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 4.48 million visitors per day and a standard deviation of 880,000 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

(a)

= 4.48

= 0.88

To find P(X<5):

Z = (5 - 4.48)/0.88 = 0.59

Table of Area Under Standard Normal Curve gives area = 0.2224

So,

P(X<5) = 0.5 + 0.2224 = 0.7224

So,

Answer is:

0.7224

(b)

= 4.48

= 0.88

To find P(X):

Z = (3 - 4.48)/0.88 = - 1.68

Table of Area Under Standard Normal Curve gives area = 0.4535

So,

P(X>3) = 0.5 + 0.4535 = 0.9535

So,

Answer is:

0.9535

(c)

= 4.48

= 0.88

To find P(3<X<4):

Case1: For X from 3 to mid value:

Z = (3 - 4.48)/0.88 = - 1.68

Table of Area Under Standard Normal Curve gives area = 0.4535

Case2: For X from 4 to mid value:

Z = (4 - 4.48)/0.88 = - 0.55

Table of Area Under Standard Normal Curve gives area = 0.2088

So,

P(3<X<4) = 0.4535 - 0.2088 = 0.2447

So,

Answer is:

0.2447

(d)

85% corresponds to area= 0.85 - 0.50 = 0.35 from mid value to Z on RHS. Table gives Z = 1.04

So,

Z = 1.04 = (X - 4.48)/0.88

So,

X = 4.48 + (1.04 X 0.88) = 5.3952

So,

Answer is:

5.3952 million