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Based on data from a dart-throwing experiment, the article "Shooting Darts" (Chance, Summer 1997, 16-19) proposed that the horizontal and vertical errors from aiming at a point target should be independent of one another, each with a normal distribution having mean 0 and variance \(\sigma^{2}\). It can then be shown that the pdf of the distance \(V\) from the target to the landing point is $$ f(v)=\frac{v}{\sigma^{2}} \cdot e^{-v^{2} / 2 \sigma^{2}} \quad v>0 $$ a. This pdf is a member of what family introduced in this chapter? b. If \(\sigma=20 \mathrm{~mm}\) (close to the value suggested in the paper), what is the probability that a dart will land within \(25 \mathrm{~mm}\) (roughly \(1 \mathrm{in}\).) of the target?

Short Answer

Expert verified
a. Rayleigh distribution b. Probability is approximately 0.5422.

Step by step solution

01

Identifying the Family of the Distribution

The probability density function (pdf) given is \( f(v)=\frac{v}{\sigma^{2}} \cdot e^{-v^{2} / 2 \sigma^{2}} \) for \( v > 0 \). This pdf represents the Rayleigh distribution, a continuous probability distribution mainly used in signal processing and is derived from two independent normally distributed variables.
02

Setting Up the Probability Calculation

With \( \sigma = 20 \mathrm{~mm} \), the pdf becomes \( f(v) = \frac{v}{400} \cdot e^{-v^{2} / 800} \). We want to find \( P(V \leq 25) \). We do this by integrating \( f(v) \) from \( 0 \) to \( 25 \).
03

Calculating the Cumulative Distribution Function

The cumulative distribution function (CDF) for the Rayleigh distribution can be used here:\[F(v) = 1 - e^{-v^{2} / 2 \sigma^{2}}\]Substitute \( v = 25 \) and \( \sigma = 20 \):\[F(25) = 1 - e^{-25^{2} / 800} = 1 - e^{-625 / 800} = 1 - e^{-0.78125}\]
04

Evaluating the Exponential Term

Calculate \( e^{-0.78125} \) using a calculator. The value is approximately \( 0.4578 \).
05

Final Probability Calculation

Subtract the exponential result from 1 to find the probability:\[F(25) = 1 - 0.4578 = 0.5422\]
06

Conclusion

Therefore, the probability that a dart lands within 25 mm of the target is approximately 0.5422.

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Key Concepts

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

Rayleigh distribution
The Rayleigh distribution arises in situations involving the magnitude of a vector sum of two orthogonal components. A classic example is when these components are independent and normally distributed, such as errors in dart throwing, which are assumed to follow a normal distribution in both the horizontal and vertical directions.

Here are some key points about the Rayleigh distribution:
  • It is a continuous distribution defined for values greater than zero.
  • The probability density function (pdf) is given by \( f(v) = \frac{v}{\sigma^{2}} \cdot e^{-v^{2} / 2 \sigma^{2}} \), where \( v \) represents the magnitude (or distance).
  • This distribution is handy in modeling phenomena that consist of independent normally distributed variables combined vectorially.
In the case of the dart-throwing exercise, the distance the dart lands from the target forms a Rayleigh distribution, owing to the independent normal errors in horizontal and vertical placements.

Studying the Rayleigh distribution helps us understand the statistical nature of certain datasets where two-dimensional Gaussian vectors are involved.
normal distribution
The normal distribution, also known as the Gaussian distribution, is one of the most essential probability distributions in statistics. It describes a symmetric, bell-shaped curve where most of the observations cluster around the mean.

Key characteristics of a normal distribution include:
  • It is defined by two parameters: mean (\(\mu\)) and variance (\(\sigma^{2}\)).
  • Its shape is perfectly symmetrical around the mean.
  • The total area under the curve is equal to 1, representing a full probability.
  • The central limit theorem confirms that with a large enough sample size, the sample mean distribution of many types of variables will be approximately normal.
In the dart throw scenario from the exercise, the horizontal and vertical errors are modeled as independent random variables that follow a normal distribution having a mean of 0 and a specified variance \(\sigma^{2}\).

