Let X_{1}, X_{2},...,X_{n} 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)\geq 0.95.

vazelinahS 2021-06-04 Answered
Let X1,X2,...,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.
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Expert Answer

unessodopunsep
Answered 2021-06-05 Author has 105 answers

Step 1
Given that μ=100 and σ=30
E(X)=E(Xi)=nμ=100n
σ(X)=σ(Xi)=n30
Step 2
Consider,
P(X>2000)0.95
P(XE(X)σ(X)>2000100nn30)0.95
P(z>20010nn3)0.95
1P(z20010nn3)0.95
10.95P(z20010nn3)
P(z20010nn3)0.05
20010nn31.645
20010n4.935n
4.935n10n+2000
n4.23
n=17.89
18
Thus, n=18.

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