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Chapter 4: Problem 27

When a dart is thrown at a circular target, consider the location of the landing point relative to the bull's eye. Let \(X\) be the angle in degrees measured from the horizontal, and assume that \(X\) is uniformly distributed on \([0,360]\). Define \(Y\) to be the transformed variable \(Y=h(X)=\) \((2 \pi / 360) X-\pi\), so \(Y\) is the angle measured in radians and \(Y\) is between \(-\pi\) and \(\pi\). Obtain \(E(Y)\) and \(\sigma_{Y}\) by first obtaining \(E(X)\) and \(\sigma_{X}\), and then using the fact that \(h(X)\) is a linear function of \(X\).

Short Answer

Expert verified
The expected value \(E(Y) = 0\) and \(\sigma_Y \approx 3.665\).

Step by step solution

01

Define the Problem Context

The problem involves a dart landing at a circular target, where the angular position relative to the bull's eye is uniformly distributed. We need to compute the expected value and standard deviation for the angle in radians, given the transformation from degrees.
02

Understand the Distribution of X

The variable \(X\) is uniformly distributed on \([0, 360]\). Hence, the mean \(E(X)\) and variance \(\sigma_X^2\) of a uniform distribution on \([a, b]\) can be calculated as follows: \(E(X) = \frac{a+b}{2}\) and \(\sigma_X^2 = \frac{(b-a)^2}{12}\).
03

Calculate Expected Value E(X)

Using the formula for the expected value of a uniform distribution: \(E(X) = \frac{0 + 360}{2} = 180\).
04

Calculate Variance and Standard Deviation of X

Using the formula for variance of a uniform distribution: \(\sigma_X^2 = \frac{(360 - 0)^2}{12} = \frac{360^2}{12} = 10800\). Thus, \(\sigma_X = \sqrt{10800} = 104.88\).
05

Express Transformation Function h(X)

The transformation is given by \(Y = h(X) = \frac{2\pi}{360}X - \pi\). This is a linear transformation of \(X\) with \(a = \frac{2\pi}{360}\) and \(b = -\pi\).
06

Calculate E(Y) Using E(X)

For linear transformations \(Y = aX + b\), the expectation is given by \(E(Y) = aE(X) + b\). Using \(a = \frac{2\pi}{360}\) and \(b = -\pi\):\[E(Y) = \frac{2\pi}{360} \times 180 - \pi = \pi - \pi = 0\].
07

Calculate \(\sigma_Y\) Using \(\sigma_X\)

For linear transformations \(Y = aX + b\), the standard deviation is \(\sigma_Y = |a| \sigma_X\). Here, \(\sigma_Y = \left|\frac{2\pi}{360}\right| \times 104.88 = \frac{\pi}{90} \times 104.88 \approx 3.665\).

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Key Concepts

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

Linear Transformation
A linear transformation is a mathematical operation applied to a variable that changes its scale and location systematically. In the context of this problem, the transformation from degrees to radians is achieved using a linear formula. Specifically, the variable \(X\) (in degrees) is transformed into \(Y\) (in radians) using the formula:
  • \(Y = \frac{2\pi}{360}X - \pi\)
This equation can be broken down into two components:
  • \(a = \frac{2\pi}{360}\), which scales the variable \(X\).
  • \(b = -\pi\), which shifts the entire distribution.
Linear transformations like this are critical because they maintain the uniform nature of the original variable. They allow us to convert measurements from one unit to another, maintaining meaningful statistical properties like the mean and standard deviation, which we'll explore further.
Expected Value
The expected value, or mean, provides a measure of the central tendency of a distribution. For a uniform distribution, the expected value \(E(X)\) is calculated as the midpoint of the interval \([a, b]\). In this case:
  • \(E(X) = \frac{0 + 360}{2} = 180\)
This value indicates the central angle, in degrees, at which a dart is equally likely to land within the range.
When applying a linear transformation \(Y = aX + b\), the expected value of \(Y\) (denoted as \(E(Y)\)) is determined using:
  • \(E(Y) = aE(X) + b\)
Plugging in the values for \(a\) and \(b\):
  • \(E(Y) = \frac{2\pi}{360} \times 180 - \pi = 0\)
This result indicates that, after the transformation, the average or expected angle in radians is 0, perfectly centering around our new interval, \([-\pi, \pi]\).
Standard Deviation
Standard deviation is a measure of the dispersion or variability within a distribution. It tells us how spread out the values are around the mean.
For a uniform distribution on \([a, b]\), the variance \(\sigma_X^2\) and standard deviation \(\sigma_X\) are given by:
  • \(\sigma_X^2 = \frac{(b-a)^2}{12}\)
  • \(\sigma_X = \sqrt{\sigma_X^2}\)
In this particular scenario, \(\sigma_X = 104.88\) degrees.
With a linear transformation like \(Y = aX + b\), the standard deviation \(\sigma_Y\) is found by adjusting \(\sigma_X\) using:
  • \(\sigma_Y = |a| \sigma_X\)
Applying the specific transformation constants gives:
  • \(\sigma_Y = \left|\frac{2\pi}{360}\right| \times 104.88 \approx 3.665\)
This shows that the variability in radians is much smaller compared to degrees, which is expected since the radian measure compresses the range of angles into a smaller interval.

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Most popular questions from this chapter

Determine \(z_{\alpha}\) for the following: a. \(\alpha=.0055\) b. \(\alpha=.09\) c. \(\alpha=.663\)

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Let \(X=\) the hourly median power (in decibels) of received radio signals transmitted between two cities. The authors of the article "Families of Distributions for Hourly Median Power and Instantaneous Power of Received Radio Signals" (J. Research National Bureau of Standards, vol. 67D, 1963: 753-762) argue that the lognormal distribution provides a reasonable probability model for \(X\). If the parameter values are \(\mu=3.5\) and \(\sigma=1.2\), calculate the following: a. The mean value and standard deviation of received power. b. The probability that received power is between 50 and \(250 \mathrm{~dB} .\) c. The probability that \(X\) is less than its mean value. Why is this probability not \(.5\) ?
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