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1 Answer
Geoff K.
Nov 15, 2017

(i) #c=1/10," " d=1/6" "# (ii) #"Var"(X)=16/3#

(iii) #"P"(X_1+X_2=10)=11/100" "# (iv) #"E"(Y)=240,#

#"Var"(Y)=256" "# (v) #"P"(220 <= Y <= 260)~~0.7888#

Explanation:

(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#

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