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Compute the mean and variance of the following discrete probability distribution. $$\begin{array}{|rr|}\hline {x} & P(x) \\\\\hline 2 & .5 \\\8 & .3 \\\10 & .2 \\\\\hline\end{array}$$

Short Answer

Expert verified
Mean: 5.4, Variance: 12.04.

Step by step solution

01

Calculate the Mean

The mean (or expected value) of a discrete probability distribution is given by \( E(X) = \sum{x_i \cdot P(x_i)} \). Using the provided distribution: \( E(X) = 2 \cdot 0.5 + 8 \cdot 0.3 + 10 \cdot 0.2 \). Calculate these values separately: \( 2 \cdot 0.5 = 1 \), \( 8 \cdot 0.3 = 2.4 \), and \( 10 \cdot 0.2 = 2 \). Finally, sum them up: \( 1 + 2.4 + 2 = 5.4 \). Thus, the mean is 5.4.
02

Calculate Each Term for Variance

Variance is calculated using \( \text{Var}(X) = \sum{(x_i - E(X))^2 \cdot P(x_i)} \). First, compute \( (x_i - E(X))^2 \) for each \( x_i \): \((2-5.4)^2 = 11.56\), \((8-5.4)^2 = 6.76\), and \((10-5.4)^2 = 21.16\).
03

Multiply Results by Corresponding Probabilities

Now multiply each squared difference by the corresponding probability: \( 11.56 \cdot 0.5 = 5.78 \), \( 6.76 \cdot 0.3 = 2.028 \), and \( 21.16 \cdot 0.2 = 4.232 \).
04

Sum the Results to Find Variance

Add the products obtained in the previous step to find the variance: \( 5.78 + 2.028 + 4.232 = 12.04 \). Therefore, the variance of the distribution is 12.04.

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Key Concepts

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

Mean Calculation
The mean of a discrete probability distribution, often called the expected value, represents the average outcome you would expect if you could repeat the process many times. To find the mean, we sum up the products of each possible outcome of the random variable and its corresponding probability. This is mathematically represented as:
\[ E(X) = \sum x_i \cdot P(x_i) \]
In simpler terms, you are multiplying each outcome by how likely it is and then adding them all up. Let’s break it down with an example:
  • For a value of 2 with a probability of 0.5, the contribution to the mean is: \( 2 \times 0.5 = 1 \).
  • For a value of 8 with a probability of 0.3, it’s: \( 8 \times 0.3 = 2.4 \).
  • Lastly, for a value of 10 with a probability of 0.2, it's: \( 10 \times 0.2 = 2 \).
Adding all these contributions gives us the mean: \[ 1 + 2.4 + 2 = 5.4 \] So, the mean (expected value) of this probability distribution is 5.4.
Variance Calculation
Variance helps us understand how much the values of the distribution differ from the mean. It tells us about the spread of the data. To calculate the variance of a discrete probability distribution, we follow these steps:
The formula for variance is given by:
\[ \text{Var}(X) = \sum (x_i - E(X))^2 \cdot P(x_i) \]
Here's how it is done step-by-step:
  • Subtract the mean from each outcome: \( 2 - 5.4 = -3.4 \), \( 8 - 5.4 = 2.6 \), \( 10 - 5.4 = 4.6 \).
  • Square each of these differences: \(-3.4^2 = 11.56 \), \(2.6^2 = 6.76 \), \(4.6^2 = 21.16 \).
  • Multiply each squared difference by the probability of its outcome: \(11.56 \times 0.5 = 5.78 \), \(6.76 \times 0.3 = 2.028 \), \(21.16 \times 0.2 = 4.232 \).
  • Finally, add up these results to get the variance: \[ 5.78 + 2.028 + 4.232 = 12.04 \]
Thus, the variance of the distribution is 12.04, indicating the degree of spread in the values around the mean.
Discrete Probability
A discrete probability distribution is a statistical function that defines the likelihood of occurrence of each value a discrete random variable can assume. It’s important for handling scenarios where the outcomes of a random experiment are counted, such as rolling a dice or counting occurrences.
Here is what you need to understand about discrete probability distributions:
  • The sum of all probabilities in a discrete distribution must equal 1. This is because the distribution accounts for all possible outcomes.
  • Probabilities assigned to each outcome must range from 0 to 1, inclusive.
  • Each probability indicates the likelihood of each distinct outcome occurring.
In our example, we had three outcomes (2, 8, and 10) with the probabilities 0.5, 0.3, and 0.2 respectively. All these probabilities need to add up to 1:
\[ 0.5 + 0.3 + 0.2 = 1 \]
This confirms that the distribution is valid, as all possible outcomes and their probabilities have been accounted for. Understanding discrete probability is crucial for interpreting data and making informed predictions based on the likelihood of different outcomes.

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Most popular questions from this chapter

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