Help me answer this question.....? (question in picture below)
1 Answer
(i)
(iii)
Explanation:
(i)
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
Similarly, the outcomes 6, 7, and 8 are equally likely, and their probabilities sum to (the other)
(ii)
#=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#
#=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
#"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
#=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)
#"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 " ~ " "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
#="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
#=0.8944-0.1056" "=0.7888#