Question

Assume that the traffic to the website of Smiley’s People, Inc., which sells customized T-shirts, follows a normal distribution, with a mean of 4.44 million visitors per day and a standard deviation of 780,000 visitors per day.

a). What is the probability that the website has fewer than 5 million visitors in a single day?

b). What is the probability that the website has 3 million or more visitors in a single day?

c). What is the probability that the website has between 3 million and 4 million visitors in a single day?

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?

Answer 1

P(X < A) = P(Z < (A - mean)/standard deviation)

Mean = 4.44 million

Standard deviation = 0.78 million

a) P(fewer than 5 million visitors) = P(X < 5)

= P(Z < (5 - 4.44)/0.78)

= P(Z < 0.72)

= 0.7642

b) P(3 million or more visitors) = 1 - P(X < 3)

= 1 - P(Z < (3 - 4.44)/0.78)

= 1 - P(Z < -1.85)

= 1 - 0.0322

= 0.9678

c) P(between 3 million and 4 million visitors) = P(X < 4) - P(X < 3)

= P(Z < (4 - 4.44)/0.78) - 0.0322

= P(Z < -0.56) - 0.0322

= 0.2877 - 0.0322

= 0.2555

d) Let the web traffic that will require Smiley’s People to purchase additional server capacity be W

P(X < W) = 0.85

P(Z < (W - 4.44)/0.78) = 0.85

(W - 4.44)/0.78 = 1.04

W = 5.25

Web traffic that will require Smiley’s People to purchase additional server capacity is 5.25 million visitors or above