The following table shows the probability distribution for the random variable B.
b
1
10
q
101
0.2
0.4
0.2
0.2
(i) Given that , show that and solve this equation. (4 marks)
(ii) Calculate the value of . (2 marks)
(i) Given "Var(Y)=1385.2"
"E(Y)=0.2(1)+0.4(10)+0.2(q)+0.2(101)"
"=24.4+0.2q"
"E(Y^2)=0.2(1)^2+0.4(10)^2+0.2(q)^2+0.2(101)^2"
"=2080.4+0.2q^2"
"Var(Y)=E(Y^2)-(E(Y))^2"
"=2080.4+0.2q^2-(24.4+0.2q)^2"
"=2080.4+0.2q^2-595.36-9.76q+0.04q^2"
"=0.16q^2-9.76q+1485.04"
Then
"0.24q^2-9.76q+99.84=0"
"q^2-61q+624=0, True"
(ii)
"D=(-61)^2-4(1)(624)=35^2""q=\\dfrac{61\\pm35}{2(1)}"
"q_1=\\dfrac{61-35}{2}=13"
"q_2=\\dfrac{61+35}{2}=48"
"E(Y)=24.4+13=37.4"
Or
The greatest value of "E(Y)" is 72.4.
Comments
Leave a comment