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Chapter 4: Problem 60

A normally distributed random variable has density function $$f(y)=\frac{1}{\sigma \sqrt{2 \pi}} e^{-(y-\mu)^{2} /\left(2 \sigma^{2}\right)}, \quad-\infty0\).

Short Answer

Expert verified
\( \sigma > 0 \) to keep the density function defined and positive.

Step by step solution

01

Understanding the Density Function

The given function \( f(y) = \frac{1}{\sigma \sqrt{2\pi}} e^{-(y-\mu)^{2} /(2 \sigma^{2})} \) is the probability density function (pdf) of a normal distribution. A pdf must be positive for all \( y \).
02

Exploring the Role of \( \sigma \) in the Denominator

The term \( \sigma \sqrt{2\pi} \) appears in the denominator of the density function's leading coefficient. Since this term is in the denominator, \( \sigma \) must take a value that keeps the function defined and positive.
03

Ensuring the Function is Defined

The expression \( \sqrt{2\pi} \) is a positive constant. Thus, \( \sigma \) must be greater than zero to avoid division by zero, ensuring that \( f(y) \) remains defined for all \( y \).
04

Maintaining Positivity of the Exponent

The exponent in the expression \( e^{-(y-\mu)^{2} /(2 \sigma^{2})} \) requires \( \sigma^2 > 0 \) to maintain meaningful computation of the function. If \( \sigma \leq 0 \), the expression would become undefinable or non-positive.

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Key Concepts

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

Normal Distribution
Normal distribution is one of the most important probability distributions in statistics, representing many natural phenomena. Its key feature is its bell-shaped curve, known as the Gaussian curve.

A normal distribution is defined by two parameters:
  • Mean (\(\mu\)): The central value of the distribution.
  • Standard Deviation (\(\sigma\)): Measures the spread or width of the distribution.
These parameters help determine the shape and position of the curve.

In the function, \(f(y) = \frac{1}{\sigma \sqrt{2\pi}} e^{-(y-\mu)^{2} /(2 \sigma^{2})}\), the exponential part captures how recorded observations might deviate from the mean, whereas the leading fraction ensures that the total area under the curve equals one. This is crucial because a probability distribution must sum to a total probability of 1.
Positivity Condition
The positivity condition ensures that a probability density function (pdf) is always positive. This condition is fundamental because probabilities cannot be negative.

For a normal distribution, the positivity of \(f(y)\) means that \(\frac{1}{\sigma \sqrt{2\pi}}\) must be positive. Here are two components to ensure this:
  • The exponential part \(e^{-(y-\mu)^{2} /(2 \sigma^{2})}\) is always positive as exponential functions never produce negative values.
  • The term \(\sigma \sqrt{2\pi}\) in the denominator needs \(\sigma > 0\) to keep \(\frac{1}{\sigma \sqrt{2\pi}}\) positive.
Thus, a positive \(\sigma\) is necessary to maintain the general positivity of the function.
Parameter Constraints
Parameters \(\mu\) and \(\sigma\) dictate specific constraints for the normal distribution to be valid and meaningful. Specifically, \(\sigma\), or the standard deviation, must satisfy certain conditions:
  • \(\sigma > 0\): To ensure the denominator \(\sigma \sqrt{2\pi}\), crucial for the probability density function, remains non-zero and keeps the function's value positive.
  • \(\sigma^2 > 0\): This maintains a meaningful computation in the exponent \(e^{-(y-\mu)^{2} /(2 \sigma^{2})}\).
If \(\sigma\) were zero or negative, the function might become undefined or non-positive, undermining its role as a probability density function. Thus, respecting these parameter constraints is essential for functioning and accurately describing the distribution.

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Most popular questions from this chapter

A telephone call arrived at a switchboard at random within a one-minute interval. The switch board was fully busy for 15 seconds into this one-minute period. What is the probability that the call arrived when the switchboard was not fully busy?

Suppose that \(Y\) is a continuous random variable with distribution function given by \(F(y)\) and probability density function \(f(y) .\) We often are interested in conditional probabilities of the form for a constant \(c\) a. Show that, for \(y \geq c\) $$ P(Y \leq y | Y \geq c)=\frac{F(y)-F(c)}{1-F(c)} $$ b. Show that the function in part (a) has all the properties of a distribution function. c. If the length of life \(Y\) for a battery has a Weibull distribution with \(m=2\) and \(\alpha=3\) (with measurements in years), find the probability that the battery will last less than four years, given that it is now two years old.

Suppose that a random variable \(Y\) has a probability density function given by $$f(y)=\left\\{\begin{array}{ll} k y^{3} e^{-y / 2}, & y>0 \\ 0, & \text { elsewhere } \end{array}\right.$$ a. Find the value of \(k\) that makes \(f(y)\) a density function. b. Does \(Y\) have a \(\chi^{2}\) distribution? If so, how many degrees of freedom? c. What are the mean and standard deviation of \(Y\) ? d. What is the probability that \(Y\) lies within 2 standard deviations of its mean?

The magnitude of earthquakes recorded in a region of North America can be modeled as having an exponential distribution with mean \(2.4,\) as measured on the Richter scale. Find the probability that an earthquake striking this region will a. exceed 3.0 on the Richter scale. b. fall between 2.0 and 3.0 on the Richter scale.

Let \(Y\) possess a density function $$f(y)=\left\\{\begin{array}{ll}c(2-y), & 0 \leq y \leq 2, \\\0, & \text { elsewhere. } \end{array}\right.$$ a. Find \(c\). b. Find \(F(y)\). c. Graph \(f(y)\) and \(F(y)\). d. Use \(F(y)\) in part (b) to find \(P(1 \leq Y \leq 2)\). e. Use geometry and the graph for \(f(y)\) to calculate \(P(1 \leq Y \leq 2)\).
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