Random Variables and Probability Distributions

LEARNING OBJECTIVES

1. To learn the concept of a random variable.
2. To learn the distinction between discrete and continuous random variables.

Definition

A random variable is a numerical quantity that is generated by a random experiment.

We will denote random variables by capital letters, such as $X$ or $Z$, and the actual values that they can take by lowercase letters, such as $x$ and $z$.

Table 4.1 "Four Random Variables" gives four examples of random variables. In the second example, the three dots indicates that every counting number is a possible value for $X$. Although it is highly unlikely, for example, that it would take 50 tosses of the coin to observe heads for the first time, nevertheless it is conceivable, hence the number 50 is a possible value. The set of possible values is infinite, but is still at least countable, in the sense that all possible values can be listed one after another. In the last two examples, by way of contrast, the possible values cannot be individually listed, but take up a whole interval of numbers. In the fourth example, since the light bulb could conceivably continue to shine indefinitely, there is no natural greatest value for its lifetime, so we simply place the symbol $\infty$ for infinity as the right endpoint of the interval of possible values.

Table 4.1 Four Random Variables

Definition

A random variable is called discrete if it has either a finite or a countable number of possible values. A random variable is called continuous if its possible values contain a whole interval of numbers.

The examples in the table are typical in that discrete random variables typically arise from a counting process, whereas continuous random variables typically arise from a measurement.

 This text was adapted by Saylor Academy under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License without attribution as requested by the work's original creator or licensor.

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LEARNING OBJECTIVES

1. To learn the concept of the probability distribution of a discrete random variable.
2. To learn the concepts of the mean, variance, and standard deviation of a discrete random variable, and how to compute them.

Probability Distributions

Associated to each possible value $x$ of a discrete random variable $X$ is the probability $P(x)$ that $X$ will take the value $x$ in one trial of the experiment.

Definition

The probability distribution of a discrete random variable $X$ is a list of each possible value of $X$ together with the probability that $X$ takes that value in one trial of the experiment.

The probabilities in the probability distribution of a random variable $X$ must satisfy the following two conditions:

1. Each probability $P(x)$ must be between o and 1: $0 \leq P(x) \leq 1$.

2. The sum of all the probabilities is $1: \Sigma P(x)=1$.

EXAMPLE 1

A fair coin is tossed twice. Let $X$ be the number of heads that are observed.

a. Construct the probability distribution of $X$.

b. Find the probability that at least one head is observed.

Solution:

$\begin{align*} P(X \geq 1)=P(1)+P(2)=0.50+0.25=0.75 \end{align*}$

Figure 4.1

Probability Distribution for Tossing a Fair Coin Twice

EXAMPLE 2

A pair of fair dice is rolled. Let $X$ denote the sum of the number of dots on the top faces.

a. Construct the probability distribution of $X$.

b. Find $P(X \geq 9)$.

c. Find the probability that $X$ takes an even value.

Solution:

The sample space of equally likely outcomes is

a. The possible values for $X$ are the numbers 2 through 12 . $X=2$ is the event $\{11\}$, so $P(2)=1 / 36 . X=3$ is the event $\{12,21\}$, so $P(3)=2 / 36$. Continuing this way we obtain the table

$\begin{array}{c|ccccccccccc} x & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline P(x) & \frac{1}{36} & \frac{2}{36} & \frac{3}{36} & \frac{4}{36} & \frac{5}{36} & \frac{6}{36} & \frac{5}{36} & \frac{4}{36} & \frac{3}{36} & \frac{2}{36} & \frac{1}{36} \end{array}$

This table is the probability distribution of $X$.

b. The event $X \geq 9$ is the union of the mutually exclusive events $X=9, X=10, X=11$, and $X=12$. Thus

$\begin{align*} P(X \geq 9)=P(9)+P(10)+P(11)+P(12)=\frac{4}{36}+\frac{3}{36}+\frac{2}{36}+\frac{1}{36}=\frac{10}{36}=0.2 \overline{7} \end{align*}$

c. Before we immediately jump to the conclusion that the probability that $x$ takes an even value must be $0.5$, note that $X$ takes six different even values but only five different odd values. We compute

 $\begin{align*} \begin{aligned} P(X \text { is even }) &=P(2)+P(4)+P(6)+P(8)+P(10)+P(12) \\ &=\frac{1}{36}+\frac{3}{36}+\frac{5}{36}+\frac{5}{36}+\frac{3}{36}+\frac{1}{36}=\frac{18}{36}=0.5 \end{aligned} \end{align*}$

A histogram that graphically illustrates the probability distribution is given in Figure 4.2

"Probability Distribution for Tossing Two Fair Dice".

Figure 4.2

Probability Distribution for Tossing Two Fair Dice

Definition

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

$\begin{align*} \mu=E(X)=\Sigma x P(x) \end{align*}$

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.

EXAMPLE 3

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

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

Solution:

The formula in the definition gives

$\begin{align*} \begin{aligned} \mu &=\Sigma x P(x) \\ &=(-2) \cdot 0.21+(1) \cdot 0.34+(2) \cdot 0.24+(3.5) \cdot 0.21=1.135 \end{aligned} \end{align*}$

EXAMPLE 4

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.

Solution:

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

$\begin{align*} P(W)=P(299)+P(199)+P(99)=0.001+0.001+0.001=0.003 \end{align*}$

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

$\begin{align*} E(X)=299 \cdot 0.001+199 \cdot 0.001+99 \cdot 0.001+(-1) \cdot 0.997=-0.4 \end{align*}$

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.

EXAMPLE 5

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.

Solution:

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:

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

Therefore

$\begin{align*} E(X)=\Sigma x P(x)=195 \cdot 0.9997+(-199,805) \cdot 0.0003=135 \end{align*}$

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.

