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The mode of a continuous distribution is the value \(x^{*}\) that maximizes \(f(x)\). a. What is the mode of a normal distribution with parameters \(\mu\) and \(\sigma\) ? b. Does the uniform distribution with parameters \(A\) and \(B\) have a single mode? Why or why not? c. What is the mode of an exponential distribution with parameter \(\lambda\) ? (Draw a picture.) d. If \(X\) has a gamma distribution with parameters \(\alpha\) and \(\beta\), and \(\alpha>1\), find the mode. [Hint: \(\ln [f(x)]\) will be maximized iff \(f(x)\) is, and it may be simpler to take the derivative of \(\ln [f(x)]\).] e. What is the mode of a chi-squared distribution having \(v\) degrees of freedom?

Short Answer

Expert verified
a. Mode is \(\mu\). b. No, it has no single mode. c. Mode is 0. d. Mode is \(\beta(\alpha-1)\). e. Mode is \(v-2\).

Step by step solution

01

Normal Distribution Mode

The probability density function (PDF) of a normal distribution is given by: \[f(x) = \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{(x-\mu)^2}{2\sigma^2}}\]For a normal distribution, the mode is simply the peak of the curve, which occurs at the mean \(\mu\). Therefore, the mode is \(\mu\).
02

Uniform Distribution Mode

The uniform distribution is constant for every point between \(A\) and \(B\). This means that every point in this interval has the same probability density. Since no single point has a higher density than others, the uniform distribution does not have a single mode.
03

Exponential Distribution Mode

The probability density function of an exponential distribution is \(f(x) = \lambda e^{-\lambda x}\), for \(x \geq 0\). The function decreases monotonically, so the mode is at the point where \(x = 0\), which is the peak of the distribution. Drawing the graph shows a sharp decline from \(x = 0\).
04

Gamma Distribution Mode

The PDF of a gamma distribution is \(f(x) = \frac{x^{\alpha-1}e^{-x/\beta}}{\beta^\alpha \Gamma(\alpha)}\). To find the mode, we differentiate \(\ln[f(x)] \), which simplifies to:\[\ln[f(x)] = (\alpha - 1)\ln(x) - \frac{x}{\beta} - \alpha\ln(\beta) - \ln[\Gamma(\alpha)]\]Differentiating with respect to \(x\) and setting the derivative to zero gives: \((\alpha-1)/x - 1/\beta = 0\), solving for \(x\) gives \(x = \beta(\alpha - 1)\). Therefore, the mode is \(\beta(\alpha - 1)\).
05

Chi-Squared Distribution Mode

The chi-squared distribution is a special case of the gamma distribution with \(\alpha = v/2\) and \(\beta = 2\). From the gamma mode, substituting these values gives \(x = \beta(\alpha - 1) = 2(v/2 - 1) = v - 2\). The mode of a chi-squared distribution with \(v\) degrees of freedom is \(v - 2\), assuming \(v > 2\).

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Key Concepts

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

Mode of Distribution
The mode of a distribution is an important concept in probability and statistics. It is the value at which the highest frequency of data points occurs, effectively representing the peak of the distribution curve. In the context of continuous probability distributions, the mode is identified as the value of x that maximizes the probability density function (PDF).

For any given distribution, the mode provides insights into the data's central tendency. It is particularly useful when you want to know the most frequently occurring data points in a distribution.

Note that a distribution can have more than one mode, making it multimodal. Distributions with one peak are known as unimodal.
Normal Distribution
The normal distribution, often referred to as a "bell curve", is one of the most frequently encountered continuous probability distributions in statistics. Its PDF is characterized by its mean (\( \mu \)) and standard deviation (\( \sigma \)).

The formula for the normal distribution's PDF is: \[ f(x) = \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{(x-\mu)^2}{2\sigma^2}}\]

The curve of a normal distribution is symmetric about its mean. Because of this symmetry, the mean, median, and mode all coincide at the same point, especially in a perfectly normal distribution without skewness. Hence, the mode of a normal distribution is its mean (\( \mu \)).

This makes normal distributions easy to analyze, as their central point provides a lot of information about the dataset. You will often find normal distribution used in real-life scenarios like test scores, heights, and any data that clusters around a central value.
Uniform Distribution
Uniform distribution is quite different from the normal distribution. It signifies that every outcome in an interval from \( A \) to \( B \) is equally likely. This makes it a constant distribution with a flat curve.

Since every value within this range has the same probability, it is impossible for one single point to have a greater frequency than the others. Consequently, a uniform distribution does not have a single mode. Instead, all the points within the interval \( [A, B] \) can be considered modes due to their equivalent density.

