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Variance problem

  • Thread starter superwolf
  • Start date
  • #1
176
0
X1, X2,...,X16 are independent and normally distributed, where mean value is 80 and variance is 18^2. Let Y = X1 + X2 + ... + X16. Calculate

i) P(X1 > 90)

ii) E(Y)

iii) Var(Y)

iv) P(Y>16*90)


i) 0.288, easy

ii) E(Y) = 16*80 = 1280

iii) Var(Y) = Var(16*Var(X)) = 16^2 * 18^2 (WRONG!)

Any suggestions?
 

Answers and Replies

  • #2
392
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X1+...+Xn is not the same as n*X

Use Var(a1*X1+...+an*Xn)=a1^2*Var(X1)+...+an^2*Var(Xn) with ai=1 for all i.
 
  • #3
176
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Brilliant! Would any of

P(X1>90) = 0.288

E(Y) = 1280

Var(Y) = 5184

P(Y>16*90) = 0.013

be correct without assuming normal distribution, but still assuming independence?
 
  • #4
392
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(ii) and (iii) don't require normal. (ii) doesn't even require independence, but (iii) does.

(i) and (iv) require normal.
 
  • #5
176
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And which of them requires independence?
 
  • #6
392
0
And which of them requires independence?
(iii) and (iv)
 

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