It's a good idea to write down what you're distribution is. You want X to be "the number of emissions in 4 seconds", you know there is 1 per second, so in 4 seconds there's going to be 4. So X~po(4)
I'm not sure what you've done there, but the poisson distribution formula is P(X=x)=x!e−λ(λ)x
It's a good idea to write down what you're distribution is. You want X to be "the number of emissions in 4 seconds", you know there is 1 per second, so in 4 seconds there's going to be 4. So X~po(4)
I'm not sure what you've done there, but the poisson distribution formula is P(X=x)=x!e−λ(λ)x
I used used the formula for P(X=0) and P(X=1) then added them which would equal 0.8?
Divide P(X=0)+P(X=1) by 4 before making it equal 0.8?
The above answer shouldnt be the right answer. Something is messed up. Try putting the given Y answer back into the formula to see if you get 0.8 which you shouldnt
The above answer shouldnt be the right answer. Something is messed up. Try putting the given Y answer back into the formula to see if you get 0.8 which you shouldnt
Yep dividing P(X<<1) by 4 gives an imaginary number. I'm not sure where I've gone wrong with the working out I've shown
Yep dividing P(X<<1) by 4 gives an imaginary number. I'm not sure where I've gone wrong with the working out I've shown
No i mean the answer of 0.206 is wrong. If you try to work backwards by putting it into the formula with x=1, it doesnt give anything close to 0.8 whether you divide/multiply by 4 or not. So their answer is most likely wrong
No i mean the answer of 0.206 is wrong. If you try to work backwards by putting it into the formula with x=1, it doesnt give anything close to 0.8 whether you divide/multiply by 4 or not. So their answer is most likely wrong
It does give 0.8 because you have to times 0.206 by 4 as the probability of 0 and 1 was for 4 seconds
I didn't see that the emission rate changes, what you have done is right up to the third line, but you can't take natural logs to one term, it must be all of it. I would suggest multiplying through by eλ , but then you get an equation with a λ term as well, which I can only think of solving right now by iterative methods. It is 0.824... , then just divide it by 4.
I didn't see that the emission rate changes, what you have done is right up to the third line, but you can't take natural logs to one term, it must be all of it. I would suggest multiplying through by eλ , but then you get an equation with a λ term as well, which I can only think of solving right now by iterative methods. It is 0.824... , then just divide it by 4.
Oh right I'm still not sure how I would go about solving it though