Problem 2
Question
Compute the mean and variance of the following discrete probability distribution. $$ \begin{array}{|rr|} \hline {}{\underline{\phantom{xx}}} {\boldsymbol{x}} & \boldsymbol{P}(\boldsymbol{x}) \\ \hline 2 & .5 \\ 8 & .3 \\ 10 & .2 \\ \hline \end{array} $$
Step-by-Step Solution
Verified Answer
The mean is 5.4 and the variance is 12.04.
1Step 1: Calculate the Mean (Expected Value)
The mean (expected value) of a discrete probability distribution is calculated using the formula \( E(X) = \sum x_i P(x_i) \). Substitute the given values: \( E(X) = 2 \times 0.5 + 8 \times 0.3 + 10 \times 0.2 \). Compute each term: \( 2 \times 0.5 = 1.0 \), \( 8 \times 0.3 = 2.4 \), \( 10 \times 0.2 = 2.0 \). Add these values up to get the mean: \( E(X) = 1.0 + 2.4 + 2.0 = 5.4 \).
2Step 2: Calculate the Variance
Variance is calculated using the formula \( Var(X) = \sum (x_i - E(X))^2 P(x_i) \). First, compute \((x_i - E(X))\) for each \(x_i\): \(2 - 5.4 = -3.4\), \(8 - 5.4 = 2.6\), \(10 - 5.4 = 4.6\). Then, compute \((x_i - E(X))^2\): \((-3.4)^2 = 11.56\), \((2.6)^2 = 6.76\), \((4.6)^2 = 21.16\). Multiply each squared term by the probability: \(11.56 \times 0.5 = 5.78\), \(6.76 \times 0.3 = 2.028\), \(21.16 \times 0.2 = 4.232\). Finally, sum these products to get the variance: \( Var(X) = 5.78 + 2.028 + 4.232 = 12.04 \).
Key Concepts
Mean (Expected Value)Variance CalculationProbability Theory
Mean (Expected Value)
In probability theory, the mean of a discrete probability distribution represents the average outcome you would expect by repeating a random experiment many times. This is calculated through the expected value, denoted as \( E(X) \). It provides a measure of central tendency, such as how centralize or "typical" the measurement is from the given outcomes. For a discrete distribution, the mean is found using the formula: \[E(X) = \sum x_i P(x_i)\] This equation sums up each possible value \( x_i \) of a random variable multiplied by its corresponding probability \( P(x_i) \). Let's break it down using an example: for the distribution with outcomes 2, 8, and 10, and probabilities 0.5, 0.3, and 0.2 respectively, the expected value is calculated as follows:
- \( 2 \times 0.5 = 1.0 \)
- \( 8 \times 0.3 = 2.4 \)
- \( 10 \times 0.2 = 2.0 \)
Variance Calculation
The variance of a discrete probability distribution measures the spread or dispersion of possible outcomes around the mean (expected value). It tells you how much the values deviate from the mean, providing an understanding of their variability. Variance is computed using the formula: \[Var(X) = \sum (x_i - E(X))^2 P(x_i)\] Here, you subtract the mean \( E(X) \) from each possible outcome \( x_i \), square this difference, multiply by the corresponding probability, and then sum it all up. Let's understand this by applying it to our example:
- For \( x = 2 \) : \((2 - 5.4)^2 = 11.56\), \( 11.56 \times 0.5 = 5.78 \)
- For \( x = 8 \) : \((8 - 5.4)^2 = 6.76\), \( 6.76 \times 0.3 = 2.028 \)
- For \( x = 10 \) : \((10 - 5.4)^2 = 21.16\), \( 21.16 \times 0.2 = 4.232 \)
Probability Theory
Probability theory is the branch of mathematics concerned with analyzing random phenomena and deciding how likely events are to occur. It provides the framework for understanding discrete probability distributions, which describe the probability of each possible outcome of a random process. Such distributions are based on two key elements: outcomes and probabilities.
- Outcomes: The possible results of a random process (like rolling a dice or drawing a card).
- Probabilities: Represent the likelihood of each outcome, summing up to 1 across all possibilities.
Other exercises in this chapter
Problem 1
Compute the mean and variance of the following discrete probability distribution. $$ \begin{array}{|cc|} \hline x & P(x) \\ \hline 0 & .2 \\ 1 & .4 \\ 2 & .3 \\
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