poisson distribution and central limit theorem

dts5044

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Mar 6, 2008
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Let X[sub:36bgr5wt]1[/sub:36bgr5wt], X[sub:36bgr5wt]2[/sub:36bgr5wt],...,X[sub:36bgr5wt]30[/sub:36bgr5wt] be a random sample of size 30 from a Poisson distribution with a mean of 2/3. Approximate:

a) P (15 < [summation from i =1 to 30] X[sub:36bgr5wt]i[/sub:36bgr5wt] <=22)

b) P(21 <= [summation from i =1 to 30] X[sub:36bgr5wt]i[/sub:36bgr5wt] < 27)

How do I convert this into approximating N(0,1)? I've worked on it for half an hour and am stuck, please help!
 
If you add up 30 independent Poissons you get a Poisson with a mean of 30 times the original.Use that and the Central Limit Theorem.
 
The central limit theorem:

W = ([summation from i = 1 to n] X[sub:2qj5y28v]i[/sub:2qj5y28v]) -nu (u = mean)
_________________________________________
SQRT(n)*o (o is STD deviation)

is N(0,1) as n --> infinity

so my new poisson distribution, summing up all 30 samples, has mean 30 *2/3 = 20, which due to the properties of the poisson dist. is also my variance

mean = variance = 20

So P(15 < [summation from i=1 to 30] X[sub:2qj5y28v]i[/sub:2qj5y28v] = Y <= 22) =

= P(15.5 <= Y <= 22.5)
= P((15.5 - nu)/(SQRT(n)*o) <= (Y - nu)/(SQRT(n)*o) <= (22.5 -nu)/(SQRT(n)*o))
= P((15.5 - 30*2/3)/(SQRT(30)*SQRT(2/3)) <= (Y - 30*2/3)/(SQRT(30)*SQRT(2/3) <= (22.5 -30*2/3)/(SQRT(30)*SQRT(2/3)))

= P( -1.00623059 <= (Y - 30*2/3)/(SQRT(30)*SQRT(2/3) <= .5590169944)

which is approximately N(0,1)

but see how this doesn't give me clean values to plug into a table for the Normal Distribution? How do I finish the problem or did I do something wrong? Please help! Thanks!
 
How good is your Normal Table? I would have my students round to two decimal places.
 
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