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Let \(\mathrm{Z}\) have a standard normal distribution and define a new rv \(Y\) by \(Y=\sigma Z+\mu\). Show that \(Y\) has a normal distribution with parameters \(\mu\) and \(\sigma\). [Hint: \(Y \leq y\) iff \(Z \leq\) ? Use this to find the cdf of \(Y\) and then differentiate it with respect to \(y\).]

Short Answer

Expert verified
The random variable \(Y\) has a normal distribution with mean \(\mu\) and standard deviation \(\sigma\).

Step by step solution

01

Understand the Transformation

Given that \(Y = \sigma Z + \mu\), we acknowledge that \(Z\) is a standard normal random variable. Our goal is to show that \(Y\) is a normal random variable with mean \(\mu\) and standard deviation \(\sigma\).
02

Express Probability Inequality

From the identity \(Y \leq y\), we can isolate \(Z\) as follows: \(Y \leq y\) iff \(\sigma Z + \mu \leq y\). Subtract \(\mu\) from both sides to get \(\sigma Z \leq y - \mu\).
03

Solve for Z

Divide the inequality by \(\sigma\) (assuming \(\sigma > 0\)) to get \(Z \leq \frac{y - \mu}{\sigma}\). This relationship allows us to connect the probabilities of \(Y\) and \(Z\).
04

Find the CDF of Y

The cumulative distribution function of \(Y\), \(F_Y(y)\), is given by \(P(Y \leq y) = P(Z \leq \frac{y - \mu}{\sigma})\). Since \(Z\) follows a standard normal distribution, this probability is \(\Phi\left(\frac{y - \mu}{\sigma}\right)\), where \(\Phi\) is the cdf of the standard normal distribution.
05

Differentiate to Find the PDF

Differentiate \(F_Y(y) = \Phi\left(\frac{y - \mu}{\sigma}\right)\) with respect to \(y\) to find the probability density function of \(Y\). Using the chain rule, the derivative is \(\frac{d}{dy} \Phi\left(\frac{y - \mu}{\sigma}\right) = \phi\left(\frac{y - \mu}{\sigma}\right) \cdot \frac{1}{\sigma}\), where \(\phi\) is the pdf of the standard normal distribution.
06

Recognize the PDF of Y

Observe that \(\phi\left(\frac{y - \mu}{\sigma}\right) \cdot \frac{1}{\sigma}\) is indeed the pdf of a normal distribution with mean \(\mu\) and standard deviation \(\sigma\). Hence, \(Y\sim N(\mu, \sigma^2)\), confirming our conclusion.

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Key Concepts

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

Standard Normal Distribution
The standard normal distribution is a special type of normal distribution. It represents a bell-shaped curve where the mean (\(\mu = 0\) ) and the standard deviation (\(\sigma = 1\) ) are both fixed at particular values.
The random variable \(Z\) , which follows a standard normal distribution, helps in simplifying and standardizing problems involving normal distributions.
Key properties include:
  • Mean of 0: It indicates that the distribution is centered around zero.
  • Standard deviation of 1: This measures the spread of the distribution.
  • The total area under the curve equals 1, showing 100% of the probabilities are covered within the curve.
The standard normal distribution is central in probability and statistics, as it provides the basis for the standardization of other normal distributions. By converting a normal variable into a standard normal variable, it becomes easier to compute probabilities and percentiles across different datasets.
Probability Density Function (PDF)
The probability density function (PDF) of a distribution provides crucial information about how the probabilities of a continuous random variable are distributed.
For continuous distributions like the normal distribution, the PDF helps determine the likelihood a random variable falls within a particular range of values.
In the context of the standard normal distribution, the PDF is denoted as \(\phi(z)\).
This function describes the shape of the standard normal curve, which has its highest point at the mean zero and rolls off symmetrically.
To get the PDF of a normal distribution with different mean \(\mu\) and standard deviation \(\sigma\), a transformation of the form \(Y = \sigma Z + \mu\) is applied. This modifies the PDF to \(\phi\left(\frac{y - \mu}{\sigma}\right)\cdot \frac{1}{\sigma}\), which accounts for the shifted and scaled version of the original distribution. Key Points:
  • The PDF is vital for determining risk and probabilities in various applications, such as finance and science.
  • The shape of the PDF affects the probability calculations and the spread of possible outcomes.
  • The area under the PDF between two points represents the probability of the variable falling within that interval.
Cumulative Distribution Function (CDF)
The cumulative distribution function (CDF) gives a complete understanding of the probabilities of a random variable being less than or equal to a certain value in a distribution.
For the standard normal distribution, the CDF is denoted as \(\Phi(z)\) . It describes the total probability from the leftmost end of the distribution up to a specific point, effectively providing the probability that the variable takes on a value less than or equal to a particular number.
Main Characteristics:
  • A CDF starts from 0 and asymptotically approaches 1, reflecting the total probability.
  • In transformations involving a normal distribution, such as \(Y = \sigma Z + \mu\), the CDF helps determine how values are distributed based on the new mean and standard deviation.
  • The CDF is particularly useful when finding percentiles or quartiles within a distribution.
When dealing with different normals, understanding the CDF helps relate to probabilities about outcomes concerning mean and variance, making it invaluable in statistics and the study of variations across datasets.

