic above the 85th percentile, which is the value of X when Z has a p-value of 0.85, so X when Z = 1.037.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.037 = \frac{X - 4.56}{0.82}[/tex]
[tex]X - 4.56 = 1.037(0.82)[/tex]
[tex] = 5.41[/tex]
For a traffic of 5.41 million visitors and above, additional server capacity has to be purchased.
A similar problem is given at https://brainly.com/question/22934264
