/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 115 Let \(I_{i}\) be the input curre... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let \(I_{i}\) be the input current to a transistor and \(I_{0}\) be the output current. Then the current gain is proportional to \(\ln \left(I_{0} / I_{i}\right)\). Suppose the constant of proportionality is 1 (which amounts to choosing a particular unit of measurement), so that current gain \(=X=\ln \left(I_{0} / I_{i}\right)\). Assume \(X\) is normally distributed with \(\mu=1\) and \(\sigma=.05\). a. What type of distribution does the ratio \(I_{0} / I_{i}\) have? b. What is the probability that the output current is more than twice the input current? c. What are the expected value and variance of the ratio of output to input current?

Short Answer

Expert verified
The ratio \( \frac{I_0}{I_i} \) is log-normally distributed. The probability of \( \frac{I_0}{I_i} > 2 \) is approximately 0. Expected value is approximately 2.72, variance is 0.0527.

Step by step solution

01

Identify the Distribution

The relationship given is that the natural logarithm of the ratio \( \ln\left( \frac{I_0}{I_i} \right) = X \) and \( X \) is normally distributed with mean \( \mu = 1 \) and standard deviation \( \sigma = 0.05 \). This implies that the ratio \( \frac{I_0}{I_i} \) is log-normally distributed.
02

Determine the Probability for Part b

The probability of \( \frac{I_0}{I_i} > 2 \) corresponds to finding \( P(X > \ln(2)) \). First, calculate \( \ln(2) \approx 0.693 \). The standard normal variable \( Z \) for \( X \) is given by \( Z = \frac{X - \mu}{\sigma} \), so for \( X = \ln(2) \) we have \( Z = \frac{0.693 - 1}{0.05} = -6.14 \). Using the standard normal distribution table, find \( P(Z > -6.14) \). Since this is very far in the tail, \( P(Z > -6.14) \approx 1 \). So, \( P\left( \frac{I_0}{I_i} > 2 \right) \approx 0 \).
03

Calculate Expected Value and Variance for Part c

For a log-normal distribution with a normally distributed log-value \( X \sim N(\mu, \sigma^2) \), the expected value is \( E\left( \frac{I_0}{I_i} \right) = e^{\mu + \frac{\sigma^2}{2}} \) and the variance is \( \text{Var}\left( \frac{I_0}{I_i} \right) = (e^{\sigma^2} - 1) e^{2\mu + \sigma^2} \). Here, \( \mu = 1 \) and \( \sigma = 0.05 \), so the expected value is \( e^{1 + \frac{0.05^2}{2}} \approx e^{1.00125} \approx 2.72 \). The variance is \((e^{0.05^2} - 1) \cdot e^{2 \cdot 1 + 0.05^2} \approx 0.0527\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Normal Distribution
The normal distribution is one of the most important concepts in statistics, often known as the bell curve due to its symmetrical, bell-shaped appearance. It's a continuous probability distribution characterized by two parameters: the mean (\(\mu\)) and the standard deviation (\(\sigma\)). When data follows a normal distribution, it implies that they are symmetrically scattered around the mean. This makes the mean the peak of the curve, and most values tend to cluster around this central spot.

The significance of the normal distribution lies in its properties:
  • Symmetry: The left and right sides of the curve are mirror images.
  • Mean, Median, Mode Coincidence: In a perfectly normal distribution, all are equal.
  • 68-95-99.7 Rule: Roughly 68% of data falls within one standard deviation from the mean, 95% within two, and 99.7% within three.
Understanding the normal distribution helps us model many real-world phenomena. For instance, in the original exercise, it's given that the current gain, denoted as \(X\), follows a normal distribution, having a mean of 1 and a standard deviation of 0.05.
Log-normal Distribution
A log-normal distribution, as its name implies, is a distribution of a variable whose natural logarithm is normally distributed. In simpler terms, if a variable \(X\) is normally distributed, then \(e^X\) is log-normally distributed. This type of distribution is ideal for modeling ratios or values that cannot be negative, such as stock prices or the ratio of output to input current in a transistor, as in our exercise.