Understanding this foundation is crucial because many real-world phenomena align closely with the normal distribution, making it one of the robust models for continuous data.
cumulative distribution function (CDF)
The cumulative distribution function (CDF) is a pivotal concept in probability and statistics. It quantifies the probability that a random variable will take a value less than or equal to some specified quantity.

For a given continuous random variable with probability density function \( f(x) \), the CDF \( F(x) \) is defined as:
  • \( F(x) = \int_{-\infty}^{x} f(t) \, dt \)
This equation represents the area under the probability density function from \(-\infty\) to \(x\).

In the context of the dart-throwing exercise, after determining the Rayleigh distribution for the distance, the CDF was used to calculate the probability that a dart lands within 25 mm of the target. This was done by evaluating \( F(v) = 1 - e^{-v^{2} / 2 \sigma^{2}} \) at \( v = 25 \) and \( \sigma = 20 \).

Understanding the CDF is beneficial since it provides a comprehensive description of the distribution of a random variable, allowing us to assess probabilities and quantify risk effectively.

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Most popular questions from this chapter

A system consists of five identical components connected in series as shown: As soon as one component fails, the entire system will fail. Suppose each component has a lifetime that is exponentially distributed with \(\lambda=.01\) and that components fail independently of one another. Define events \(A_{i}=\) \(\\{i\) th component lasts at least \(t\) hours \(\\}, i=1, \ldots, 5\), so that the \(A_{i} \mathrm{~s}\) are independent events. Let \(X=\) the time at which the system fails-that is, the shortest (minimum) lifetime among the five components. a. The event \(\\{X \geq t\\}\) is equivalent to what event involving \(A_{1}, \ldots, A_{5}\) ? b. Using the independence of the \(A_{i}{ }^{\prime}\) s, compute \(P(X \geq t)\). Then obtain \(F(t)=P(X \leq t)\) and the pdf of \(X\). What type of distribution does \(X\) have? c. Suppose there are \(n\) components, each having exponential lifetime with parameter \(\lambda\). What type of distribution does \(X\) have?

The lifetime \(X\) (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters \(\alpha=2\) and \(\beta=3\). Compute the following: a. \(E(X)\) and \(V(X)\) b. \(P(X \leq 6)\) c. \(P(1.5 \leq X \leq 6)\) (This Weibull distribution is suggested as a model for time in service in "On the Assessment of Equipment Reliability: Trading Data Collection Costs for Precision," J. of Engr. Manuf., 1991: 105-109.)

An individual's credit score is a number calculated based on that person's credit history that helps a lender determine how much he/she should be loaned or what credit limit should be established for a credit card. An article in the Los Angeles Times gave data which suggested that a beta distribution with parameters \(A=150, B=850\), \(\alpha=8, \beta=2\) would provide a reasonable approximation to the distribution of American credit scores. a. Let \(X\) represent a randomly selected American credit score. What are the mean value and standard deviation of this random variable? What is the probability that \(X\) is within 1 standard deviation of its mean value? b. What is the approximate probability that a randomly selected score will exceed 750 (which lenders consider a very good score)?

Suppose only \(75 \%\) of all drivers in a certain state regularly wear a seat belt. A random sample of 500 drivers is selected. What is the probability that a. Between 360 and 400 (inclusive) of the drivers in the sample regularly wear a seat belt? b. Fewer than 400 of those in the sample regularly wear a seat belt?

Suppose the proportion \(X\) of surface area in a randomly selected quadrat that is covered by a certain plant has a standard beta distribution with \(\alpha=5\) and \(\beta=2\). a. Compute \(E(X)\) and \(V(X)\). b. Compute \(P(X \leq .2)\). c. Compute \(P(.2 \leq X \leq .4)\). d. What is the expected proportion of the sampling region not covered by the plant?

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