91Ó°ÊÓ

A random variable is a concept used in probability and statistics to represent a potential outcome of a random occurrence. It's a variable whose possible values are numerical outcomes of a random phenomenon. In the context of our problem, the random variable is denoted by \( X \). It represents the possible sums obtained from drawing two balls from the bag without replacement. Each pair of balls provides a different value for \( X \), depending on the numbers drawn according to the rules—namely, doubling the value of the other ball if one is a \(*\). This is what allows us to calculate the probability distribution, each value of \( X \) corresponding to a specific combination of balls.

The expected value, often represented by \( \mu \), is essentially the mean of a probability distribution. It provides a measure of the "center" or average outcome when considering all possible results weighted by their probabilities. For the exercise, the expected value \( \mu \) is calculated by summing up each possible value of \( X \) multiplied by its probability:

• \( \mu = 3 \times \frac{1}{10} + 5 \times \frac{1}{10} + 8 \times \frac{2}{10} + 2 \times \frac{1}{10} + 6 \times \frac{1}{10} + 9 \times \frac{1}{10} + 4 \times \frac{1}{10} + 11 \times \frac{1}{10} + 14 \times \frac{1}{10} = 6.5 \).

This means that, on average, you can expect a value close to 6.5 when drawing two balls according to the described rules.

Variance is a measure of how spread out the values of a random variable are. It quantifies the degree to which each number in a set differs from the mean of the set. For our problem, the variance \( \sigma^2 \) is calculated by determining the squared difference from the mean for each possible \( X \) value, then multiplying each by its probability:

• \( \sigma^2 = \sum [(X_i - \mu)^2 \times P(X_i)] \).

Plugging in our values, the variance is approximately \( 14.65 \). This tells us that there's a noticeable spread in the values around the mean of 6.5.

Standard deviation is the square root of the variance and provides a measure of the average distance each value in a probability distribution is from the expected value. It helps interpret the variability of the data in the same units as the data itself, making it more intuitive. In our exercise, the standard deviation \( \sigma \) is given by

• \( \sigma = \sqrt{\sigma^2} \).

When calculated as \( \sqrt{14.65} \), it results in approximately \( 3.83 \). This means the values can typically differ from the expected value by about 3.83, providing insight into their spread around the mean.