Question #100403
8. Two 10p coins are tossed. The random variable X represents the total value of each coin lands heads up.
(a)Find E(X) and Var(X).

The random variables S and T are defined as follows:
S = X-10 and T = (1/2)X-5
(b)Show that E(S) = E(T).
(c)Find Var(S) and Var (T).
1
Expert's answer
2019-12-16T10:19:33-0500

a) Find the distribution of XX


P(X=0)=1212=14P(X=0)={1 \over 2}\cdot{1 \over 2}={1 \over4}

P(X=1)=1212+1212=12P(X=1)={1 \over 2}\cdot{1 \over 2}+{1 \over 2}\cdot{1 \over 2}={1 \over2}

P(X=2)=1212=14P(X=2)={1 \over 2}\cdot{1 \over 2}={1 \over4}

              x:012P(X=x):141214\begin{array}{cc} \ \ \ \ \ \ \ \ \ \ \ \ \ \ x: & 0 & 1 & 2 \\ P(X=x): & {1 \over4} & {1 \over2} & {1 \over4} \end{array}

Use symmetry E(X)=10E(X)=10

Check


E(X)=014+1012+2014=10E(X)=0\cdot{1 \over4}+10\cdot{1 \over2}+20\cdot{1 \over4}=10

Var(X)=0214+10212+20214102=50Var(X)=0^2 \cdot{1 \over4}+10^2\cdot{1 \over2}+20^2\cdot{1 \over4}-10^2=50

b)


E(S)=E(X10)=E(X)10=1010=0E(S)=E(X-10)=E(X)-10=10-10=0

E(T)=E(12X5)=E(12X)5=12105=0E(T)=E({1 \over2}X-5)=E({1 \over2}X)-5={1 \over2}\cdot10-5=0

c)


Var(S)=Var(X)=50Var(S)=Var(X)=50

Var(T)=(12)2Var(X)=1450=12.5Var(T)=({1 \over2})^2Var(X)={1 \over4}\cdot50=12.5




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