Question

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.

Answer 1

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

(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.