Question

The number of hits to a website follows a Poisson process. Hits occur at the rate of 1.3 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 8:17 P.M. and 8:22 P.M. Interpret each result. (a) exactly five (b) fewer than five (c) at least five (a) P(5)= (Round to four decimal places as needed.) On about of every 100 time intervals between 8:17 P.M. and 8:22 P.M, the website will receive exactly 115 hits). (Round to the nearest whole number as needed.) (b) P(x<5)=0 (Round to four decimal places as needed.) On about of every 100 time intervals between 8:17 P.M. and 8-22 P.M. the website will receive fewer than 6 hits). (Round to the nearest whole number as needed.) (c) P( x5)= (Round to four decimal places as needed.) On about 97 of every 100 time intervals between 8:17 P.M. and 8:22 P.M, the website will receive at least 5 hits). (Round to the nearest whole number as needed.)

Answer 1

The probability of exactly five hits between 8:17 P.M. and 8:22 P.M. is approximately 0.0847. The probability of fewer than five hits is approximately 0.0512. The probability of at least five hits is approximately 0.9488.

Step-by-step explanation:

To compute the probability of each scenario, we can use the Poisson distribution formula.

(a) To find the probability of exactly five hits between 8:17 P.M. and 8:22 P.M., we can use the formula P(X = k) = (e^(-λ) * λ^k) / k!, where λ is the average rate of hits per minute and k is the desired number of hits. In this case, λ = (1.3 hits/minute) * (5 minutes) = 6.5. Plugging the values into the formula, we get P(X = 5) = (e^(-6.5) * 6.5^5) / 5! ≈ 0.0847. So, the probability of exactly five hits is approximately 0.0847.

(b) To find the probability of fewer than five hits, we can sum the probabilities of getting 0, 1, 2, 3, or 4 hits. Using the Poisson distribution formula, we can calculate each individual probability and sum them up. Plugging the values into the formula, we get P(X < 5) ≈ (e^(-6.5) * 6.5^0) / 0! + (e^(-6.5) * 6.5^1) / 1! + (e^(-6.5) * 6.5^2) / 2! + (e^(-6.5) * 6.5^3) / 3! + (e^(-6.5) * 6.5^4) / 4! ≈ 0.0512. So, the probability of fewer than five hits is approximately 0.0512.

(c) To find the probability of at least five hits, we can subtract the probability of getting fewer than five hits from 1. Using the Poisson distribution formula, we can calculate the probability of getting fewer than five hits (as calculated in part b) and subtract it from 1. Plugging the values into the formula, we get P(X ≥ 5) ≈ 1 - P(X < 5) ≈ 1 - 0.0512 ≈ 0.9488. So, the probability of at least five hits is approximately 0.9488.