Help me answer this question.....? (question in picture below)

(i) `"P"(X < 6) = "P"(X >= 6)`

and by total probability

`"P"(X < 6) + "P"(X >= 6)=1`

so

`"P"(X < 6) = 1/2 = "P"(X >= 6)`

Since there are 5 outcomes less than 6, and all 5 outcomes have the same probability, we divide that `1/2` into 5 equal pieces to give to each of the 5 outcomes: `c = (1/5)(1/2) = 1/10`.

Similarly, the outcomes 6, 7, and 8 are equally likely, and their probabilities sum to (the other) `1/2`, so `d = (1/3)(1/2)=1/6.`

(ii) `"E"(X)=sum_(x=1)^8[x * "P"(X=x)]`

`=sum_(x=1)^5[x * "P"(X=x)]+sum_(x=6)^8[x * "P"(X=x)]`
`=1/10sum_(x=1)^5x+1/6sum_(x=6)^8x`

`=15/10+21/6`

`=(45+105)/30=150/30=5`

`"Var"(X)="E"(X^2)-{"E"(X)}^2`

`=sum_(x=1)^8[x^2"P"(X=x)] - 5^2`
`=1/10sum_(x=1)^5x^2+1/6sum_(x=6)^8x^2-25`

`=55/10+149/6-25`

`=(165+745-750)/30=160/30=16/3`

(iii) Let `X_1` and `X_2` be the scores for the first roll and second roll, respectively. Then

`"P"(X_1+X_2=10)`

`=(("P"(X_1=2, X_2=8)),(+"P"(X_1=3, X_2=7)),(+"P"(X_1=4, X_2=6)))+"P"(X_1=5, X_2=5)+(("P"(X_1=6, X_2=4)),(+"P"(X_1=7, X_2=3)),(+"P"(X_1=8, X_2=2)))`

SInce all 6 terms in the big brackets have the same probability (due to independence of `X_1` and `X_2`), we have

`=6["P"(X_1=2)"P"(X_2=8)] + ["P"(X_1=5)"P"(X_2=5)]`

`=6[1/10 xx 1/6]+[1/10xx1/10]`

`=1/10 + 1/100`

`=(10+1)/100 = 11/100`

(iv) `Y=sum_(i=1)^48 X_i`, where the `X_i"'s"` are independent. So

`"E"(Y)="E"(sum_(i=1)^48 X_i)`
`color(white)("E"(Y))=sum_(i=1)^48 "E"(X_i)" "` (by independence)
`color(white)("E"(Y))=sum_(i=1)^48 5`

`color(white)("E"(Y))=48 xx 5=240`

Similarly,

`"Var"(Y)="Var"(sum_(i=1)^48 X_i)`
`color(white)("Var"(Y))=sum_(i=1)^48 "Var"(X_i)" "` (by independence)
`color(white)("Var"(Y))=sum_(i=1)^48 16/3`

`color(white)("Var"(Y))=48 xx 16/3=256`

(v) Assuming `Y` is normally distributed, we have

`Y " ~ " "N"(mu=240, sigma^2=256)`

So

`"P"(220 <= Y <= 260)`

`="P"((220-mu)/sigma <= (Y-mu)/sigma <= (260-mu)/sigma)`

Using the standard normal `(Z)` equivalence, this is

`="P"((220-240)/16 <= Z <= (260-240)/16)`

`="P"("–"1.25 <= Z <= 1.25)`

`="P"(Z <= 1.25) - "P"(Z < "–"1.25)`

`=Phi(1.25)-Phi("–"1.25)`

These values can be found by lookup in a `z`-table (or using software). By table lookup:

`=0.8944-0.1056" "=0.7888`