Answer to Question #111162 in Statistics and Probability for judith

Question

QUESTION 13
A roulette wheel is divided into 4 sectors of equal area numbered 1, 0, 3 and 6. Suppose the
wheel is spun twice and the random variable X is the sum from the two spins.
(a) Find the probability mass function of X. (3)
(b) Find the expected value of X, E.X/. (3)
(c) Find variance of X, V.X

Answer 1

a)Possible values of $X$: 0=0+0, 1=1+0=0+1, 2=1+1, 3=3+0=0+3, 4=1+3=3+1, 6=6+0=0+6=3+3, 7=6+1=1+6, 9=6+3=3+6, 12=6+6. All 16 variants are listed.

So $P(X=0)=\frac{1}{16}, P(X=1)=\frac{1}{8}, P(X=2)=\frac{1}{16},$

$P(X=3)=\frac{1}{8}, P(X=4)=\frac{1}{8}, P(X=6)=\frac{3}{16},$

$P(X=7)=\frac{1}{8}, P(X=9)=\frac{1}{8}, P(X=12)=\frac{1}{16}$

b)$EX=0\cdot\frac{1}{16}+1\cdot\frac{1}{8}+2\cdot\frac{1}{16}+3\cdot\frac{1}{8}+4\cdot\frac{1}{8}+6\cdot\frac{3}{16}+$

$+7\cdot\frac{1}{8}+9\cdot\frac{1}{8}+12\cdot\frac{1}{16}=5$

c)$EX^2=EX=0^2\cdot\frac{1}{16}+1^2\cdot\frac{1}{8}+2^2\cdot\frac{1}{16}+3^2\cdot\frac{1}{8}+$

$+4^2\cdot\frac{1}{8}+6^2\cdot\frac{3}{16}+7^2\cdot\frac{1}{8}+9^2\cdot\frac{1}{8}+12^2\cdot\frac{1}{16}=\frac{71}{2}$

$VX=EX^2-(EX)^2=\frac{71}{2}-5^2=\frac{21}{2}$

Answer: a)$\begin{array}{ccccccccc} 0&1&2&3&4&6&7&9&12\\ \frac{1}{16}&\frac{1}{8}&\frac{1}{16}&\frac{1}{8}&\frac{1}{8}&\frac{3}{16}&\frac{1}{8}&\frac{1}{8}&\frac{1}{16} \end{array}$

b)$5$, c)$\frac{21}{2}$