Question

5.1 basics of probability distributions. compute the mean and standard deviation of a discrete random variable. emilio plays a game with a biased coin with p(heads) = 0.71 and p(tails) = 0.29. he plays each game by tossing the coin once. if he tosses a head, he pays $6. if he tosses a tail, he wins $12. complete the probability distribution table. probability distribution table x p(x) 12 what is your long - term expected profit per game? $ what is the standard deviation? $ question help: message instructor post to forum

Explanation:

Step1: Define the random - variable values and probabilities

Let $x_1=- 6$ (a loss of $6$ when getting a head) with probability $P(x_1)=0.71$, and $x_2 = 12$ (a win of $12$ when getting a tail) with probability $P(x_2)=0.29$.

Step2: Calculate the mean (expected value)

The formula for the mean $\mu$ of a discrete random variable is $\mu=\sum_{i}x_iP(x_i)$.
So, $\mu=x_1P(x_1)+x_2P(x_2)=(-6)\times0.71 + 12\times0.29=-4.26+3.48=-0.78$.

Step3: Calculate the variance

The formula for the variance $\sigma^{2}=\sum_{i}(x_i - \mu)^2P(x_i)$.
First, $(x_1-\mu)=(-6)-(-0.78)=-6 + 0.78=-5.22$ and $(x_2-\mu)=12-(-0.78)=12 + 0.78 = 12.78$.
Then, $(x_1 - \mu)^2P(x_1)=(-5.22)^2\times0.71 = 27.2484\times0.71=19.346364$.
$(x_2 - \mu)^2P(x_2)=(12.78)^2\times0.29=163.3284\times0.29 = 47.365236$.
$\sigma^{2}=19.346364 + 47.365236=66.7116$.

Step4: Calculate the standard - deviation

The standard - deviation $\sigma=\sqrt{\sigma^{2}}=\sqrt{66.7116}\approx8.17$.

Answer:

The long - term expected profit per game is $-0.78$ dollars.
The standard deviation is approximately $8.17$ dollars.