Definition

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

$\begin{align*} \sigma^{2}=\Sigma(x-\mu)^{2} P(x) \end{align*}$

which by algebra is equivalent to the formula

$\begin{align*} \sigma^{2}=\left[\Sigma x^{2} P(x)\right]-\mu^{2} \end{align*}$

Definition

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

$\begin{align*} \sigma=\sqrt{\Sigma(x-\mu)^{2} P(x)}=\sqrt{\left[\Sigma x^{2} P(x)\right]-\mu^{2}} \end{align*}$

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 $X$.

EXAMPLE 6

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 $X$.

Solution:

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 $\mu$,

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

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

• The probability distribution of a discrete random variable $X$ is a listing of each possible value $x$ taken by $X$ along with the probability $P(x)$ that $X$ takes that value in one trial of the experiment. 
• The mean $\mu$ of a discrete random variable $X$ is a number that indicates the average value of $X$ over numerous trials of the experiment. It is computed using the formula $\mu=\Sigma x P(x)$. 
• The variance $\sigma^{2}$ and standard deviation $\sigma$ of a discrete random variable $X$ are numbers that indicate the variability of $X$ over numerous trials of the experiment. They may be computed using the formula $\sigma^{2}=\left[\Sigma x^{2} P(x)\right]-\mu^{2}$, taking the square root to obtain $\sigma$.

BASIC

1. Determine whether or not the table is a valid probability distribution of a discrete random variable. Explain fully.

3. A discrete random variable X has the following probability distribution:

$\begin{align*} \begin{array}{c|ccccc} x & 77 & 78 & 79 & 80 & 81 \\ \hline P(x) & 0.15 & 0.15 & 0.20 & 0.40 & 0.10 \end{array} \end{align*}$

Compute each of the following quantities.

5. If each die in a pair is "loaded" so that one comes up half as often as it should, six comes up half again as often as it should, and the probabilities of the other faces are unaltered, then the probability distribution for the sum $X$ of the number of dots on the top faces when the two are rolled is

$\begin{aligned} &\begin{array}{c|cccccc} x & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline P(x) & \frac{1}{144} & \frac{4}{144} & \frac{8}{144} & \frac{12}{144} & \frac{16}{144} & \frac{22}{144} \end{array}\\ &\begin{array}{c|ccccc} x & 8 & 9 & 10 & 11 & 12 \\ \hline P(x) & \frac{24}{144} & \frac{20}{144} & \frac{16}{144} & \frac{12}{144} & \frac{9}{144} \end{array} \end{aligned}$

Compute each of the following.

APPLICATIONS

7. In a hamster breeder's experience the number $X$ of live pups in a litter of a female not over twelve months in age who has not borne a litter in the past six weeks has the probability distribution

$\begin{align*} \begin{array}{c|ccccccc} x & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline P(x) & 0.04 & 0.10 & 0.26 & 0.31 & 0.22 & 0.05 & 0.02 \end{array} \end{align*}$

9. Let $X$ denote the number of boys in a randomly selected three-child family. Assuming that boys and girls are equally likely, construct the probability distribution of $X$.

11. Five thousand lottery tickets are sold for $1 each. One ticket will win $1,000, two tickets will win $500 each, and ten tickets will win $100 each. Let $X$ denote the net gain from the purchase of a randomly selected ticket.

13. An insurance company will sell a $90,000 one-year term life insurance policy to an individual in a particular risk group for a premium of $478. Find the expected value to the company of a single policy if a person in this risk group has a 99.62% chance of surviving one year.

15. An insurance company estimates that the probability that an individual in a particular risk group will survive one year is 0.9825. Such a person wishes to buy a $150,000 one-year term life insurance policy. Let $C$ denote how much the insurance company charges such a person for such a policy.

17. A roulette wheel has 38 slots. Thirty-six slots are numbered from 1 to 36 ; half of them are red and half are black. The remaining two slots are numbered 0 and 00 and are green. In a $1 bet on red, the bettor pays $1 to play. If the ball lands in a red slot, he receives back the dollar he bet plus an additional dollar. If the ball does not land on red he loses his dollar. Let $X$ denote the net gain to the bettor on one play of the game.

19. The time, to the nearest whole minute, that a city bus takes to go from one end of its route to the other has the probability distribution shown. As sometimes happens with probabilities computed as empirical relative frequencies, probabilities in the table add up only to a value other than 1.00 because of round-off error.

$\begin{align*} \begin{array}{c|cccccc} x & 42 & 43 & 44 & 45 & 46 & 47 \\ \hline P(x) & 0.10 & 0.23 & 0.34 & 0.25 & 0.05 & 0.02 \end{array} \end{align*}$

ADDITIONAL EXERCISES

21. The number $X$ of nails in a randomly selected 1-pound box has the probability distribution shown. Find the average number of nails per pound.

$\begin{align*} \begin{array}{c|ccc} x & 100 & 101 & 102 \\ \hline P(x) & 0.01 & 0.96 & 0.03 \end{array} \end{align*}$

23. Two fair dice are rolled at once. Let $X$ denote the difference in the number of dots that appear on the top faces of the two dice. Thus for example if a one and a five are rolled, $X=4$, and if two sixes are rolled, $X=$ $0 .$

25. A manufacturer receives a certain component from a supplier in shipments of 100 units. Two units in each shipment are selected at random and tested. If either one of the units is defective the shipment is rejected. Suppose a shipment has 5 defective units.

27. The owner of a proposed outdoor theater must decide whether to include a cover that will allow shows to be performed in all weather conditions. Based on projected audience sizes and weather conditions, the probability distribution for the revenue $X$ per night if the cover is not installed is

The additional cost of the cover is $410,000. The owner will have it built if this cost can be recovered from the increased revenue the cover affords in the first ten 90-night seasons.

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