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 4.52 million visitors per day and a standard deviation of 840,000 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. _ million visitors per day

Answer 1

Solution:
Given in the question
Traffic to the website of smiley's people follows a normal distribution with
Mean() = 4.52
Standard deviation () = 0.84
Solution(a)
In this question, we need to calculate the web site has fewer than 5 million visitors in a single day
P(X<5) =?
Here we will use the standard normal distribution, first, we will calculate Z-score which can be calculated as
Z-score = (X-)/ = (5-4.52)/0.84 = 0.57
From Z table we found a p-value = 0.7161
So there is a 71.61% probability that web site has fewer than 5 million visitors in a single day.
Solution(b)
Here we need to calculate P(X>=3) = 1- P(X<3)
Z = (3-4.52)/0.84 = -1.81
From Z table we found p-value
P(X>=3) = 1- P(X<3) = 1 - 0.0351 = 0.9649
So there is 96.49% probability that web site has 3 million or more visiors in a single day
Solution(c)
Here we need to calculate P(3<X<4) = P(X<4) - P(X<3)
Z= (3-4.52)/0.84 = -1.81
Z= (4-4.52)/0.84 = -0.62
From Z table we found p-value
P(3<X<4) = P(X<4) - P(X<3) = 0.2676 - 0.0351 = 0.2325
So there is 23.25% probability that website has b/w 3 and 4 million visitors in a single day.
Solution(d)
Here Given P-value = 0.85
from Z-table we found z-score = 1.0364
Web site tracffic that will be require smiley's people to purchase additional server capacity is
X = + Z-score* = 4.52 + 1.0364*0.84 = 4.52 + 0.87 = 5.39
So required capacity is 5.39 million per day.