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Calculate the standard deviation of \(X\) for each probability distribution. (You calculated the expected values in the last exercise set. Round all answers to two decimal places.) $$ \begin{array}{|c|c|c|c|c|} \hline \boldsymbol{x} & 10 & 20 & 30 & 40 \\ \hline \boldsymbol{P}(\boldsymbol{X}=\boldsymbol{x}) & \frac{3}{10} & \frac{2}{5} & \frac{1}{5} & \frac{1}{10} \\ \hline \end{array} $$

Short Answer

Expert verified
The standard deviation of the random variable X, to two decimal places, is approximately 9.43.

Step by step solution

01

Find the Expected Value (Mean) of the Random Variable X

The expected value (mean) of a random variable, say X, denoted by E(X) or μ is obtained by the formula: \( E(X) = \sum_{all x} x \cdot P(X = x) \). According to previous calculations this gives us: \( E(X) = 10 \cdot \frac{3}{10} + 20 \cdot \frac{2}{5} + 30 \cdot \frac{1}{5} + 40 \cdot \frac{1}{10} = 3 + 8 + 6 + 4 = 21 \).
02

Calculate the Variance

The variance, often denoted as \( Var(X) \) or \( \sigma^2 \), is calculated using the formula: \( Var(X) = E(X^2) - [E(X)]^2 = \sum_{all x} x^2 \cdot P(X = x) - [E(X)]^2 \). This formula means that we calculate the expected value of the squares of each possible value of X, then subtract the square of the expected value calculated in Step 1. Calculating \( E(X^2) \): \( E(X^2) = 10^2 \cdot \frac{3}{10} + 20^2 \cdot \frac{2}{5} + 30^2 \cdot \frac{1}{5} + 40^2 \cdot \frac{1}{10} = 30 + 160 + 180 + 160 = 530 \). Next apply the variance formula: \( Var(X) = E(X^2) - [E(X)]^2 = 530 - 21^2 = 530 - 441 = 89 \).
03

Calculate the Standard Deviation

The standard deviation of X, denoted as \( \sigma \) or sd(X), is the square root of the variance: \( \sigma = \sqrt{Var(X)} = \sqrt{\sigma^2} \). So, \( \sigma = \sqrt{89} \approx 9.43 \). Therefore, the standard deviation of the random variable X, to two decimal places, is 9.43.

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Key Concepts

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

Variance Calculation
Variance is a measure of how much a set of values varies. In the context of probability distributions, it gives us insight into how much we expect the values to spread out from the expected value or mean. To calculate variance, we follow a series of steps:
  • First, find the expected value of the square of each potential value of the random variable. This involves squaring each value of the random variable and multiplying it by the probability of its occurrence.
  • Next, sum these squared results to get the expected value of the squares. For example, if our values are 10, 20, 30, and 40 with respective probabilities, we calculate it as: \[ E(X^2) = 10^2 \cdot \frac{3}{10} + 20^2 \cdot \frac{2}{5} + 30^2 \cdot \frac{1}{5} + 40^2 \cdot \frac{1}{10} = 30 + 160 + 180 + 160 = 530 \].
  • Finally, to find the variance, subtract the square of the expected value from the expected value of the squares: \[ Var(X) = E(X^2) - [E(X)]^2 = 530 - 21^2 = 530 - 441 = 89 \].
Understanding variance helps us grasp the consistency of our probability distribution. A higher variance means more spread out data, while a lower variance indicates the data is closer to the mean.
Expected Value
The expected value, often referred to as the mean, is a key concept in probability and statistics. It provides a single summarizing figure representing the average outcome of a probability distribution if the process could be repeated infinitely or across a very large number of trials. To calculate the expected value:
  • Multiply each value by its corresponding probability.
  • Add these products together to get the expected value.
For instance, with the given values and probabilities \(x\), \(P(X=x)\):\[ E(X) = 10 \cdot \frac{3}{10} + 20 \cdot \frac{2}{5} + 30 \cdot \frac{1}{5} + 40 \cdot \frac{1}{10} = 3 + 8 + 6 + 4 = 21 \].
This number, 21, tells us the center of the distribution or where the majority of our probable outcomes lie on average. The expected value does not necessarily need to be one of your actual outcomes; it represents a theoretical long-term average.
Probability Distributions
Probability distributions give a mathematical description of distributions of outcomes for a certain random variable. They lay out all possible values a random variable can take and their associated probabilities. Common types include binomial, normal, and uniform distributions, each with unique patterns:
  • For our specific table of values and probabilities, it represents a discrete probability distribution, as it encompasses finite values (10, 20, 30, and 40) each with set probabilities.
  • The probabilities for all possible values should sum up to 1, ensuring that one of the outcomes will happen: \( \frac{3}{10} + \frac{2}{5} + \frac{1}{5} + \frac{1}{10} = 1 \).
  • These distributions help anticipate the variability in random processes and guide statistical inference methods.
Using probability distributions, we gain predictive power, understanding not only potential outcomes but also their likelihood and potential impacts.

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