Question

For the past

112112

​years, a certain state suffered

2828

direct hits from major​ (category 3 to​ 5) hurricanes. Assume that this was typical and the number of hits per year follows a Poisson distribution. Complete parts​ (a) through​ (d).

​(a) What is the probability that the state will not be hit by any major hurricanes in a single​ year?

The probability is

1-The number of hits to a website follows a Poisson process. Hits occur at the rate of

1.0 per minute1.0 per minute

between​ 7:00 P.M. and

99​:00

P.M. Given below are three scenarios for the number of hits to the website. Compute the probability of each scenario between

8 : 27 P.M.8:27 P.M.

and

88​:3535

P.M. Interpret each result.

​(a) exactly fivefive

​(b) fewer than fivefive

​(c) at least fivefive

2-Determine the required value of the missing probability to make the distribution a discrete probability distribution. x ​P(x) 3 0.35 0.35 4 ​? 5 0.16 0.16 6 0.27 0.27 ​P(4) =

Answer 1

Solution:

1) We are given that:

For the past 112 ​years, a certain state suffered 28 direct hits from major​ (category 3 to​ 5) hurricanes.

This was typical and the number of hits per year follows a Poisson distribution.

Thus parameter of Poisson distribution is:

per year.

We have to find:

the probability that the state will not be hit by any major hurricanes in a single​ year.

That is: P( X = 0) = .........?

Thus using probability mass function of Poisson distribution:

2) The number of hits to a website follows a Poisson process. Hits occur at the rate of 1.0 per minute1.0 per minute between​ 7:00 P.M. and 9​:00 P.M.

iven below are three scenarios for the number of hits to the website. Compute the probability of each scenario between 8 : 27 P.M and 8​:35 P.M. Interpret each result.

Part a) exactly five.

X = number of of hits to the website follows Poisson distribution with parameter = per minute.

Then between 8 : 27 P.M and 8​:35 P.M, we have 8 minutes.

Hence parameter will change to per 8 minutes.

Thus using Poisson probability distribution, we get:

Part b) fewer than five

P( X < 5 ) = ........?

P( X < 5) = P( X= 0) + P(X = 1) + P( X =2 ) + P( X = 3 ) + P( X = 4)

Part c) P( At least five) = ..........?

3) Determine the required value of the missing probability to make the distribution a discrete probability distribution:

x	P(x)
3	0.35
4	-
5	0.16
6	0.27

We have to find: P(4)

We use following rule: