Let X1, . . . , X20 be a sequence of i.i.d. observation from N(5, 16).?

(a) Since X1, ...., X20 are iid and N(5, 16), we see that
W = X1 + ... + X10 - X11 - ... - X20 is also normal with
mean = 5 + ... + 5 - 5 - ... - 5 = 0 and variance 20 * 16 = 320.

So, P(X1 + ... + X10 ≥ X11 + ... + X20)
= P(W ≥ 0)
= P(Z ≥ (0 - 0)/√320), converting to Z
= 0.5.
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(b) Similarly, U = X1 - X2 + ... - X20 is normal with mean 0 and variance 320.

So, we want P(U > 1) = P(Z > (1 - 0)/√320) ≈ 0.4777.

I hope this helps!

N(5,16) is mean 5 and variance 16
z1 = x1+x2+...+x10 is N (50, 1600)
z2= x11+x12+....+x20 is N(50,1600)
P(z1=z2) = 1


P( x1-x2+x3-x4+...+x19-x20 >1 )
S = Σ(-1)^i-1 Xi has mean 0 and variance 20*16 = 3200

μ = 0
σ = 3200
standardize x to z = (x - μ) / σ
P(S > 1) = P( z > (1-0) / 3200)
= P(z > 0.0003) = 0.5
(From Normal probability table)