# Variance problem

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?

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.

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?

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

(i) and (iv) require normal.

And which of them requires independence?

And which of them requires independence?
(iii) and (iv)