Poisson Distribution Question

[King_Silver]

Homework Statement

The number of tornadoes per year, in Georgia, has a Poisson distribution with a mean of 2.4 tornadoes. Calculate the probability that in any given year, there will be:
(i) At most 2 cases.
(ii) At least one case.
(iii) Calculate the probability that there will be exactly 10 tornadoes in the next seven years.

Homework Equations

λ = 2.4 (Mean),

Formula: P(X = x) = e-λ (λx/x!)

Where X = number of events at given internal

e = ~2.71

x = 0,1,2,3,4……..n (where n = any number)

The Attempt at a Solution

(i)At most 2, Therefore we need to examine P(X=0),P(X=1),P(X=2)

P(X=0) = e-2.4(2.40/0!) = 0.090717953

P(X=1) = e-2.4(2.41/1!) = 0.217723087

P(X=2) = e-2.4(2.42/2!) = 0.261267705

0.0907+0.2177+0.2612 = 0.5697 (56.97%)



(ii) At least 1, Therefore we need to examine P(X=0). Then 1-P(X=0)

P(X=0) = e-2.4(2.40/0!) = 0.090717953

1-0.090717953 = 0.90929 (~90.93%)



(iii) At least 1, Therefore we need to examine P(X=10). Over 7 years

P(X=0) = e-2.4(2.40/0!) = 0.090717953

P(X=1) = e-2.4(2.41/1!) = 0.217723087

P(X=2) = e-2.4(2.42/2!) = 0.261267705

P(X=3) = e-2.4(2.43/3!) = 0.209014164

P(X=4) = e-2.4(2.44/4!) = 0.125408498

P(X=5) = e-2.4(2.45/5!) = 0.060196079

P(X=6) = e-2.4(2.46/6!) = 0.024078431

P(X=7) = e-2.4(2.47/7!) = 0.008255462

P(X=8) = e-2.4(2.48/8!) = 0.002476638

P(X=9) = e-2.4(2.49/9!) = 0.000660436

P(X=10) = e-2.4(2.410/10!) = 0.000158504

Part (iii) is where I run into an issue. The question states it must be exactly 10 tornadoes in the space of 7 years. Meaning there could be any combination of tornadoes in the years. For example; the first year could have all 10 tornadoes, with the following years not having none. Or the first year could have none, the second year could have 3, then the remaining 6 tornadoes in the following years.
How I was going to attempt this final part was by summing these values up. This would give the maximum probability per year, then multiply this by 7 (as 7 years) I doubt this is correct.

Where does the 7 years come into the formula? and is it correct to assume that calculating P(X= 1 to 10) is relevant to the question?

 

[mjc123]

First, please use the superscript function to make your expressions clear (on grey bar at the top of the text window). Thus
P(X=x) = e^-λλ^x/x!
P(X=9) = e^-2.4(2.4⁹/9!) etc.
Now do you know, or can you work out, a formula for the 7 year case? You know that the mean for 1 year is 2.4. What do you think is the mean for 7 years? Using this, can you suggest a formula for P(X=x) in 7 years?

 

[King_Silver]

First, please use the superscript function to make your expressions clear (on grey bar at the top of the text window). Thus
P(X=x) = e^-λλ^x/x!
P(X=9) = e^-2.4(2.4⁹/9!) etc.
Now do you know, or can you work out, a formula for the 7 year case? You know that the mean for 1 year is 2.4. What do you think is the mean for 7 years? Using this, can you suggest a formula for P(X=x) in 7 years?

At least 10, Therefore we need to examine P(X=10). Over 7 years

λ = (2.4)(7yrs) = 16.8

P(X=10) = e-2.4(16.810/10!) = 0.02495 (~2.5%)

is this correct?

 

[Charles Link]

At least 10, Therefore we need to examine P(X=10). Over 7 years

λ = (2.4)(7yrs) = 16.8

P(X=10) = e-2.4(16.810/10!) = 0.02495 (~2.5%)

is this correct?

If $\lambda$ has changed to 16.8, you need to use $e^{-16.8}$ in the formula. Also, use ^ to get an exponent in Latex. The formula reads $\lambda^k$.

 

[Ray Vickson]

At least 10, Therefore we need to examine P(X=10). Over 7 years

λ = (2.4)(7yrs) = 16.8

P(X=10) = e-2.4(16.810/10!) = 0.02495 (~2.5%)

is this correct?

If your e-2.4 means e^(-2.4) ("e to the power -2.4") then yes, it is correct.

As mjc123 points out in #2, you MUST distinguish a power from another type of operation. So, either use the "superscript" button (labelled "$x^2$" in the grey ribbon at the top of the input panel) or else use "^", to write a^b instead of a b (when you mean $a^b$). (If I were marking your work I would mark it wrong as you have submitted it.)

 

[Charles Link]

If your e-2.4 means e^(-2.4) ("e to the power -2.4") then yes, it is correct.

As mjc123 points out in #2, you MUST distinguish a power from another type of operation. So, either use the "superscript" button (labelled "$x^2$" in the grey ribbon at the top of the input panel) or else use "^", to write a^b instead of a b (when you mean $a^b$). (If I were marking your work I would mark it wrong as you have submitted it.)

@Ray Vickson It's very early in the morning, but shouldn't it (the normalizing factor of the Poisson distribution) be $e^{-16.8}$? I believe the OP got it incorrect.

 

[King_Silver]

@Ray Vickson It's very early in the morning, but shouldn't it be $e^{-16.8}$? I believe the OP got it incorrect.

My bad! I calculated it as if it was e^-16.8 just left it as 2.4 when I wrote it! my bad. Thanks for the help though!

If your e-2.4 means e^(-2.4) ("e to the power -2.4") then yes, it is correct.

As mjc123 points out in #2, you MUST distinguish a power from another type of operation. So, either use the "superscript" button (labelled "$x^2$" in the grey ribbon at the top of the input panel) or else use "^", to write a^b instead of a b (when you mean $a^b$). (If I were marking your work I would mark it wrong as you have submitted it.)

My bad, again; not sure why I did that. Was in a rush responding. Appreciate the feedback. Cheers!

 

[Ray Vickson]

@Ray Vickson It's very early in the morning, but shouldn't it (the normalizing factor of the Poisson distribution) be $e^{-16.8}$? I believe the OP got it incorrect.

Yes, of course: I meant that the final numerical answer in (iii) was correct.