# Thread: central limit theorm help

1. ## central limit theorm help

Z1,Z2,.... are identically independently distributed with uniform distribution [-20,10]. How do i use the central limit theorem to estimate the the probability
P(summation (900, i=1) : Zi >= -4470)

My attempt
E(Zi)=-5
Var(Zi)=80
P(summation Zi <= -4470) = P([summation Zi]/900 <= [-4470]/900)
=P(sample Z (900) <= 80)
=P([sample Z (900) + 5]/sqrt(80/900) <= [80+5]/sqrt(80/900))
~=P(Z<=285.1) where Z~Normal(0,1)

2. Originally Posted by Sneaky
Z1,Z2,.... are identically independently distributed with uniform distribution [-20,10]. How do i use the central limit theorem to estimate the the probability
P(summation (900, i=1) : Zi >= -4470)

My attempt
E(Zi)=-5
Var(Zi)=80
P(summation Zi <= -4470) = P([summation Zi]/900 <= [-4470]/900)
=P(sample Z (900) <= 80)
=P([sample Z (900) + 5]/sqrt(80/900) <= [80+5]/sqrt(80/900))
~=P(Z<=285.1) where Z~Normal(0,1)
The distribution of the sum on $\displaystyle $$n iid RVs with mean \displaystyle$$ m$ and variance $\displaystyle$$v$ has distribution $\displaystyle N(n \times m,n \times v)$ (approximatly, by the CLT)

The mean of $\displaystyle U(a,b)$ is $\displaystyle (a+b)/2$ and its variance is $\displaystyle (b-a)^2/12$

CB

3. so i redid it and now i get

1-P(Z<=0.115)

where Z ~ N(0,1)

Is this right?

4. Originally Posted by Sneaky
so i redid it and now i get

1-P(Z<=0.115)

where Z ~ N(0,1)

Is this right?
Yes

CB