The characteristics of a log-normal distribution include:
  • Positivity: Values are strictly positive, meaning they never dip below zero.
  • Right-skewed: It is not symmetrical like the normal distribution but rather stretches out indefinitely to the right.
  • Multiplicative Processes: It often models growth processes or multiplicative random variables.
In the given context, knowing the distribution type of the current ratio \(\frac{I_0}{I_i}\) helps in calculating probabilities and statistical measures like the expected value.
Expected Value and Variance
Expected value and variance are crucial statistical measures. The expected value, often considered the mean, provides a summary of a distribution's central tendency. It is essentially the average value one might expect from a stochastic process. For a log-normal distributed \(\frac{I_0}{I_i}\) ratio, the formula for expected value is\[E\left( \frac{I_0}{I_i} \right) = e^{\mu + \frac{\sigma^2}{2}}\]where \(\mu\) is the mean of the logarithmic values and \(\sigma^2\) is their variance.

Variance provides insight into the dispersion or spread of the data around the expected value. A higher variance means a wider spread. For a log-normal distribution, the formula for variance is\[\text{Var}\left( \frac{I_0}{I_i} \right) = (e^{\sigma^2} - 1) e^{2\mu + \sigma^2}\]In the exercise, with \(\mu = 1\) and \(\sigma = 0.05\), these formulas help us find that the expected value is approximately 2.72, and the variance is around 0.0527. These calculations provide a sense of what the average gain of the current might be and how it varies from measurement to measurement.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Sales delay is the elapsed time between the manufacture of a product and its sale. According to the article "Warranty Claims Data Analysis Considering Sales Delay" (Quality and Reliability Engr. Intl., 2013: 113-123), it is quite common for investigators to model sales delay using a lognormal distribution. For a particular product, the cited article proposes this distribution with parameter values \(\mu=2.05\) and \(\sigma^{2}=.06\) (here the unit for delay is months). a. What are the variance and standard deviation of delay time? b. What is the probability that delay time exceeds 12 months? c. What is the probability that delay time is within one standard deviation of its mean value? d. What is the median of the delay time distribution? e. What is the 99 th percentile of the delay time distribution? f. Among 10 randomly selected such items, how many would you expect to have a delay time exceeding 8 months?

The authors of the article "A Probabilistic Insulation Life Model for Combined Thermal-Electrical Stresses" (IEEE Trans. on Elect. Insulation, 1985: 519-522) state that "the Weibull distribution is widely used in statistical problems relating to aging of solid insulating materials subjected to aging and stress." They propose the use of the distribution as a model for time (in hours) to failure of solid insulating specimens subjected to \(\mathrm{AC}\) voltage. The values of the parameters depend on the voltage and temperature; suppose \(\alpha=2.5\) and \(\beta=200\) (values suggested by data in the article). a. What is the probability that a specimen's lifetime is at most 250? Less than 250? More than 300 ? b. What is the probability that a specimen's lifetime is between 100 and 250 ? c. What value is such that exactly \(50 \%\) of all specimens have lifetimes exceeding that value?

The error involved in making a certain measurement is a continuous rv \(X\) with pdf $$ f(x)=\left\\{\begin{array}{cc} .09375\left(4-x^{2}\right) & -2 \leq x \leq 2 \\ 0 & \text { otherwise } \end{array}\right. $$ a. Sketch the graph of \(f(x)\). b. Compute \(P(X>0)\). c. Compute \(P(-1.5)\).

If bolt thread length is normally distributed, what is the probability that the thread length of a randomly selected bolt is a. Within 1.5 SDs of its mean value? b. Farther than \(2.5\) SDs from its mean value? c. Between 1 and 2 SDs from its mean value?

Spray drift is a constant concern for pesticide applicators and agricultural producers. The inverse relationship between droplet size and drift potential is well known. The paper "Effects of 2,4 -D Formulation and Quinclorac on Spray Droplet Size and Deposition" (Weed Technology, 2005: 1030-1036) investigated the effects of herbicide formulation on spray atomization. A figure in the paper suggested the normal distribution with mean \(1050 \mu \mathrm{m}\) and standard deviation \(150 \mu \mathrm{m}\) was a reasonable model for droplet size for water (the "control treatment") sprayed through a \(760 \mathrm{ml} / \mathrm{min}\) nozzle. a. What is the probability that the size of a single droplet is less than \(1500 \mu \mathrm{m}\) ? At least \(1000 \mu \mathrm{m}\) ? b. What is the probability that the size of a single droplet is between 1000 and \(1500 \mu \mathrm{m}\) ? c. How would you characterize the smallest \(2 \%\) of all droplets? d. If the sizes of five independently selected droplets are measured, what is the probability that exactly two of them exceed \(1500 \mu \mathrm{m}\) ?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.