Let X1,X2,s˙,Xn be n independent random variables each with mean 100 and standard deviation 30....

Yasmin

Yasmin

Answered

2021-09-03

Let X1,X2,s˙,Xn be n independent random variables each with mean 100 and standard deviation 30. Let X be the sum of these random variables. Find n such that Pr(X>2000)0.95

Answer & Explanation

Faiza Fuller

Faiza Fuller

Expert

2021-09-04Added 108 answers

Step 1
X1,X2,s˙,Xn are independent ranodom variables with mean 100 and standard deviation 30
So, Var(Xi)=302=900
Now given that,
X=iXi
So E(X)=E(iXi)=iE(Xi)=100n
and Var(X)=Var(iXi)=iVar(Xi) [ Covariances are zero due to independence]
So, Var(X) = 900n
Step 2
By Central Limit Theorem,
XμσX100n900nN(0,1)
Now, P(X>2000)>0,95
implies 1P(X2000)>0,95
P(X2000)<0,05
P(X100n900n2000100n30n)<0,05
implies Φ[2000100n30n]<0,05=Φ(1,645)
implies 2000100n30n<1,645
Here you can solve this in two methods,
Trial and error method
Numerically using root finding method
Using trial and error method,
For n = 22
2000100n30n=1,4213
For n=23
2000100n30n=2,085
So, for n = 23 it is crossing -1.645. Hence n = 23

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?