Question

In Exercises \(\mathrm{P} .78\) to \(\mathrm{P} .81,\) use the probability function given in the table to calculate: (a) The mean of the random variable (b) The standard deviation of the random variable $$ \begin{array}{lcccc} \hline x & 10 & 12 & 14 & 16 \\ \hline p(x) & 0.25 & 0.25 & 0.25 & 0.25 \\ \hline \end{array} $$

Short answer

The mean of the random variable is 13.0 and the standard deviation is approximately 2.24.

Step-by-step solution

Calculate Mean (Expected Value)

The mean also called the expected value of a random variable, is calculated by multiplying each possible value of the variable by their probabilities and then summing all these products. Following this rules, we get: \(μ = 10*(0.25) + 12*(0.25) + 14*(0.25) + 16*(0.25) = 13.0 \)

Calculate Variance

The variance (V) is the expected value of the squared deviation from the mean. The squared deviation is found by subtracting the mean from each x and squaring the result, then multiplying by the corresponding p(x). We have: \( V = [(10-13)^2*(0.25)] + [(12-13)^2*(0.25)] + [(14-13)^2*(0.25)] + [(16-13)^2*(0.25)] = 5 \)

Calculate Standard Deviation

The standard deviation (SD) is the square root of the variance, thus we find \(\sqrt{V} = \sqrt{5} \approx 2.24 \)

Key concepts

Mean of a Random Variable

Understanding the mean of a random variable is crucial for grasping the behavior of data in probability and statistics. The mean, often referred to as the expected value, serves as a measure of the central tendency of a random variable. In simpler terms, it tells us where the center of the data distribution is located.

To find the mean of a discrete random variable, we multiply each possible outcome by its probability and then add all these products together. For continuous random variables, the process involves integration, but the idea is fundamentally the same: to find an average value weighted by probability.

Application to Our Exercise

As shown in the textbook exercise, the random variable can take on the values of 10, 12, 14, and 16, each with an equal probability of 0.25. When we perform the calculation, following the formula, we get a mean value of 13. This tells us that, on average, we can expect the random variable to be around the value of 13 given the distribution specified.

Standard Deviation of a Random Variable

The standard deviation is a measure of how spread out the values of a random variable are. In essence, it quantifies the amount of variability or dispersion from the mean. A low standard deviation indicates that the data points tend to be close to the mean, while a high standard deviation suggests the data are spread out over a wider range of values.

The way to calculate the standard deviation is by taking the square root of the variance. This step brings the measure of spread back to the original unit of measurement, as variance is expressed in the units squared.

Interpretation in Our Exercise

Once we know the variance, as calculated in the exercise, is 5, we take the square root to find the standard deviation, which is approximately 2.24. This suggests a moderate dispersion of the random variable around the mean, implying that although the likelihood of each value is the same, there is variability in the outcomes around the expected average.

Variance of a Random Variable

Variance is a powerful statistical concept that measures the spread of a random variable's possible values. It’s calculated as the average of the squared differences from the mean, which gives us a clear picture of data dispersion. If the variance is high, the values of the random variable are widely scattered around the mean. Conversely, if the variance is low, the values are more clustered closely around the mean.

To calculate variance for a discrete random variable, we square the deviation of each possible outcome from the mean (the squared difference), multiply by its probability, and sum all these values. For a continuous random variable, this involves an integral.

Details from the Textbook Exercise

According to the step-by-step solution in the textbook exercise, after subtracting the mean from each value of the random variable, squaring the differences, and weighting by the probabilities, the variance is obtained as 5. This numeric value becomes a baseline for understanding how much the possible outcomes deviate from the expected mean, offering insight into the variability inherent in the probability distribution of the exercise.