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Let a random variable \(X\) have probability density function $$\begin{array}{l|c|c|c|c}x & 1 & 2 & 6 & 8 \\\\\hline p(X=x) & 0.4 & 0.1 & 0.3 & 0.2\end{array}$$ Compute the variance and standard deviation of \(X\) with \(\mu=4\).

Short Answer

Expert verified
The variance is 8.4 and the standard deviation is approximately 2.9.

Step by step solution

01

Determine the Variance Formula

The variance of a random variable \(X\) with mean \(\mu\) is given by the formula: \(\text{Var}(X) = E[(X - \mu)^2] = \sum_{i=1}^{n} (x_i - \mu)^2 \cdot p(x_i)\). In this case, \(\mu = 4\).
02

Compute Each Squared Difference

First, calculate each \((x_i - \mu)^2\) for the given values of \(x\):- For \(x = 1\), \((1 - 4)^2 = 9\).- For \(x = 2\), \((2 - 4)^2 = 4\).- For \(x = 6\), \((6 - 4)^2 = 4\).- For \(x = 8\), \((8 - 4)^2 = 16\).
03

Multiply by Their Probabilities

Now, multiply each squared difference by its corresponding probability:- For \(x = 1\), \(9 \cdot 0.4 = 3.6\).- For \(x = 2\), \(4 \cdot 0.1 = 0.4\).- For \(x = 6\), \(4 \cdot 0.3 = 1.2\).- For \(x = 8\), \(16 \cdot 0.2 = 3.2\).
04

Sum the Results for Variance

Add all the results from Step 3 to find the variance:\[\text{Var}(X) = 3.6 + 0.4 + 1.2 + 3.2 = 8.4\]
05

Compute the Standard Deviation

The standard deviation is the square root of the variance. Thus, calculate \(\text{SD}(X) = \sqrt{8.4} \approx 2.9\).

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Key Concepts

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

Random Variables
A random variable, often symbolized as \(X\), represents a numerical outcome of a probability experiment. It's called 'random' because the outcome is not deterministic, meaning it can fluctuate each time the experiment is carried out. Random variables can be discrete, taking on specific separate values, like rolling a die, or continuous, taking any value within a range, like measuring height.
  • Discrete Random Variables: Take distinct, separate values. For example, the roll of a standard six-sided die results in a discrete value between 1 to 6.
  • Continuous Random Variables: Can take any value in a given interval. Think of measuring time or temperature, where the result could be any number along a continuum.
In the context of the problem you are studying, \(X\) is a discrete random variable since it takes on specified values: 1, 2, 6, and 8. Each value has an associated probability, indicating how likely it is to occur.
Probability Density Function
A probability density function (pdf) is a function that describes the probability of a random variable taking on particular values. For discrete random variables, this function assigns probabilities to each discrete outcome. The sum of all probabilities for the random variable must equal 1, ensuring that the total probability covers all possible outcomes. In the provided exercise, the pdf is presented in table form, showing each discrete value of \(X\) along with its probability:
  • \(p(X=1) = 0.4\)
  • \(p(X=2) = 0.1\)
  • \(p(X=6) = 0.3\)
  • \(p(X=8) = 0.2\)
Checking the sum of these probabilities (0.4 + 0.1 + 0.3 + 0.2 = 1) confirms they are proportional outcomes. This ensures that every possible result of experiment has been accounted for. Each probability indicates the likelihood of occurrence for its corresponding \(x\). If the random variable were continuous, we'd be dealing with a probability density function, which requires integration to find the probability of landing within any interval.
Standard Deviation
Standard deviation is a measure of how spread out numbers are around the mean (average) in a dataset or set of probabilities. In terms of random variables, it gives insights into the variability or dispersion from the expected value, represented by the mean \(\mu\). Here's how the standard deviation is determined from variance:
  • Variance: First, calculate the variance, which is the average squared deviation from the mean. It provides a measure of the overall spread of the data.
  • Standard Deviation: Take the square root of the variance. This step allows the measure to return to the same units as the original data, offering a clearer understanding of distribution.
In your exercise, after calculating that the variance \( \text{Var}(X) = 8.4 \), the standard deviation \( \text{SD}(X) \) is found by taking the square root of 8.4, resulting in approximately 2.9. Thus, this value signifies how much values of \(X\) tend to deviate from the mean value of 4. A higher standard deviation means more spread out data, while a lower standard deviation indicates data clustered around the mean.

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Most popular questions from this chapter

Suppose we draw three balls from an urn containing two red balls and three black balls. We do not replace the balls after we draw them. In terms of the hypergeometric distribution, what is the probability of getting two red balls? Compute this probability.

Suppose \(A\) and \(B\) are disjoint events in a sample space \(\Omega\). Is it possible that \(A\) and \(B\) could be independent? Explain your answer.

Define a random variable \(X\) on the sample space \(\Omega\) by setting \(X(\omega)=3\) for all \(\omega \in \Omega\). What is \(E(X) ? \operatorname{Var}(X) ?\)

Compute the expectation \(E(X)\) of the random variable \(X\) that counts the number of heads in four flips of a coin that lands heads with frequency \(1 / 3\).

Consider a game based on the days of a 31 -day month. A day is chosen at randomsay, by spinning a spinner. The prize is a number of dollars equal to the sum of the digits in the date of the chosen day, For example, choosing the 31 st of the month pays \(\$ 3+\$ 1=\$ 4,\) as does choosing the fourth day of the month. (a) Set up the underlying sample space \(\Omega\) and its probability density, the value of which at \(\omega\) gives the reward associated with \(\omega\). (b) Define a random variable \(X(\omega)\) on \(\Omega\) with a value at \(\omega\) that gives the reward associated with \(\omega\). (c) Set up a sample space \(\Omega_{X}\) consisting of the elements in the range of \(X,\) and give the probability distribution \(p x\) on \(\Omega_{X}\) arising from \(X\). (d) Determine \(P(X=6)\). (e) Determine \(P(2 \leq X \leq 4)=P(\omega: 2 \leq X(\omega) \leq 4)\). (f) Determine \(P(X>10)=P(\omega: X(\omega)>10)\).

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