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The probability distribution of a random variable \(X\) is given. Compute the mean, variance, and standard deviation of \(X\). $$ \begin{array}{lllll} \hline \boldsymbol{x} & 1 & 2 & 3 & 4 \\ \hline \boldsymbol{P}(\boldsymbol{X}=\boldsymbol{x}) & .4 & .3 & .2 & .1 \\ \hline \end{array} $$

Short Answer

Expert verified
The mean (expected value) of the random variable \(X\) is \(E(X) = 2\). The variance of \(X\) is \(Var(X) = 1.6\), and the standard deviation of \(X\) is \(\sigma(X) \approx 1.26\).

Step by step solution

01

Compute the mean (expected value)

To compute the mean, we use the following formula for the expected value of a discrete random variable: \[E(X) = \sum_{i} x_i P(X=x_i)\] Using the given probabilities, we get: \[E(X) = (1 \cdot .4) + (2 \cdot .3) + (3 \cdot .2) + (4 \cdot .1)\] Now, calculate the sum: \[E(X) = 0.4 + 0.6 + 0.6 + 0.4 = 2\]
02

Compute the variance

To compute the variance, we use the following formula for variance of a discrete random variable: \[Var(X) = E(X^2) - [E(X)]^2\] First, we need to find \(E(X^2)\), which can be computed using the same formula for expected value: \[E(X^2) = \sum_{i} x_i^2 P(X=x_i)\] Thus, we have: \[E(X^2) = (1^2 \cdot .4) + (2^2 \cdot .3) + (3^2 \cdot .2) + (4^2 \cdot .1)\] Now, calculate the sum: \[E(X^2) = 1 + 1.2 + 1.8 + 1.6 = 5.6\] Now, calculate the variance using the formula: \[Var(X) = E(X^2) - [E(X)]^2 = 5.6 - 2^2 = 5.6 - 4 = 1.6\]
03

Compute the standard deviation

The standard deviation is the square root of the variance. Therefore: \[\sigma(X) = \sqrt{Var(X)} = \sqrt{1.6} \approx 1.26\] Summary of the results: 1. Mean (expected value): \(E(X) = 2\) 2. Variance: \(Var(X) = 1.6\) 3. Standard Deviation: \(\sigma(X) \approx 1.26\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Expected Value
The expected value of a random variable is a fundamental concept in probability and statistics. It gives us a measure of the central tendency, or the "average" value, that we can expect from a probability distribution. Think of it as a way to summarize the distribution of possible outcomes for a random variable. To find the expected value for a discrete random variable, you multiply each possible outcome by its probability and then sum these products.In our exercise, the random variable, denoted as \(X\), can take on the values 1, 2, 3, and 4 with respective probabilities of 0.4, 0.3, 0.2, and 0.1. The formula to compute the expected value is:
  • \(E(X) = \sum x_i P(X = x_i)\)
Here, \(E(X)\) is calculated as:
  • \(E(X) = (1 \cdot 0.4) + (2 \cdot 0.3) + (3 \cdot 0.2) + (4 \cdot 0.1) = 2\)
This means that, on average, we would expect the value of \(X\) to be 2 over many trials.
Variance
Variance provides a measure of how much the values in a probability distribution spread out from the expected value. It tells us how much variance, or "spread," there is from the mean. A higher variance indicates that the numbers are more spread out.To compute variance, we first must find \(E(X^2)\), which represents the expected value of the square of the random variable. This calculation involves squaring each outcome, multiplying by its probability, and summing the results:
  • \(E(X^2) = (1^2 \cdot 0.4) + (2^2 \cdot 0.3) + (3^2 \cdot 0.2) + (4^2 \cdot 0.1) = 5.6\)
Then, the variance \(Var(X)\) is calculated using:
  • \(Var(X) = E(X^2) - [E(X)]^2 = 5.6 - 2^2 = 1.6\)
This means that \(X\)'s values vary by approximately 1.6 from the square of the mean.
Standard Deviation
The standard deviation is a widely used measure of variability that quantifies the amount of variation or dispersion in a set of values. It is the square root of the variance and provides a measure that is in the same units as the data, making it more interpretable.To find the standard deviation from the variance, you simply take the square root:
  • \( \sigma(X) = \sqrt{Var(X)} \)
  • \( \sigma(X) = \sqrt{1.6} \approx 1.26\)
Thus, the standard deviation gives us a sense of how much the values of \(X\) typically deviate from the mean value of 2. In practical terms, a standard deviation of approximately 1.26 means that the values of \(X\) tend to vary by about 1.26 units from the mean.

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