Let x equal the number observed on the throw of a single bal

Let x equal the number observed on the throw of a single balanced die. (a) Find and graph the probability distribution for x. p(x)

a) On a single throw of a dice , probability of occurence of any number = 1/6 =0.167 ( constant)

So, option 3

b) (X) = 1×P(X = 1) + 2×P(X = 2) + 3×P(X = 3) + 4×P(X=4) + 5×P(X=5) + 6×P(X=6)

Therefore E(X) = 1/6 + 2/6 + 3/6 + 4/6 + 5/6 + 6/6 = 7/2

So the expectation is 3.50

c) E(X2) = 1/6 + 4/6 + 9/6 + 16/6 + 25/6 + 36/6 = 91/6 = 15.167

Var(X) = E(X2) – m2 = 15.167 -3.5^2 = 2.917

Std deviation = sqrt(Var(x) = 1.70

d) mu + /- 2sigma = 3.5 +/- 2*1.70 = 6.9 - 0.1

Intervcal of x :: 0.1 --- 6.9