Random Variables and Probability Distributions


The mean (also called the expected value) of a discrete random variable X is the number

\mu=E(X)=\Sigma x P(x)

The mean of a random variable may be interpreted as the average of the values assumed by the random variable in repeated trials of the experiment.


Find the mean of the discrete random variable X whose probability distribution is

\begin{align*}\begin{array}{c|cccc}x & -2 & 1 & 2 & 3.5 \\\hline P(x) & 0.21 & 0.34 & 0.24 & 0.21\end{array}\end{align*}


The formula in the definition gives

\mu &=\Sigma x P(x) \\
&=(-2) \cdot 0.21+(1) \cdot 0.34+(2) \cdot 0.24+(3.5) \cdot 0.21=1.135


A service organization in a large town organizes a raffle each month. One thousand raffle tickets are sold for $1 each. Each has an equal chance of winning. First prize is $300, second prize is $200, and third prize is $100. Let X denote the net gain from the purchase of one ticket.

a. Construct the probability distribution of X.

b. Find the probability of winning any money in the purchase of one ticket.

c. Find the expected value of X, and interpret its meaning.


a. If a ticket is selected as the first prize winner, the net gain to the purchaser is the $300 prize less the $1 that was paid for the ticket, hence X=300-1=299. There is one such ticket, so P(299)=0.001. Applying the same "income minus outgo" principle to the second and third prize winners and to the 997 losing tickets yields the probability distribution:

\begin{align*}\begin{array}{c|cccc}x & 299 & 199 & 99 & -1 \\\hline P(x) & 0.001 & 0.001 & 0.001 & 0.997\end{array}\end{align*}

b. Let W denote the event that a ticket is selected to win one of the prizes. Using the table


c. Using the formula in the definition of expected value,

E(X)=299 \cdot 0.001+199 \cdot 0.001+99 \cdot 0.001+(-1) \cdot 0.997=-0.4

The negative value means that one loses money on the average. In particular, if someone were to buy tickets repeatedly, then although he would win now and then, on average he would lose 40 cents per ticket purchased.

The concept of expected value is also basic to the insurance industry, as the following simplified example illustrates.


A life insurance company will sell a $200,000 one-year term life insurance policy to an individual in a particular risk group for a premium of $195. Find the expected value to the company of a single policy if a person in this risk group has a 99.97% chance of surviving one year.


Let X denote the net gain to the company from the sale of one such policy. There are two possibilities: the insured person lives the whole year or the insured person dies before the year is up. Applying the { }^{\prime \prime} income minus outgo" principle, in the former case the value of X is 195-0; in the latter case it is 195-200,000=-199,805. Since the probability in the first case is 0.9997 and in the second case is 1-0.9997=0.0003, the probability distribution for X is:

\begin{align*}\begin{array}{c|cc}x & 195 & -199,805 \\\hline P(x) & 0.9997 & 0.0003\end{array}\end{align*}


E(X)=\Sigma x P(x)=195 \cdot 0.9997+(-199,805) \cdot 0.0003=135

Occasionally (in fact, 3 times in 10,000) the company loses a large amount of money on a policy, but typically it gains $195, which by our computation of E(X) works out to a net gain of $135 per policy sold, on average.


The variance, \sigma^{2}, of a discrete random variable X is the number

\sigma^{2}=\Sigma(x-\mu)^{2} P(x)

which by algebra is equivalent to the formula

\sigma^{2}=\left[\Sigma x^{2} P(x)\right]-\mu^{2}


The standard deviation, \sigma, of a discrete random variable X is the square root of its variance, hence is given by the formulas

\sigma=\sqrt{\Sigma(x-\mu)^{2} P(x)}=\sqrt{\left[\Sigma x^{2} P(x)\right]-\mu^{2}}

The variance and standard deviation of a discrete random variable X may be interpreted as measures of the variability of the values assumed by the random variable in repeated trials of the experiment. The units on the standard deviation match those of X.


A discrete random variable X has the following probability distribution:

\begin{align*}\begin{array}{c|cccc}x & -1 & 0 & 1 & 4 \\\hline P(x) & 0.2 & 0.5 & a & 0.1\end{array}\end{align*}

A histogram that graphically illustrates the probability distribution is given in Figure 4.3 "Probability Distribution of a Discrete Random Variable".

Figure 4.3

Probability Distribution of a Discrete Random Variable

Compute each of the following quantities.

a. a.

b. P(0).

c. P(X>0).

d. P(X \geq 0).

e. P(X \leq-2)

f. The mean \mu of X.

g. The variance \sigma^{2} of X.

h. The standard deviation \sigma of X.


a. Since all probabilities must add up to 1, a=1-(0.2+0.5+0.1)=0.2.

b. Directly from the table, P(0)=0.5.

c. From the table, P(X>0)=P(1)+P(4)=0.2+0.1=0.3.

d. From the table, P(X \geq 0)=P(0)+P(1)+P(4)=0.5+0.2+0.1=0.8.

e. Since none of the numbers listed as possible values for X is less than or equal to -2, the event X \leq -2 is impossible, so P(X \leq-2)=0

f. Using the formula in the definition of \mu,

\begin{align*}\mu=\Sigma x P(x)=(-1) \cdot 0.2+0 \cdot 0.5+1 \cdot 0.2+4 \cdot 0.1=0.4\end{align*}

g. Using the formula in the definition of \sigma^{2} and the value of \mu that was just computed,

\begin{align*}\begin{aligned}\sigma^{2} &=\Sigma(x-\mu)^{2} P(x) \\&=(-1-0.4)^{2} \cdot 0.2+(0-0.4)^{2} \cdot 0.5+(1-0.4)^{2} \cdot 0.2+(4-0.4)^{2} \cdot 0.1 \\&=1.84\end{aligned}\end{align*}

h. Using the result of part (g), \sigma=\sqrt{1.84}=1.3565.