Finding variance of sum of normal variables

[superwolf]

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?

 

[Billy Bob]

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.

 

[superwolf]

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?

 

[Billy Bob]

(ii) and (iii) don't require normal. (ii) doesn't even require independence, but (iii) does.

(i) and (iv) require normal.

 

[superwolf]

And which of them requires independence?

 

[Billy Bob]

And which of them requires independence?

(iii) and (iv)