Let X_1,X_2 , dots , X_n be n independent random variables each with mean 100 and standard deviation 30. Let X be the sum of these random variables.

Yasmin

Yasmin

Answered question

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

Skilled2021-09-04Added 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

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