In order to solve this problem, we need to discuss probabilities and the generation of probability distribution models. A probability is generally defined as the chances or likelihood of an event occurring. It is calculated by identifying two components: the event and the sample space. The event is defined as the favorable outcome or success that we wish to observe. On the other hand, the sample space is defined as the set of all possible outcomes for the event. Mathematically we calculate probabilities by dividing the event by the sample space:

\(\displaystyle P=\frac{\textup{event: particular phenomenon we wish to observe}}{\textup{sample space: total number of possible outcomes}}\)

Let's use a simple example: the rolling of a die. We want to know the probability of rolling a one. We know that the sample space is six because there are six sides or outcomes to the die. Also, we know that there is only a single side with a value of one; therefore,

\(\displaystyle P=\frac{1}{6}\)

Now, let's convert this into a percentage:

\(\displaystyle \frac{1}{6}=0.1666\)

\(\displaystyle 0.1666\times100\%=16.66\%\)

Probabilities expressed in fraction form will have values between zero and one. One indicates that an event will definitely occur, while zero indicates that an event will not occur. Likewise, probabilities expressed as percentages possess values between zero and one hundred percent where probabilities closer to zero are unlikely to occur and those close to one hundred percent are more likely to occur. 

\(\displaystyle 0< P< 1 \textup{ or }0\%< P< 100\%\)

Now that we understand probabilities in a general sense, we need to determine how we can create a probability distribution graph for a probability model. We will use the following equation to calculate the probabilities to be used in this graphical display:

\(\displaystyle P(x=a)=_{n}C_{a}\times p^{a}\times q^{n-a}\)

Remember that in combinations and permutations a combination is calculated using the following formula:

\(\displaystyle _{n}C_{a}=\frac{n!}{a!\times (n-a)!}\)

Now, we can write the following formula.

\(\displaystyle P(x=a)=\frac{n!}{a!\times (n-a)!} \times p^{a}\times q^{n-a}\)

In this formula variables are defined in the following manner:

Let's investigate this standard through the use of an example. Suppose a researcher rolls a die twelve times and notes whether the die rolls on an even or an odd number. If the die is fair (i.e. every number has an equal probability of being rolled or each roll is random), then would the probability distribution graph follow the pattern of a normal distribution? Lets create a table and solve for each variable. The probability of rolling an even number—a two, four, or a six— is three out of six or fifty percent. Likewise, the probability of failure (i.e. rolling an odd number) is three out of six or fifty percent. Next, we need to list the number of successes for each event—variable \(\displaystyle a\). The researcher can roll an even number every time he rolls and he may not even roll an even number in all twelve trial. We need to calculate this probability for every possible number of successful events; therefore, the number of successes ranges from zero to twelve. Last, we know that there are a total of twelve trials. We have solved for the probability of each of these variables for every possible number of successes in the trials.

Once this data is tabulated we can graph the probability of rolling an even number. If we look at the graph then we can see if it follows the bell shape curve of a normal distribution.

A bell curve is shaped like the following image:

We can quickly tell that the graph of the probability distribution does follow the shape of a normal or "bell" curve.

Let's use this information to solve the question. If we look at the question, then we know that the probability of turning right or left in a random situation is fifty percent; therefore, the probability of success (i.e. the right path) or failure (i.e. the left path) is one half or fifty percent. Next we know that there were ten people questioned—or ten trials—and the number of successes per trial ranged from zero to ten. This information has been tabulated and the probability of choosing the left or right has been calculated.

We can now graph the probabilities. 

We can see that the graph follows a normal distribution with a characteristic bell shape.