/*! 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 83 Suppose that \(X\) has a discret... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose that \(X\) has a discrete uniform distribution on the integers 0 through \(9 .\) Determine the mean, variance, and standard deviation of the random variable \(Y=5 X\) and compare to the corresponding results for \(X\).

Short Answer

Expert verified
The mean of Y is 22.5, variance is 206.25, and standard deviation is 14.36.

Step by step solution

01

Calculate the Mean of X

Since variable \(X\) follows a discrete uniform distribution on integers from 0 to 9, each integer has equal probability. The mean \(\mu_X\) of \(X\) is calculated by:\[ \mu_X = \frac{0 + 1 + 2 + \, ... \, + 9}{10} = \frac{45}{10} = 4.5 \]
02

Compute the Variance of X

The variance \(\sigma_X^2\) is calculated using the formula for a discrete uniform distribution:\[ \sigma_X^2 = \frac{(n^2 - 1)}{12} \]where \(n\) is the number of integers, which is 10:\[ \sigma_X^2 = \frac{(10^2 - 1)}{12} = \frac{99}{12} = 8.25 \]
03

Determine the Standard Deviation of X

The standard deviation \(\sigma_X\) is the square root of the variance:\[ \sigma_X = \sqrt{8.25} \approx 2.87 \]
04

Calculate the Mean of Y

Given \(Y = 5X\), the mean \(\mu_Y\) is obtained by the linear transformation of the mean of \(X\):\[ \mu_Y = 5 \times \mu_X = 5 \times 4.5 = 22.5 \]
05

Compute the Variance of Y

For variance, the transformation \(Var(aX) = a^2 Var(X)\) is used, where \(a=5\):\[ \sigma_Y^2 = 5^2 \times \sigma_X^2 = 25 \times 8.25 = 206.25 \]
06

Determine the Standard Deviation of Y

The standard deviation \(\sigma_Y\) of \(Y\) is the square root of the variance:\[ \sigma_Y = \sqrt{206.25} = 14.36 \]
07

Comparison of Results

The mean of \(X\) is 4.5, while the mean of \(Y\) is 22.5. The variance of \(X\) is 8.25, and the variance of \(Y\) is 206.25. The standard deviation of \(X\) is approximately 2.87, and the standard deviation of \(Y\) is approximately 14.36.

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Key Concepts

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

Discrete Uniform Distribution
The discrete uniform distribution is a simple yet important concept in descriptive statistics. In this distribution, every integer within a specified range has an equal probability of occurring.
Thus, if a random variable \(X\) is uniformly distributed over a range such as 0 to 9, it implies that each integer from 0 through 9 has a probability of \(\frac{1}{10}\).
This is because there are 10 integers in total and each one is equally likely to occur. Essentially, the discrete uniform distribution is straightforward, making it an excellent model for understanding randomness within a limited set of numbers.
This type of distribution is useful in scenarios where outcomes are limited and equiprobable. A good analogy is a fair die, which uniformly distributes numbers 1 through 6.
Mean and Variance Calculations
Calculating the mean and variance is crucial for understanding random variables within a distribution.
For a discrete uniform distribution, the mean (or expected value) provides a central point around which the integers are distributed.
This can be calculated by taking the sum of all integers in the range and dividing by the total number of integers. For our example, the mean \(\mu_X\) is:\[\mu_X = \frac{0 + 1 + 2 + \dots + 9}{10} = 4.5\]
Variance measures how much the numbers in the distribution deviate from the mean, offering insights into the distribution's spread.
For the discrete uniform distribution, the formula for variance is:\[\sigma_X^2 = \frac{n^2 - 1}{12}\]In our case, with \(n = 10\), it computes to:\[\sigma_X^2 = \frac{99}{12} = 8.25\]
The standard deviation is simply the square root of the variance and provides a measure of spread in the same units as the data. For our distribution:\[\sigma_X \approx 2.87\]Understanding these metrics helps gauge the likelihood and distribution of outcomes.
Linear Transformations of Random Variables
Linear transformations are essential in modifying random variables, allowing us to see how changes affect the variable's distribution.
Suppose we have a transformation expressed as \(Y = aX + b\), where \(a\) and \(b\) are constants.
For simplicity, focus on the case where \(b = 0\), such as \(Y = 5X\).The mean of \(Y\), \(\mu_Y\), results from the linear transformation and is calculated by multiplying the mean of \(X\) by \(a\):\[\mu_Y = a \cdot \mu_X = 5 \times 4.5 = 22.5\]Variance transformations follow a different rule: the variance of \(Y\), \(\sigma_Y^2\), is determined by multiplying the variance of \(X\) by \(a^2\):\[\sigma_Y^2 = a^2 \cdot \sigma_X^2 = 25 \times 8.25 = 206.25\]This means variance scales by the square of the multiplying factor. Consequently, the standard deviation, being the square root of variance, would be:\[\sigma_Y \approx 14.36\]Such transformations allow for control over randomness and the overall structure of data, useful in various statistical applications and simulations.

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