# Thread: Central limit Theorem problem (exponential(2))

1. ## Central limit Theorem problem (exponential(2))

Let X1,X2,... be i.i.d. with distribution Exponential(2). Use the central limit theorem to estimate the probability.

S is a sumation 1000 = n k = 1 <= is greater or less then. (sorry I don't know latex and I can't transfer word 07 equations over)

1000
P(S Xn <= 505)
k=1

This is what i can do.

E(x) = 1/2 var(x)= 1/4 Sd(x) = 1/2
so,
P( (S1000 - 1000*.5) /(sqrt(1000)*.5 ) <= (505 - 1000) /(sqrt(1000) * .5 ) )
I don't know where to go from here nor do I know if i set the problem up correctly. Sorry about the notation and stuff I don't know how to right equations in this. If I start to come here often I will learn latex.

2. Originally Posted by JeremyH
Let X1,X2,... be i.i.d. with distribution Exponential(2). Use the central limit theorem to estimate the probability.

S is a sumation 1000 = n k = 1 <= is greater or less then. (sorry I don't know latex and I can't transfer word 07 equations over)

1000
P(S Xn <= 505)
k=1

This is what i can do.

E(x) = 1/2 var(x)= 1/4 Sd(x) = 1/2
so,
P( (S1000 - 1000*.5) /(sqrt(1000)*.5 ) <= (505 - 1000) /(sqrt(1000) * .5 ) )
I don't know where to go from here nor do I know if i set the problem up correctly. Sorry about the notation and stuff I don't know how to right equations in this. If I start to come here often I will learn latex.
$Z_n = \frac{S_n - n \mu}{\sigma \sqrt{n}} = \frac{S_n - 500}{5 \sqrt{10}}$.

Substitute $S_n = 505$: $Z_n = \frac{1}{\sqrt{10}} \approx 0.3162$.

So calculate $\Pr(Z \leq 0.3162)$ where Z ~ Normal(0, 1).