(a) What is the probability that the website 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 website 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 website 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:
Given Data:
The Population Mean $(\mu)=4.6 $ million
The population Standard deviation $(\sigma) = \frac{720000}{1000000} = 0.72 $ million
Solution:
(A) The probability that the website has fewer than 5 million visitors in a single day:
$P(x < 5) = P\left( \frac{x – \mu}{\sigma} < \frac{5 – 4.6}{0.72} \right)$
$P(x < 5) = P(z < 0.556)$
$P(x < 5) = 0.7109$
(B) The probability that the website has 3 million or more visitors in a single day:
$P(x \geq 3) = P\left( \frac{x – \mu}{\sigma} \geq \frac{3 – 4.6}{0.72} \right)$
$P(x \geq 3) = P(z \geq -2.222)$
$P(x \geq 3) = 1 – P(z < -2.222)$
$P(x \geq 3) = 1 – 0.0131$
$P(x \geq 3) = 0.9869$
(C) The probability that the website has between 3 million and 4 million visitors in a single day:
$P(3 < x < 4) = P\left( \frac{3 – 4.6}{0.72} < \frac{x – \mu}{\sigma} < \frac{4 – 4.6}{0.72} \right)$
$P(3 < x < 4) = P(-2.222 < z < -0.833)$
$P(3 < x < 4) = P(z < -0.833) – P(z < -2.222)$
$P(3 < x < 4) = 0.2024 – 0.0131$
$P(3 < x < 4) = 0.1893$
(D) The amount of web traffic that will require Smiley’s People to purchase additional server capacity:
$P(x < P_{85}) = 0.85$
$P\left( z < \frac{P_{85} – \mu}{\sigma} \right) = 0.85$
From the Excel function,
$\frac{P_{85} – \mu}{\sigma} = 1.036$
$\frac{P_{85} – 4.6}{0.72} = 1.036$
$P_{85} – 4.6 = 1.036 \times 0.72$
$P_{85} – 4.6 = 0.746$
$P_{85} = 5.3459$
$P_{85} \approx 5.35$
Final Answer:
(A) The probability that the web site has fewer than 5 million visitors in a single day is 0.7109
(B) The probability that the web site has 3 million or more visitors in a single day is 0.9869
(C) The probability that the web site has between 3 million and 4 million visitors in a single day is 0.1893
(D) 5.35 million visitors per day is the amount of web traffic that will require Smiley’s People to purchase additional server capacity