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Most popular questions from this chapter

Nonpoint source loads are chemical masses that travel to the main stem of a river and its tributaries in flows that are distributed over relatively long stream reaches, in contrast to those that enter at well-defined and regulated points. The article "Assessing Uncertainty in Mass Balance Calculation of River Nonpoint Source Loads" (J. of Envir. Engr., 2008: 247-258) suggested that for a certain time period and location, \(X=\) nonpoint source load of total dissolved solids could be modeled with a lognormal distribution having mean value 10,281 \(\mathrm{kg} / \mathrm{day} / \mathrm{km}\) and a coefficient of variation \(C V=.40(C V=\) \(\left.\sigma_{X} / \mu_{X}\right)\). a. What are the mean value and standard deviation of \(\ln (X) ?\) b. What is the probability that \(X\) is at most 15,000 \(\mathrm{kg} / \mathrm{day} / \mathrm{km}\) ? c. What is the probability that \(X\) exceeds its mean value, and why is this probability not .5? d. Is 17,000 the 95 th percentile of the distribution?

The article "Response of \(\mathrm{SiC}_{\mathrm{I}} / \mathrm{Si}_{3} \mathbf{N}_{4}\) Composites Under Static and Cyclic Loading-An Experimental and Statistical Analysis" \((J\). of Engr. Materials and Technology, 1997: 186-193) suggests that tensile strength (MPa) of composites under specified conditions can be modeled by a Weibull distribution with \(\alpha=9\) and \(\beta=180\). a. Sketch a graph of the density function. b. What is the probability that the strength of a randomly selected specimen will exceed 175 ? Will be between 150 and 175 ? c. If two randomly selected specimens are chosen and their strengths are independent of one another, what is the probability that at least one has a strength between 150 and 175 ? d. What strength value separates the weakest \(10 \%\) of all specimens from the remaining \(90 \%\) ?

A 12 -in. bar that is clamped at both ends is to be subjected to an increasing amount of stress until it snaps. Let \(Y=\) the distance from the left end at which the break occurs. Suppose \(Y\) has pdf $$ f(y)=\left\\{\begin{array}{cc} \left(\frac{1}{24}\right) y\left(1-\frac{y}{12}\right) & 0 \leq y \leq 12 \\ 0 & \text { otherwise } \end{array}\right. $$ Compute the following: a. The cdf of \(Y\), and graph it. b. \(P(Y \leq 4), P(Y>6)\), and \(P(4 \leq Y \leq 6)\) c. \(E(Y), E\left(Y^{2}\right)\), and \(V(Y)\) d. The probability that the break point occurs more than 2 in. from the expected break point. e. The expected length of the shorter segment when the break occurs.

The weekly demand for propane gas (in 1000s of gallons) from a particular facility is an rv \(X\) with pdf $$ f(x)=\left\\{\begin{array}{cl} 2\left(1-\frac{1}{x^{2}}\right) & 1 \leq x \leq 2 \\ 0 & \text { otherwise } \end{array}\right. $$ a. Compute the cdf of \(X\). b. Obtain an expression for the \((100 p)\) th percentile. What is the value of \(\tilde{\mu}\) ? c. Compute \(E(X)\) and \(V(X)\). d. If \(1.5\) thousand gallons are in stock at the beginning of the week and no new supply is due in during the week, how much of the \(1.5\) thousand gallons is expected to be left at the end of the week? [Hint: Let \(h(x)=\) amount left when demand \(=x\).]

Let \(X\) have a binomial distribution with parameters \(n=25\) and \(p\). Calculate each of the following probabilities using the normal approximation (with the continuity correction) for the cases \(p=.5, .6\), and \(.8\) and compare to the exact probabilities calculated from Appendix Table A.1. a. \(P(15 \leq X \leq 20)\) b. \(P(X \leq 15)\) c. \(P(20 \leq X)\)

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