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.

(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

(a)

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

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:

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:

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;

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

Then the value of x is:

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

In this problem:

Item a:

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

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:

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:

has a p-value of 0.2483.

X = 3:

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.

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

A similar problem is given at link

if x = 1, then the answer is 3.5