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CIE statistic2 doubts! Help please
Tohru Kyo Sohma:
anyone who did CIE statistics 2 pls help me with qns 7 part (b) and qns 8 of chapter one:the poisson distribution. pls help me
Romeesa-Chan:
Please try uploading the image of the question yuh have doubt in! =]
Most UniQue™:
--- Quote from: Tohru Kyo Sohma on July 25, 2011, 08:52:36 pm ---anyone who did CIE statistics 2 pls help me with qns 7 part (b) and qns 8 of chapter one:the poisson distribution. pls help me
--- End quote ---
Please give the link to the question paper or mention it's of which year so that others can be able to answer.
Thank You
Tohru Kyo Sohma:
its not from pastpapers..its from the text book...i told u guys im doing self study for S2
here is the question;
Q.7)Assume that a car pass under a bridge at a rate of 100 per hour and that a Poisson distribution is appropriate.
(a)what is the probability that during a 3 min period no car will pass under the bridge?
(b)what time interval is such that the probability is atleast 0.25 that no car will pass under the bridge during that interval
Q.8)A radioactive source emits particles at an average rate of one per second. Assume that the number of emissions follows a Poisson distribution.
(a)calculate the probability that 0 or 1 particle will be emitted in 4 seconds.
(b)the emission rates changes such that the probability of 0 or 1 emission in 4 seconds becomes 0.8. What is the new emission rate?
Alpha:
--- Quote from: Tohru Kyo Sohma on July 29, 2011, 02:19:36 pm ---its not from pastpapers..its from the text book...i told u guys im doing self study for S2
here is the question;
Q.7)Assume that a car pass under a bridge at a rate of 100 per hour and that a Poisson distribution is appropriate.
(a)what is the probability that during a 3 min period no car will pass under the bridge?
(b)what time interval is such that the probability is atleast 0.25 that no car will pass under the bridge during that interval
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"...at a rate of 100 per hour"
60 mins => 100 cars
(a) Find the mean, we denote it by for a Poisson Distribution. [Equivalent of E(X)]
3 mins => 100/60*3 = 5 cars.
X is the random variable denoting the number of cars passing under the bridge during a 3 minutes interval.
Therefore, X ~ Po (5).
No cars -> X=0.
P (X=0) = e^(-5)*5^0/ 0!
= e^(-5)
Give your answer to 3 significant figures.
(b) Let the required time be n minutes.
Now, X is the random variable denoting the number of cars passing under the bridge during an n minutes interval.
= 100n/60 = 10n/6.
"...probability is atleast 0.25 that no car will pass under the bridge during that interval"
I will use 'G' to denote the greater or equal to sign: ,
and 'L' to denote the less or equal to sign: .
X ~ Po (10n/6).
P(x=0) G 0.25
e^(-10n/6) G 0.25
Take ln on both sides.
-10n/6 G ln 0.25
n L ln 0.25* (-6/10) [G changes to L because you are multiplying by a negative number]
n L 0.831
"At least", prob. should keep on increasing. Therefore, the value of e^(-10n/6) should be greater and greater, which will happen the smaller -10n/6 gets. e^(-no.)= 1/e^(no.)
Take the greatest value of n you can from the answer you obtained, i.e. n= 0.831.
Check by substituting n= 0.831 in e^(-10n/6). You will get a probability which is greater than 0.25, answer is correct.
It's advisable to check your answers for such types of questions in probability.
Question 8 is similar to question 7. Since you are self-studying, I would suggest you try question 8 by yourself by first understanding my procedures in question 7. Post your answer here so that I can check it for you. If you are still unable to do it, let me know, I'll do it then.
Hint for question 8: probability that 0 or 1 => Add P(x=0) and P(X=1).
I hope it's clear. If not, lemme know. :)
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