Question

Normal Distribution question that is based on a Poisson Random Variable. Attached is the photo of the question, and a posting of my work (which is incorrect)

Answer 1

I arrived at 0.829 (which is incorrect)
I did z <= to 10030, as such
\[\frac{ 10030 - (10*1000) }{ \sqrt{1000} } = 0.95\]

Answer 2

The z value of 0.95 reflects to be 0.8289 rounded to 0.829

Answer 3

"contains more than"

Answer 4

why are you dividing by \(\sqrt{1000}\)

Answer 5

i have to divide by the standard deviation as such it is the sqrt{1000}, i think...

Answer 6

ok...what is the standard deviation

Answer 7

and how does the mean and standard deviation of a Poisson r.v. relate?

Answer 8

i thought it was sqrt of 1000, but actually, the 1000 is the mean....i see...

Answer 9

why did you use 10*1000 in the numerator (it is correct...but why)

Answer 10

because the lambda is 10 * the mean....

Answer 11

and that becomes your new mean

Answer 12

yes.

Answer 13

so I ask again ...and how does the mean and standard deviation of a Poisson r.v. relate?

Answer 14

i am reviewing my notes...

Answer 15

the variance is equal to the mean.
as such the standard deviation is equal to the sqrt of the mean.

Answer 16

and this will be the sqrt of my new mean, which is the 10*1000 or is it the original?

Answer 17

sqrt of the new mean

Answer 18

also since it is more than, is my total number = (10030 + 0.5)?

Answer 19

for a continuity correct...yes

Answer 20

\[\frac{ (10030+0.5)-(10*1000) }{\sqrt{10*1000}}\]

Answer 21

yes

Answer 22

this changes my z value.
z is 0.305 which equates to 0.61982
or wait... do i subtract? so it would be 1-0.61982?

Answer 23

you do subtract

Answer 24

https://www.wolframalpha.com/input/?i=sum+10000%5En*exp%28-10000%29%2Fn%21%2C+n%3D10031+to+infinity

Answer 25

your method was correct, thanks for walking me through.

Answer 26

no problem