Mathematics, 12.02.2020 03:12 vbucks72

Assume that the traffic to the web site of Smiley’s People, Inc., which sells customized T-shirts, follows a normal distribution, with a mean of 4.56 million visitors per day and a standard deviation of 820,000 visitors per day.
(a) What is the probability that the web site has fewer than 5 million visitors in a single day? If needed, round your answer to four decimal digits.(b) What is the probability that the web site has 3 million or more visitors in a single day? If needed, round your answer to four decimal digits.(c) What is the probability that the web site has between 3 million and 4 million visitors in a single day? If needed, round your answer to four decimal digits.(d) Assume that 85% of the time, the Smiley’s People web servers can handle the daily web traffic volume without purchasing additional server capacity. What is the amount of web traffic that will require Smiley’s People to purchase additional server capacity? If needed, round your answer to two decimal digits.

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Answer from: mckadams02

(a) 0.7054

(b) 0.9713

(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 $X\sim N(\mu = 4.56 mn, \sigma = 0.82mn)$

(a)

Determine the value of P (X < 5) as follows:

P (X

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:

$P (X \geq 3)=P(\frac{X-\mu}{\sigma}\geq \frac{3-4.56}{0.82})\\ =P(Z\geq -1.9024)\\=P(Z$

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:

$P 93\leq X\leq 4)=P(X\leq 4) - P(X\leq 3)\\=P(\frac{X-\mu}{\sigma}\leq \frac{4-4.56}{0.82})-P(\frac{X-\mu}{\sigma}\leq \frac{3-4.56}{0.82}) \\=P(Z\leq -0.683)-P(Z\leq -1.9024)\\=0.2483-0.0287\\=0.2196$

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;

P(X

Using the z tables the value of z such that P (Z < z) = 0.15 is -1.03.

Then the value of x is:

$\frac{x-4.56}{0.82}=-1.03\\ x=4.56-(1.03\times0.82)\\=3.7154\\\approx3.72$

Thus, the amount of web traffic that will require Smiley’s People to purchase additional server capacity is 3.72.

Answer from: khambay

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 $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = 4.56$ Standard deviation of 820,000 visitors, our unit is millions, thus $\sigma = 0.82$

Item a:

This probability is the p-value of Z when X = 5, thus:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5 - 4.56}{0.82}$

Z = 0.54

Z = 0.54 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:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{3 - 4.56}{0.82}$

Z = -1.9

Z = -1.9 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:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{4 - 4.56}{0.82}$

Z = -0.68

Z = -0.68 has a p-value of 0.2483.

X = 3:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{3 - 4.56}{0.82}$

Z = -1.9

Z = -1.9 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.

$Z = \frac{X - \mu}{\sigma}$

$1.037 = \frac{X - 4.56}{0.82}$

X - 4.56 = 1.037(0.82)

= 5.41

For a traffic of 5.41 million visitors and above, additional server capacity has to be purchased.

A similar problem is given at link

Answer from: Quest

if the eqn of the parabola is f(x) = (x-4)(x+2), we need to mult. this out to find the coordinates of the vertex.

then f(x) = x^2 + 2x - 4x - 8, and f(x) = x^2 - 2x - 8.

one way in which you can find the vertex is to calculate x = -b / (2a). here, that comes to

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x = = 1. f(1) = 1^2 - 2(1) - 8 = - 9. vertex is at (1, -9).

2(1)

find another point on the curve: let x=0, for which y=-8. then (0, -8) is on the curve. plot the vertex and this point and draw a smooth parabola thru both.

hope this

Use the parabola tool to graph the quadratic function f(x)=(x-5)^2+1 graph the parabola by first plo

Answer from: Quest

the results from a pre-test for students for the year 2000 and the year 2010 are illustrated in the box plot. what do these results tell us about how students performed on the 29 question pre-test for the two years?

if we compare only the lowest and highest scores between the two years, we might conclude that the students in 2010 did better than the students in 2010. this conclusion seems to follow since the lowest score of 8 in 2010 is greater in value than the lowest score of 6 in 2000. also, the highest score of 28 in 2010 is greater in value than the highest score of 27 in 2000.

but the box portion of the illustration gives us more detailed information. the middle bar in each box shows us that the median score of 20 in 2000 is greater in value than the median score of 17 in 2010. further, we note that the box and whiskers divide the illustration into four pieces. each of these four pieces represents the same portion of students. so, the upper half of the students in 2000 scored in the same score range as the upper one-fourth of the students in 2010, see the illustration at a score of 20.

by considering the upper one-fourth, upper half, and upper three-fourths instead of just the lowest and highest scores, we would conclude that the students as a whole did much better in 2000 than in 2010. we would conclude that as a whole the students in 2010 are less prepared than the students in 2000.

in this section, we discuss box-and-whisker plots and the five key values used in constructing a box-and-whisker plot. the key values are called a five-number summary, which consists of the minimum, first quartile, median, third quartile, and maximum.

step-by-step explanation: