Answer to Question #264354 in Statistics and Probability for ellie

Question #264354

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)



1
Expert's answer
2021-11-12T07:04:56-0500
"\\def\\arraystretch{1.5}\n \\begin{array}{c:c:c:c:c}\n y & 1 & 10 & q & 101 \\\\ \\hline\n P(Y=y) & 0.2 & 0.4 & 0.2 & 0.2 \\\\\n \n\\end{array}"

(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


"0.16q^2-9.76q+1485.04=1385.2"

"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


"E(Y)=24.4+48=72.4"

The greatest value of "E(Y)" is 72.4.



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