/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 167 Let \(Y\) be a random variable w... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(Y\) be a random variable with mean 11 and variance \(9 .\) Using Tchebysheff's theorem, find a. a lower bound for \(P(6

Short Answer

Expert verified
a. Lower bound is 0.64. b. Value of C is 10.

Step by step solution

01

Understanding Tchebysheff's Theorem

Tchebysheff's theorem states that for any random variable with mean \(\mu\) and variance \(\sigma^2\), the probability that the variable is within \(k\) standard deviations of the mean is at least \(1 - \frac{1}{k^2}\). This gives us a way to find probabilities over intervals centered at the mean.
02

Calculate standard deviation

Variance is given as \(9\), so the standard deviation \(\sigma\) is the square root of variance: \(\sigma = \sqrt{9} = 3\). This will help us determine \(k\) in subsequent steps.
03

Determine k for P(6 < Y < 16)

The interval \(6 < Y < 16\) is symmetric around the mean \(\mu = 11\). Calculate the distance from \(11\) to the bounds \(6\) and \(16\). This distance is \(5\), so \(k = \frac{5}{3}\) since 5 is the number of standard deviations away from the mean.
04

Apply Tchebysheff's Theorem to find lower bound

Use Tchebysheff's inequality, \(P(|Y-\mu| < k\sigma) \geq 1 - \frac{1}{k^2}\). With \(k = \frac{5}{3}\), the inequality becomes \(P(6 < Y < 16) \geq 1 - \frac{1}{(\frac{5}{3})^2} \). Calculate \(1 - \frac{1}{(\frac{5}{3})^2} = 1 - \frac{9}{25} = \frac{16}{25} = 0.64.\)
05

Calculate k for P(|Y-11| ≥ C) ≤ 0.09

To satisfy \(P(|Y-11| \geq C) \leq 0.09\), Tchebysheff's theorem suggests \(P(|Y-\mu| < k\sigma) \geq 0.91\). Hence, \(1 - \frac{1}{k^2} = 0.91\). Solving this, we get \(\frac{1}{k^2} = 0.09\), or \(k^2 = \frac{1}{0.09} = \frac{100}{9}\). Thus, \(k = \frac{10}{3}\).
06

Find value of C

Since \(k = \frac{C}{\sigma}\), where \(\sigma = 3\), and from Step 5, \(k = \frac{10}{3}\), we set up the equation \(C = k \times \sigma = \frac{10}{3} \times 3 = 10\). Therefore, \(C = 10\).

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Key Concepts

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

Random Variable
A random variable is a fundamental concept in probability and statistics. It represents a quantity whose possible values are numerical outcomes of a random phenomenon. Think of a random variable as a way to assign numbers to the outcomes of a random process. For example, if you roll a die, the number that comes up is a value of the random variable.
There are two types of random variables: discrete and continuous. Discrete random variables can take on a finite number of values, like the result of rolling a die. Continuous random variables can take on any value within a given range, such as the height of people.
In the context of this exercise, the random variable in question is denoted as \( Y \). It has a specific mean and variance, which are key to using Tchebysheff's Theorem effectively. Understanding random variables helps form the basis for analyzing expected outcomes and the variability of those outcomes.
Mean and Variance
The mean and variance are statistical measures that describe the characteristics of a random variable. The mean, often represented by \( \mu \), is the average value of all possible outcomes of the random variable. It tells us the central point around which the values of the random variable cluster.
The variance, denoted by \( \sigma^2 \), measures the spread of the random variable's values around the mean. It is the average of the squared differences from the mean. The larger the variance, the more spread out the values are.
In this exercise, the random variable \( Y \) has a mean of 11 and a variance of 9. These parameters are crucial as they guide us in applying Tchebysheff's Theorem. Specifically, knowing the variance allows us to compute the standard deviation, which is the square root of the variance, essential for further steps in the problem.
Probability Bounds
Probability bounds refer to the limits within which the outcome of a random variable is expected to lie, with a certain level of confidence. Tchebysheff's Theorem provides a way to calculate these bounds by relating them to the mean and standard deviation of the variable.
With Tchebysheff's Theorem, we can determine how likely it is for a random variable to fall within a specific interval around its mean, using the formula \[ P(|Y - \mu| < k\sigma) \geq 1 - \frac{1}{k^2} \]This theorem is powerful because it applies to any random distribution as long as we know the mean and variance.
In our problem, we use the theorem to find a lower bound for the probability that \( Y \) is between 6 and 16. Using the calculated \( k = \frac{5}{3} \), we figured out that the probability is at least 0.64. This means we can assert with at least 64% confidence that \( Y \) falls within that range.
Standard Deviation
The standard deviation is a measure of the amount of variation or dispersion of a set of values. It is the square root of the variance and provides a metric for how much a random variable deviates from its mean on average. Generally, a smaller standard deviation means the values are tightly clustered around the mean, while a larger standard deviation indicates the values are more spread out.
In mathematical terms, if the variance of \( Y \) is 9, then the standard deviation \( \sigma \) is calculated as:\[ \sigma = \sqrt{9} = 3 \]This is a crucial step because the standard deviation directly influences the calculation of \( k \), the number of standard deviations used in Tchebysheff's Theorem.
For instance, when determining the value of \( C \) such that \( P(|Y-11| \geq C) \leq 0.09 \), we calculated \( k \) and then applied it to find that \( C = 10 \). The standard deviation helps scale \( k \) into the actual distance in the original units of \( Y \).

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