This characteristic is useful in scenarios where outcomes are equally distributed, such as rolling a fair die or picking random numbers from a specified range.
Gamma Distribution
Gamma distribution is a continuous probability distribution characterized by parameters \( \alpha \) (shape) and \( \beta \) (scale). It is frequently used in waiting time problems or to model skewed distributions.

The PDF of a gamma distribution is: \[ f(x) = \frac{x^{\alpha-1}e^{-x/\beta}}{\beta^\alpha \Gamma(\alpha)}\]

To find the mode of a gamma distribution, we look at the peak point by differentiating the log of its PDF. Through calculations, it results in the mode being \( \beta(\alpha - 1) \), provided \( \alpha > 1 \).

Understanding the mode of a gamma distribution helps identify the most frequent or likely events, especially useful in fields like hydrology, finance, and queuing models.
Chi-Squared Distribution
The chi-squared distribution is widely used in statistical tests of independence and in checks of goodness of fit. It is essentially a special case of the gamma distribution where the shape \( \alpha \) is equal to half the degrees of freedom \( v \), and the scale \( \beta \) is 2.

The mode for the chi-squared distribution can be derived similarly to that of the gamma distribution. For a chi-squared distribution with \( v \) degrees of freedom, the mode is found at \( v - 2 \), assuming \( v > 2 \).

This means that in a chi-squared distribution with more than two degrees of freedom, the mode shifts to the right as the number of degrees of freedom increases. This distribution is crucial in tests like ANOVA, where it helps determine statistical differences between group means.

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Most popular questions from this chapter

Suppose that \(10 \%\) of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to be scrapped). Consider a random sample of 200 shafts, and let \(X\) denote the number among these that are nonconforming and can be reworked. What is the (approximate) probability that \(X\) is a. At most 30? b. Less than 30 ? c. Between 15 and 25 (inclusive)?

Let \(X\) be the total medical expenses (in 1000 s of dollars) incurred by a particular individual during a given year. Although \(X\) is a discrete random variable, suppose its distribution is quite well approximated by a continuous distribution with pdf \(f(x)=k(1+x / 2.5)^{-7}\) for \(x \geq 0\). a. What is the value of \(k\) ? b. Graph the pdf of \(X\). c. What are the expected value and standard deviation of total medical expenses? d. This individual is covered by an insurance plan that entails a \(\$ 500\) deductible provision (so the first \(\$ 500\) worth of expenses are paid by the individual). Then the plan will pay \(80 \%\) of any additional expenses exceeding \(\$ 500\), and the maximum payment by the individual (including the deductible amount) is \(\$ 2500\). Let \(Y\) denote the amount of this individual's medical expenses paid by the insurance company. What is the expected value of \(Y\) ? [Hint: First figure out what value of \(X\) corresponds to the maximum out-of- pocket expense of \(\$ 2500\). Then write an expression for \(Y\) as a function of \(X\) (which involves several different pieces) and calculate the expected value of this function.]

When circuit boards used in the manufacture of compact dise players are tested, the long-run percentage of defectives is \(5 \%\). Suppose that a batch of 250 boards has been received and that the condition of any particular board is independent of that of any other board. a. What is the approximate probability that at least \(10 \%\) of the boards in the batch are defective? b. What is the approximate probability that there are exactly 10 defectives in the batch?

Let \(X\) be a continuous rv with cdf $$ F(x)=\left\\{\begin{array}{cl} 0 & x \leq 0 \\ \frac{x}{4}\left[1+\ln \left(\frac{4}{x}\right)\right] & 04 \end{array}\right. $$ [This type of cdf is suggested in the article "Variability in Measured Bedload-Transport Rates" (Water 91Ó°ÊÓ Bull., 1985: 39 -48) as a model for a certain hydrologic variable.] What is a. \(P(X \leq 1)\) ? b. \(P(1 \leq X \leq 3)\) ? c. The pdf of \(X\) ?

Let \(Z\) be a standard normal random variable and calculate the following probabilities, drawing pictures wherever appropriate. a. \(P(0 \leq Z \leq 2.17)\) b. \(P(0 \leq Z \leq 1)\) c. \(P(-2.50 \leq Z \leq 0)\) d. \(P(-2.50 \leq Z \leq 2.50)\) e. \(P(Z \leq 1.37)\) f. \(P(-1.75 \leq Z)\) g. \(P(-1.50 \leq Z \leq 2.00)\) h. \(P(1.37 \leq Z \leq 2.50)\) i. \(P(1.50 \leq Z)\) j. \(P(|Z| \leq 2.50)\)

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