91Ó°ÊÓ

A probability distribution assigns a probability to each possible outcome of a random variable. In this case, the variable \(X\) is the number of games played, and its possible values are 4, 5, 6, and 7.
First, we need to calculate the probabilities for these values using the given frequency data.
The total number of observations is calculated by summing the frequencies:
\( N = 18 + 18 + 20 + 35 = 91 \).
The probabilities for each value of \(X\) are as follows:

• \( P(4) = \frac{18}{91} \)
• \( P(5) = \frac{18}{91} \)
• \( P(6) = \frac{20}{91} \)
• \( P(7) = \frac{35}{91} \)

Create a probability distribution table to display these probabilities:
\( \begin{array}{|c|c|} \hline x & P(x) \ \hline 4 & \frac{18}{91} \ \hline 5 & \frac{18}{91} \ \hline 6 & \frac{20}{91} \ \hline 7 & \frac{35}{91} \ \hline \end{array} \)
Making probability distributions helps in understanding how likely different outcomes are and is crucial for performing further statistical analysis.

The mean and standard deviation are key metrics to summarize the data.
The mean represents the average value of the data and can be calculated using:

• \( \mu = \sum x_i P(x_i) \)

For our data:
\( \mu = 4 \cdot \frac{18}{91} + 5 \cdot \frac{18}{91} + 6 \cdot \frac{20}{91} + 7 \cdot \frac{35}{91} \approx 5.67 \).
This means that, on average, the number of games played is about 5.67.

Standard deviation measures how much the data varies from the mean.
It is calculated using:

• \( \sigma = \sqrt{ \sum (x_i - \mu)^2 P(x_i) } \)

First, find each squared difference multiplied by its probability:
\( (4 - 5.67)^2 \cdot \frac{18}{91} \approx 0.307 \)
\( (5 - 5.67)^2 \cdot \frac{18}{91} \approx 0.099 \)
\( (6 - 5.67)^2 \cdot \frac{20}{91} \approx 0.022 \)
\( (7 - 5.67)^2 \cdot \frac{35}{91} \approx 0.544 \)
Add these values and take the square root to get the standard deviation:
\( \sigma = \sqrt{0.307 + 0.099 + 0.022 + 0.544} \approx 1.04 \)
This indicates that the number of games played typically deviates by about 1.04 games from the mean.

Understanding finite data frequency is essential when dealing with discrete probability distributions.
In our example, the frequencies are the counts of how often each value of \(X\) (number of games played) appears in the data set.
The frequencies given are:

• 4 games: 18 times
• 5 games: 18 times
• 6 games: 20 times
• 7 games: 35 times

These frequencies help us calculate the probabilities, as shown earlier.
The sum of these frequencies gives us the total number of observations \(N\), which is 91 in this case.
Using finite data frequency allows us to transform raw data into a meaningful probability distribution.
This process is crucial for statistical analysis, as it provides insights into the likelihood and variability of different outcomes.