Question

Stath Question..

Answer 1

You really don't know?

Answer 2

The exponential distribution is \[
P(X=x) = \lambda e^{-\lambda x}=5e^{-5x}
\]

Answer 3

Since there are \(35\) rings and they want a mean of \(4\) then that means the total of the diameters is \(140\).

Answer 4

that is my problem, we are using \[P(X=x) = \frac{1}{\theta} e^{-\frac{1}{\theta} x}\]
and I do not know why, it is confusing me..

Answer 5

\[\theta=\frac{1}
{\lambda}\\P(X>4) = \frac{1}{\frac{1}{5}} e^{-\frac{1}{\frac{1}{5}} x}=5e^{-5x}\]

Answer 6

when we take integral from 4 to inf. it is almost zero..

Answer 7

Aren't you supposed to find the probability that the sum of the independent exponential distributions are greater than \(140\)?

Answer 8

we are using central limit theorem. since it is not normal, but we have n=35>30 we can assume it is normal and X~N(1/5,1/5/(sqrt35))

by using TI-89 Normalcdf(4,inf,1/5,1/5/(sqrt35))=0

Answer 9

for exp. dist. theta=mu and theta=sigma

Answer 10

lol, the probability is 0?

Answer 11

not exactly almost, so small..

Answer 12

2.061153622*10^-9

Answer 13

I used mathematica and got that..

Answer 14

Let me think, the \(z\) score would be \[
z = \frac{4-5^{-1}}{5^{-1}}=20-1= 19
\]19 standard deviations away from the mean.

Answer 15

You could get the precise value using Erlang CDF actually.

Answer 16

what is that?

Answer 17

Erlang PDF is the sum of identical independent exponential distributions.

Answer 18

So you'd use \(X\sim Erlang(k=35,\lambda =5)\) and you'd want \(\Pr(X>140)\)

Answer 19

http://en.wikipedia.org/wiki/Erlang_distribution

Answer 20

thanks for extra info (: