Question

Most states and Canadian provinces have government-sponsored lotteries. Here is a simple lottery wager, from the Tri-State Pick $3$game that New Hampshire shares with Maine and Vermont. You choose a number with $3$digits from $0$to $9$; the state chooses a three-digit winning number at random and pays you $\backslash (500$if your number is chosen. Because there are $1000$numbers with three digits, you have a probability of $1/1000$of winning. Taking $X$to be the amount your ticket pays you, the probability distribution of $X$is

(a) Show that the mean and standard deviation of $X$are ${\mu }_{\mathrm{X}}=\backslash )0.50$and ${\sigma }_{\mathrm{X}}=\backslash (15.80$.

(b) If you buy a Pick 3 ticket, your winnings are $W=X-1$, because it costs $\backslash )1$to play. Find the mean and standard deviation of $W$. Interpret each of these values in context.

Short answer

(a) $\mu =\$0.50$

$\sigma \approx \$15.80$

(b) ${\mu }_W=-\$0.50$

${\sigma }_W=\$15.80$

Step-by-step solution

Part (a) Step 1: Given Information 

Given in the question that a following table

Part (a) Step 2: Explanation 

The expected value is calculated by multiplying each possibility by its probability:

$$\begin{gathered} \mu =\sum xP\left(x\right) \\ =\$0\times 0.999+\$500\times 0.001 \\ =\$0.50 \end{gathered}$$

The expected value of the squared variation from the mean is the variance:

$$\begin{gathered} {\sigma }^2=\sum (x-\mu {)}^{2}P(x) \\ =(\$0-\$0.50{)}^2\times 0.999+(\$500-\$0.50{)}^2\times 0.001 \\ =249.75 \end{gathered}$$

The square root of the variance is the standard deviation:

$$\begin{gathered} \sigma =\sqrt{{\sigma }^2} \\ =\sqrt{249.75} \\ \approx \$15.80 \end{gathered}$$

Part (b) Step 1: Given Information 

Consider the Result from exercise 42 a

${\mu }_X=\$0.50$

${\sigma }_X\approx \$15.80$

Part (b) Step 2: Explanation 

Properties mean and standard deviation:

$$\begin{gathered} {\mu }_{aX+b}=a{\mu }_X+b \\ {\sigma }_{aX+b}=a{\sigma }_X \end{gathered}$$

Then we obtain for $W=X-1$:

localid="1649908788228" $$\begin{gathered} {\mu }_W={\mu }_{X-1} \\ ={\mu }_X-1 \\ =\$0.50-\$1 \\ =-\$0.50 \end{gathered}$$

localid="1649908805592" $$\begin{gathered} {\sigma }_W={\sigma }_{X-1} \\ ={\sigma }_X \\ =\$15.80 \end{gathered}$$