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Show that the maximum value of the normal density with parameters \(\mu\) and \(\sigma\) is \(1 /(\sigma \sqrt{2 \pi})\) and occurs when \(y=\mu\).

Short Answer

Expert verified
The maximum value is \(\frac{1}{\sigma \sqrt{2\pi}}\) and occurs at \(y = \mu\).

Step by step solution

01

Understand the Normal Density Function

The normal density function, also known as the probability density function (PDF) of a normal distribution, is given by:\[ f(y) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(y-\mu)^2}{2\sigma^2}} \]This function describes the likelihood of the random variable taking on a value close to \(y\), where \(\mu\) is the mean and \(\sigma\) is the standard deviation.
02

Identify the Point of Maximum

To find where the function \(f(y)\) reaches its maximum value, observe that the exponential component \(e^{-\frac{(y-\mu)^2}{2\sigma^2}}\) is maximized when the expression \(y-\mu\) is minimized, meaning \(y = \mu\). This yields the highest value because the exponential function, \(e^x\), is largest when its exponent is zero.
03

Calculate the Maximum Value

Substitute \(y = \mu\) into the normal density function:\[ f(\mu) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(\mu-\mu)^2}{2\sigma^2}} = \frac{1}{\sigma \sqrt{2\pi}} e^{0} = \frac{1}{\sigma \sqrt{2\pi}} \]The exponential term simplifies to 1 because \(e^0 = 1\). Thus, the maximum value of the normal density function is \(\frac{1}{\sigma \sqrt{2\pi}}\).
04

Conclusion: Summary of Results

The maximum value of the probability density function of a normal distribution with mean \(\mu\) and standard deviation \(\sigma\) occurs at \(y = \mu\) and is \(\frac{1}{\sigma \sqrt{2\pi}}\). This is because the exponential term reaches its maximum value at this point.

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Key Concepts

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

Probability Density Function
The probability density function, or PDF, is a fundamental concept in statistics. It plays a crucial role in defining the probabilities of continuous random variables. In the case of the normal distribution, the PDF is often referred to as the normal density function. This function is depicted by the equation:
\[ f(y) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(y-\mu)^2}{2\sigma^2}} \]
Here, the PDF uses the parameters \( \mu \) and \( \sigma \), where \( \mu \) represents the mean of the distribution, and \( \sigma \) is the standard deviation.
The PDF gives us a way to find the likelihood of the variable \( y \) being close to any specific value. The higher the PDF value at \( y \), the more likely \( y \) is to occur.
This function is symmetric around the mean \( \mu \), where it achieves its maximum value. Understanding the PDF is essential for applications involving statistics, data analysis, and probability theory.
Standard Deviation
Standard deviation, denoted by \( \sigma \), is a measure of how spread out the values in a data set are around the mean. In a normal distribution, \( \sigma \) plays a crucial role in determining the shape of the bell curve.
In practical terms, standard deviation helps us understand the level of variability or diversity from the mean. A smaller \( \sigma \) signifies that the data points are closely clustered around the mean, resulting in a narrow peak at the center of the normal distribution.
Conversely, a larger standard deviation indicates that the data points are more spread out, leading to a flatter and wider curve.
  • The curve is precisely centered at the mean \( \mu \).
  • One standard deviation from \( \mu \) covers approximately 68% of the data in a normal distribution.
  • About 95% of data falls within two standard deviations, and 99.7% within three.
Understanding the standard deviation helps in making statistically informed predictions and decisions based on data distribution.
Mean
The mean, often symbolized by \( \mu \), is another core concept of the normal distribution, serving as a measure of central tendency. In the context of the normal distribution, it represents the peak of the bell-shaped curve, where most of the data points aggregate.
The mean provides a single value that describes the center of the data set, effectively summarizing the entire distribution into one representative value.
In the normal distribution, the mean has several significant properties:
  • The distribution is symmetric about the mean, meaning it is a mirror image on either side.
  • The mean, median, and mode of a normal distribution are all equal, located at \( \mu \).
  • The mean signifies the point at which the PDF attains its maximum value, precisely where \( y = \mu \).
By understanding the mean, one gains insights into the distribution's central position and peak probability, aiding in interpreting datasets accurately.

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Most popular questions from this chapter

Calls for dial-in connections to a computer center arrive at an average rate of four per minute. The calls follow a Poisson distribution. If a call arrives at the beginning of a one-minute interval, what is the probability that a second call will not arrive in the next 20 seconds?

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