a) is quite simple actually.


if P(X = x) = kx,

Then P(X = 1) = k(1) = k
P(X = 3) = k(3) = 3k
P(X = 5) = 5k
P(X = 7) = 7k
P(X = 9) = 9k
P(X = 11) = 11k


Now total probability = 1

So k + 3k + 5k + 7k + 9k + 11k = 1

36k = 1

k = 1/36

E(X)

= ∑ xP(X = x)

= (1)P(X = 1) + 3P(X = 3) + 5P(X = 5) + 7P(X = 7) + 9P(X = 9) + 11P(X = 11)

= (1)k + 3(3k) + 5(5k) + 7(7k) + 9(9k) + 11(11k)

= 286k

= 286/36

= 7.944444444444444.....

= 7.94 (3s.f)


Var(X)

= E(X²) - [E(X)]²

= ∑ x²P(X = x) - (286/36)²

= (1²)k + 3²(3k) + 5²(5k) + 7²(7k) + 9²(9k) + 11²(11k) - (286/36)²

= 2556k - (286/36)²

= 2556/36 - (286/36)²

71 - 63 37/324

= 7 287/324

≈ 7.88580247

= 7.89 (3 s.f)

Expectation, E(X) = μ ≈ 7.944
Variance, Var(X) = σ² = 7.886

S.D , σ = √7.886
= 2.8081671
≈ 2.808

μ + σ = 7.944 + 2.808 = 10.752
μ - σ = 7.944 - 2.808 = 5.136

P(μ - σ < X < μ + σ)

= P(5.136 < X < 10.752)

= P(X = 7 or X = 9)

= P(X = 7) + P(X = 9)

= 7k + 9k

= 16k

= 16/36

= 4/9

P(X1 + X2 < 7) = P(X1 + X2 ≤ 6)

Possibilities :

①Both observations have a score of 1 , total = 2

②Both observations have a score of 3, total = 6

③One score is 1 , the other 5, total = 6

④One score is 3, the other 1, total = 4

So,

P(X1 + X2 < 7) = P(X1 + X2 ≤ 6)

= P(X1 = 5) x P(X2 = 1)
+ P(X1 = 1) x P(X2 = 5)
+ P(X1 = 3) x P(X2 = 3)
+ P(X1 = 3) x P(X2 = 1)
+ P(X1 = 1) x P(X2 = 3)
+ P(X1 = 1) x P(X2 = 1)

5k(k) + k(5k) + 3k(3k) + 3k(k) + k(3k) +k(k)

= 26k²

= 26/36²

= 13/648