Question

the number of hits to a website follows a poisson process. hits occur at the rate of 0.2 per minute between 7:00 p.m. and 9:00 p.m. given below are three scenarios for the number of hits to the website. compute the probability of each scenario between 8:38 p.m. and 8:44 p.m. interpret each result.
(a) exactly four
(b) fewer than four
(c) at least four

(a) ( p(4)=0.0260 )
(round to four decimal places as needed.)
on about 3 of every 100 time intervals between 8:38 p.m. and 8:44 p.m, the website will receive exactly 4 hit(s).
(round to the nearest whole number as needed.)
(b) ( p(x<4)=0.9663 )
(round to four decimal places as needed.)
on about 97 of every 100 time intervals between 8:38 p.m. and 8:44 p.m, the website will receive fewer than 4 hit(s).
(round to the nearest whole number as needed.)
(c) ( p(xgeq4)=0.0337 )
(round to four decimal places as needed.)
on about ( square ) of every 100 time intervals between 8:38 p.m. and 8:44 p.m, the website will receive hit(s).
(round to the nearest whole number as needed.)

Explanation:

Step1: Calculate the number of minutes between 8:38 P.M. and 8:44 P.M.

The time interval is \(8:44 - 8:38=6\) minutes.

Step2: Find the mean \(\lambda\) for the Poisson distribution.

Given the rate is \(0.2\) hits per minute. So \(\lambda = 0.2\times6 = 1.2\)

Step3: Recall the Poisson probability formula

The Poisson probability formula is \(P(X = k)=\frac{e^{-\lambda}\lambda^{k}}{k!}\), where \(X\) is the random variable representing the number of hits, \(\lambda\) is the mean, \(k\) is the number of events, and \(e\approx2.71828\)

Part (a)

Step4: Calculate \(P(4)\)

$$
LATEXBLOCK0
$$

Interpretation: \(P(X = 4)=0.0260\) means that on about \(0.0260\times100 = 2.6\approx3\) of every 100 time - intervals between 8:38 P.M. and 8:44 P.M., the website will receive exactly 4 hits.

Part (b)

Step5: Calculate \(P(X\lt4)=P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)\)

$$
LATEXBLOCK1
$$

$$
LATEXBLOCK2
$$

Interpretation: \(P(X\lt4) = 0.9663\) means that on about \(0.9663\times100=96.63\approx97\) of every 100 time - intervals between 8:38 P.M. and 8:44 P.M., the website will receive fewer than 4 hits.

Part (c)

Step6: Use the complement rule \(P(X\geq4)=1 - P(X\lt4)\)

Since \(P(X\lt4)=0.9663\), then \(P(X\geq4)=1 - 0.9663=0.0337\)
Interpretation: \(P(X\geq4)=0.0337\) means that on about \(0.0337\times100 = 3.37\approx3\) of every 100 time - intervals between 8:38 P.M. and 8:44 P.M., the website will receive at least 4 hits.

Answer:

• (a) \(P(4)=0.0260\). On about \(3\) of every \(100\) time intervals between \(8:38\) P.M. and \(8:44\) P.M, the website will receive exactly \(4\) hit(s).
• (b) \(P(x\lt4)=0.9663\). On about \(97\) of every \(100\) time intervals between \(8:38\) P.M. and \(8:44\) P.M, the website will receive fewer than \(4\) hit(s).
• (c) \(P(x\geq4)=0.0337\). On about \(3\) of every \(100\) time intervals between \(8:38\) P.M. and \(8:44\) P.M, the website will receive at least \(4\) hit(s).